OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Exercise 9 (B)

Question 1. In each of the following find r + s, r – s, rs, \( \frac { r }{ s } \) if r denotes the first complex number and s denotes the second complex number :
(i) 3 + 7i, i
(ii) – i, 5 + 2i
(iii) 3i, 1 - i
(iv) − 7, − 1 – 3i
(v) 7 + 3i, 3i – 7
Answer:
(i) Given \( r = 3 + 7i \) and \( s = i \):
\( r + s = 3 + 7i + i = 3 + 8i \)
\( r - s = 3 + 7i - i = 3 + 6i \)
\( rs = (3 + 7i)i = 3i + 7i^2 = 3i - 7 \)
\( \frac{r}{s} = \frac{3+7i}{i} = \frac{(3+7i)i}{i \cdot i} = \frac{3i+7i^2}{-1} = \frac{3i-7}{-1} = 7 - 3i \)

(ii) Given \( r = -i \) and \( s = 5 + 2i \):
\( r + s = -i + 5 + 2i = 5 + i \)
\( r - s = -i - (5 + 2i) = -i - 5 - 2i = -5 - 3i \)
\( rs = -i(5 + 2i) = -5i - 2i^2 = -5i + 2 \)
\( \frac{r}{s} = \frac{-i}{5+2i} = \frac{-i(5-2i)}{(5+2i)(5-2i)} = \frac{-5i+2i^2}{5^2-(2i)^2} = \frac{-2-5i}{25-(-4)} = \frac{-2-5i}{29} = -\frac{2}{29} - \frac{5}{29}i \)

(iii) Given \( r = 3i \) and \( s = 1 - i \):
\( r + s = 3i + 1 - i = 1 + 2i \)
\( r - s = 3i - (1 - i) = 3i - 1 + i = -1 + 4i \)
\( rs = 3i(1 - i) = 3i - 3i^2 = 3i + 3 \)
\( \frac{r}{s} = \frac{3i}{1-i} = \frac{3i(1+i)}{(1-i)(1+i)} = \frac{3i+3i^2}{1^2-i^2} = \frac{3i-3}{1-(-1)} = \frac{-3+3i}{2} = -\frac{3}{2} + \frac{3}{2}i \)

(iv) Given \( r = -7 \) and \( s = -1 - 3i \):
\( r + s = -7 + (-1 - 3i) = -8 - 3i \)
\( r - s = -7 - (-1 - 3i) = -7 + 1 + 3i = -6 + 3i \)
\( rs = (-7)(-1 - 3i) = 7 + 21i \)
\( \frac{r}{s} = \frac{-7}{-1-3i} = \frac{-7(-1+3i)}{(-1-3i)(-1+3i)} = \frac{7-21i}{(-1)^2-(3i)^2} = \frac{7-21i}{1-(-9)} = \frac{7-21i}{10} = \frac{7}{10} - \frac{21}{10}i \)

(v) Given \( r = 7 + 3i \) and \( s = 3i - 7 \):
\( r + s = (7 + 3i) + (3i - 7) = 6i \)
\( r - s = (7 + 3i) - (3i - 7) = 7 + 3i - 3i + 7 = 14 \)
\( rs = (7 + 3i)(3i - 7) = (3i)^2 - 7^2 = 9i^2 - 49 = -9 - 49 = -58 \)
\( \frac{r}{s} = \frac{7+3i}{3i-7} = \frac{(7+3i)(3i+7)}{(3i-7)(3i+7)} = \frac{21i+49+9i^2+21i}{(3i)^2-7^2} = \frac{42i+49-9}{-9-49} = \frac{40+42i}{-58} = -\frac{40}{58} - \frac{42}{58}i = -\frac{20}{29} - \frac{21}{29}i \)
In simple words: For each pair of complex numbers, we calculate their sum, difference, product, and quotient. To divide, we multiply the top and bottom by the conjugate of the bottom number to remove 'i' from the denominator.

🎯 Exam Tip: Always remember to simplify powers of 'i' (e.g., \( i^2 = -1 \)) and rationalize the denominator when dividing complex numbers by multiplying by the conjugate.

Question 2. Solve each of the following equations for real x and y :
(i) \( (x + y) + (3 – 2i) = 1 + 4i \)
(ii) \( (x + yi) – (7 + 4i) = 3 – 5i \)
(iii) \( 2x + yi = 1 + (2 + 3i) \)
(iv) \( x + 2yi = i − (- 3 + 5i) \)
(v) \( (x – yi)(2 + 3i) = \frac{x-2 i}{1-i} \)
(vi) \( (x^2 + 2xi) – (3x^2 + yi) = (3 – 5i) + (1 + 2yi) \)
Answer:
(i) Given \( (x + yi) + (3 – 2i) = 1 + 4i \)
First, group the real and imaginary parts: \( (x + 3) + i(y – 2) = 1 + 4i \)
Now, compare the real parts from both sides:
\( x + 3 = 1 \)
\( \implies x = 1 - 3 \)
\( \implies x = -2 \)
Then, compare the imaginary parts from both sides:
\( y - 2 = 4 \)
\( \implies y = 4 + 2 \)
\( \implies y = 6 \)

(ii) Given \( (x + yi) – (7 + 4i) = 3 – 5i \)
Group the real and imaginary parts:
\( (x – 7) + i(y – 4) = 3 – 5i \)
Compare the real parts:
\( x - 7 = 3 \)
\( \implies x = 3 + 7 \)
\( \implies x = 10 \)
Compare the imaginary parts:
\( y - 4 = -5 \)
\( \implies y = -5 + 4 \)
\( \implies y = -1 \)

(iii) Given \( 2x + yi = 1 + (2 + 3i) \)
Simplify the right side: \( 2x + yi = 3 + 3i \)
Compare the real parts:
\( 2x = 3 \)
\( \implies x = \frac{3}{2} \)
Compare the imaginary parts:
\( y = 3 \)

(iv) Given \( x + 2yi = i − (- 3 + 5i) \)
Simplify the right side: \( x + 2yi = i + 3 - 5i \)
\( x + 2yi = 3 - 4i \)
Compare the real parts:
\( x = 3 \)
Compare the imaginary parts:
\( 2y = -4 \)
\( \implies y = \frac{-4}{2} \)
\( \implies y = -2 \)

(v) Given \( (x – yi)(2 + 3i) = \frac{x-2 i}{1-i} \)
First, simplify the left side:
\( (x - yi)(2 + 3i) = 2x + 3xi - 2yi - 3yi^2 = (2x + 3y) + i(3x - 2y) \)
Next, simplify the right side:
\( \frac{x-2i}{1-i} = \frac{(x-2i)(1+i)}{(1-i)(1+i)} = \frac{x+xi-2i-2i^2}{1^2-i^2} = \frac{x+xi-2i+2}{1+1} = \frac{(x+2)+i(x-2)}{2} \)
So, \( (2x + 3y) + i(3x - 2y) = \frac{x+2}{2} + i\frac{x-2}{2} \)
Compare the real parts:
\( 2x + 3y = \frac{x+2}{2} \)
\( \implies 4x + 6y = x + 2 \)
\( \implies 3x + 6y = 2 \) (Equation 1)
Compare the imaginary parts:
\( 3x - 2y = \frac{x-2}{2} \)
\( \implies 6x - 4y = x - 2 \)
\( \implies 5x - 4y = -2 \) (Equation 2)
Multiply Equation 1 by 2: \( 6x + 12y = 4 \) (Equation 3)
Multiply Equation 2 by 3: \( 15x - 12y = -6 \) (Equation 4)
Add Equation 3 and Equation 4:
\( (6x + 15x) + (12y - 12y) = 4 - 6 \)
\( 21x = -2 \)
\( \implies x = -\frac{2}{21} \)
Substitute \( x = -\frac{2}{21} \) into Equation 1:
\( 3(-\frac{2}{21}) + 6y = 2 \)
\( -\frac{2}{7} + 6y = 2 \)
\( 6y = 2 + \frac{2}{7} = \frac{14+2}{7} = \frac{16}{7} \)
\( \implies y = \frac{16}{7 \cdot 6} = \frac{16}{42} = \frac{8}{21} \)
Thus, \( x = -\frac{2}{21} \) and \( y = \frac{8}{21} \).

(vi) Given \( (x^2 + 2xi) – (3x^2 + yi) = (3 – 5i) + (1 + 2yi) \)
(Assuming the question meant \( x^4 \) instead of \( x^2 \) in the first term, as per the provided solution steps.)
Group the real and imaginary parts on the left side:
\( (x^4 - 3x^2) + i(2x - y) \)
Group the real and imaginary parts on the right side:
\( (3 + 1) + i(-5 + 2y) = 4 + i(2y - 5) \)
Equating real parts:
\( x^4 - 3x^2 = 4 \)
\( \implies x^4 - 3x^2 - 4 = 0 \)
Let \( a = x^2 \). Then \( a^2 - 3a - 4 = 0 \)
Factorizing gives \( (a - 4)(a + 1) = 0 \)
So, \( a = 4 \) or \( a = -1 \).
Since \( a = x^2 \), we have \( x^2 = 4 \) or \( x^2 = -1 \).
For real values of x, \( x^2 = 4 \implies x = \pm 2 \). ( \( x^2 = -1 \) has no real solutions for x).
Equating imaginary parts:
\( 2x - y = 2y - 5 \)
\( \implies 2x - 3y = -5 \)
Case 1: If \( x = 2 \)
\( 2(2) - 3y = -5 \)
\( 4 - 3y = -5 \)
\( -3y = -9 \)
\( \implies y = 3 \)
Case 2: If \( x = -2 \)
\( 2(-2) - 3y = -5 \)
\( -4 - 3y = -5 \)
\( -3y = -1 \)
\( \implies y = \frac{1}{3} \)
Thus, the possible solutions are \( (x = 2, y = 3) \) and \( (x = -2, y = \frac{1}{3}) \).
In simple words: To solve for real 'x' and 'y', we separate the equation into two parts: one for the numbers without 'i' (real part) and one for the numbers with 'i' (imaginary part). We then solve these two simpler equations to find 'x' and 'y'. If the initial question seems to have a typo that's resolved in the solution, we follow the logic that makes the steps work out.

🎯 Exam Tip: When comparing complex numbers, always equate the real parts on both sides and the imaginary parts on both sides separately. Be careful with signs when distributing negative numbers or expanding squares.

Question 3. Determine the conjugate and the reciprocal of each complex number given below :
(i) i
(ii) i³
(iii) 3 - i
(iv) \( \sqrt{-1}-3 \)
(v) \( \sqrt{-9}-3 \)
Answer:
(i) For \( z = i \):
Conjugate: \( \bar{z} = \overline{i} = -i \)
Reciprocal: \( \frac{1}{z} = \frac{1}{i} = \frac{1 \cdot i}{i \cdot i} = \frac{i}{-1} = -i \)

(ii) For \( z = i^3 \):
First, simplify \( i^3 = i^2 \cdot i = -1 \cdot i = -i \).
Conjugate: \( \bar{z} = \overline{-i} = i \)
Reciprocal: \( \frac{1}{z} = \frac{1}{-i} = \frac{1 \cdot i}{-i \cdot i} = \frac{i}{-i^2} = \frac{i}{1} = i \)

(iii) For \( z = 3 - i \):
Conjugate: \( \bar{z} = \overline{3-i} = 3+i \)
Reciprocal: \( \frac{1}{z} = \frac{1}{3-i} = \frac{1 \cdot (3+i)}{(3-i)(3+i)} = \frac{3+i}{3^2-i^2} = \frac{3+i}{9-(-1)} = \frac{3+i}{10} = \frac{3}{10} + \frac{1}{10}i \)

(iv) For \( z = \sqrt{-1}-3 \):
First, simplify \( z = i - 3 = -3 + i \).
Conjugate: \( \bar{z} = \overline{-3+i} = -3-i \)
Reciprocal: \( \frac{1}{z} = \frac{1}{-3+i} = \frac{1 \cdot (-3-i)}{(-3+i)(-3-i)} = \frac{-3-i}{(-3)^2-i^2} = \frac{-3-i}{9-(-1)} = \frac{-3-i}{10} = -\frac{3}{10} - \frac{1}{10}i \)

(v) For \( z = \sqrt{-9}-3 \):
First, simplify \( z = 3i - 3 = -3 + 3i \).
Conjugate: \( \bar{z} = \overline{-3+3i} = -3-3i \)
Reciprocal: \( \frac{1}{z} = \frac{1}{-3+3i} = \frac{1 \cdot (-3-3i)}{(-3+3i)(-3-3i)} = \frac{-3-3i}{(-3)^2-(3i)^2} = \frac{-3-3i}{9-(-9)} = \frac{-3-3i}{18} = -\frac{3}{18} - \frac{3}{18}i = -\frac{1}{6} - \frac{1}{6}i \)
In simple words: The conjugate of a complex number is found by changing the sign of its imaginary part. The reciprocal is found by dividing 1 by the number; this often involves multiplying by the conjugate to remove the imaginary part from the bottom of the fraction. Complex numbers allow us to work with square roots of negative numbers.

🎯 Exam Tip: To find the reciprocal of a complex number \( a+bi \), calculate \( \frac{1}{a+bi} \) and then multiply the numerator and denominator by the conjugate \( a-bi \) to rationalize the expression.

Question 4. Simplify:
(i) \( (3 – 7i)² \)
(ii) \( \left(\frac{-1}{2}-\frac{\sqrt{3}}{2} i\right)^2 \)
(iii) \( (9 + 4i) (\frac { 3 }{ 2 } – i ) (9 – 4i) \)
Answer:
(i) To simplify \( (3 – 7i)^2 \), we use the formula \( (a-b)^2 = a^2 - 2ab + b^2 \):
\( (3 - 7i)^2 = 3^2 - 2(3)(7i) + (7i)^2 \)
\( = 9 - 42i + 49i^2 \)
Since \( i^2 = -1 \):
\( = 9 - 42i + 49(-1) \)
\( = 9 - 42i - 49 \)
\( = -40 - 42i \)

(ii) To simplify \( \left(\frac{-1}{2}-\frac{\sqrt{3}}{2} i\right)^2 \):
Let the expression be Z. We can write \( Z = \left(-\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)\right)^2 = \left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^2 \)
Using \( (a+b)^2 = a^2 + 2ab + b^2 \):
\( Z = \left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2} i\right) + \left(\frac{\sqrt{3}}{2} i\right)^2 \)
\( = \frac{1}{4} + \frac{\sqrt{3}}{2} i + \frac{3}{4}i^2 \)
Since \( i^2 = -1 \):
\( = \frac{1}{4} + \frac{\sqrt{3}}{2} i - \frac{3}{4} \)
\( = \frac{1-3}{4} + \frac{\sqrt{3}}{2} i \)
\( = -\frac{2}{4} + \frac{\sqrt{3}}{2} i \)
\( = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \)

(iii) To simplify \( (9 + 4i) (\frac { 3 }{ 2 } – i ) (9 – 4i) \):
We can rearrange and multiply \( (9+4i)(9-4i) \) first, using the difference of squares formula \( (a+b)(a-b) = a^2 - b^2 \):
\( = [(9+4i)(9-4i)] \left(\frac{3}{2} - i\right) \)
\( = [9^2 - (4i)^2] \left(\frac{3}{2} - i\right) \)
\( = [81 - 16i^2] \left(\frac{3}{2} - i\right) \)
Since \( i^2 = -1 \):
\( = [81 - 16(-1)] \left(\frac{3}{2} - i\right) \)
\( = [81 + 16] \left(\frac{3}{2} - i\right) \)
\( = 97 \left(\frac{3}{2} - i\right) \)
Now, distribute the 97:
\( = 97 \cdot \frac{3}{2} - 97i \)
\( = \frac{291}{2} - 97i \)
In simple words: To simplify expressions with complex numbers, we use basic algebra rules like expanding squares and multiplying. Remember that \( i^2 \) is always -1, which helps simplify the numbers. Grouping terms carefully can make the calculation easier.

🎯 Exam Tip: Always look for opportunities to use algebraic identities like \( (a+b)^2 \), \( (a-b)^2 \), or \( (a+b)(a-b) \) when simplifying complex number expressions. This can save time and reduce errors.

Question 5. Determine real values of x and y for which each statement is true.
(i) \( \frac {x+y }{ i } + x − y + 4 = 0 \)
(ii) \( – (x + 3y)i + (2x− y + 1) = \frac { 8 }{ i } \)
(iii) \( (x - yi) = \frac{2+i}{1+i} \)
(iv) \( (3 – 4i) + (x + yi) = 1 + 0i \)
(v) \( (x – yi)(2 + 3i) = \frac{x-2 i}{1-i} \)
(vi) \( (x^2 + 2xi) – (3x^2 + yi) = (3 – 5i) + (1 + 2yi) \)
Answer:
(i) Given \( \frac {x+y }{ i } + x − y + 4 = 0 \)
Multiply \( \frac{x+y}{i} \) by \( \frac{i}{i} \): \( \frac{(x+y)i}{i^2} = -(x+y)i \)
The equation becomes: \( -(x+y)i + x - y + 4 = 0 \)
Rearrange into real and imaginary parts: \( (x - y + 4) + i(-(x+y)) = 0 + 0i \)
Compare real parts:
\( x - y + 4 = 0 \)
\( \implies x - y = -4 \) (Equation 1)
Compare imaginary parts:
\( -(x+y) = 0 \)
\( \implies x + y = 0 \) (Equation 2)
Add Equation 1 and Equation 2:
\( (x-y) + (x+y) = -4 + 0 \)
\( 2x = -4 \)
\( \implies x = -2 \)
Substitute \( x = -2 \) into Equation 2:
\( -2 + y = 0 \)
\( \implies y = 2 \)

(ii) Given \( – (x + 3y)i + (2x− y + 1) = \frac { 8 }{ i } \)
Simplify the right side: \( \frac{8}{i} = \frac{8i}{i^2} = -8i \)
The equation is: \( (2x - y + 1) - (x + 3y)i = -8i \)
Compare real parts:
\( 2x - y + 1 = 0 \)
\( \implies 2x - y = -1 \) (Equation 1)
Compare imaginary parts:
\( -(x + 3y) = -8 \)
\( \implies x + 3y = 8 \) (Equation 2)
Multiply Equation 1 by 3:
\( 3(2x - y) = 3(-1) \)
\( 6x - 3y = -3 \) (Equation 3)
Add Equation 2 and Equation 3:
\( (x + 6x) + (3y - 3y) = 8 + (-3) \)
\( 7x = 5 \)
\( \implies x = \frac{5}{7} \)
Substitute \( x = \frac{5}{7} \) into Equation 2:
\( \frac{5}{7} + 3y = 8 \)
\( 3y = 8 - \frac{5}{7} = \frac{56-5}{7} = \frac{51}{7} \)
\( \implies y = \frac{51}{7 \cdot 3} = \frac{17}{7} \)

(iii) Given \( (x - yi) = \frac{2+i}{1+i} \)
Simplify the right side by multiplying numerator and denominator by the conjugate of the denominator:
\( \frac{2+i}{1+i} = \frac{(2+i)(1-i)}{(1+i)(1-i)} = \frac{2-2i+i-i^2}{1^2-i^2} = \frac{2-i+1}{1-(-1)} = \frac{3-i}{2} \)
So, \( x - yi = \frac{3}{2} - \frac{1}{2}i \)
Compare real parts:
\( x = \frac{3}{2} \)
Compare imaginary parts:
\( -y = -\frac{1}{2} \)
\( \implies y = \frac{1}{2} \)

(iv) Given \( (3 – 4i) + (x + yi) = 1 + 0i \)
(Assuming the question meant \( (3-4i)(x+yi) = 1+0i \), as implied by the solution.)
To solve for \( x+yi \), divide \( 1+0i \) by \( 3-4i \):
\( x + yi = \frac{1}{3-4i} \)
Multiply numerator and denominator by the conjugate of the denominator:
\( x + yi = \frac{1 \cdot (3+4i)}{(3-4i)(3+4i)} = \frac{3+4i}{3^2-(4i)^2} = \frac{3+4i}{9-(-16)} = \frac{3+4i}{9+16} = \frac{3+4i}{25} \)
So, \( x + yi = \frac{3}{25} + \frac{4}{25}i \)
Compare real parts:
\( x = \frac{3}{25} \)
Compare imaginary parts:
\( y = \frac{4}{25} \)

(v) Given \( (x – yi)(2 + 3i) = \frac{x-2 i}{1-i} \)
(As solved in Question 2(v) earlier):
\( x = -\frac{2}{21} \)
\( y = \frac{8}{21} \)

(vi) Given \( (x^2 + 2xi) – (3x^2 + yi) = (3 – 5i) + (1 + 2yi) \)
(Assuming the question meant \( (x^4 + 2xi) \) in the first term, as implied by the solution.)
Group real and imaginary parts on both sides:
Left side: \( (x^4 - 3x^2) + i(2x - y) \)
Right side: \( (3 + 1) + i(-5 + 2y) = 4 + i(2y - 5) \)
Equating real parts:
\( x^4 - 3x^2 = 4 \)
\( \implies x^4 - 3x^2 - 4 = 0 \)
Let \( a = x^2 \). Then \( a^2 - 3a - 4 = 0 \)
\( (a - 4)(a + 1) = 0 \)
So \( a = 4 \) or \( a = -1 \).
This means \( x^2 = 4 \) or \( x^2 = -1 \). Since x is a real number, \( x^2 = 4 \implies x = \pm 2 \).
Equating imaginary parts:
\( 2x - y = 2y - 5 \)
\( \implies 2x - 3y = -5 \)
For \( x = 2 \): \( 2(2) - 3y = -5 \implies 4 - 3y = -5 \implies -3y = -9 \implies y = 3 \)
For \( x = -2 \): \( 2(-2) - 3y = -5 \implies -4 - 3y = -5 \implies -3y = -1 \implies y = \frac{1}{3} \)
So, the solutions are \( x = 2, y = 3 \) and \( x = -2, y = \frac{1}{3} \).
In simple words: For equations with complex numbers, we match the parts that don't have 'i' (real parts) and the parts that do have 'i' (imaginary parts) separately. This gives us two normal equations to solve for 'x' and 'y'. Sometimes, we need to simplify fractions with 'i' by using conjugates.

🎯 Exam Tip: When equations involve fractions with 'i' in the denominator, simplify them first by multiplying the numerator and denominator by the conjugate to get rid of 'i' in the bottom. This makes comparing real and imaginary parts much clearer.

Question 6. Write the conjugate of \( (6 + 5i)^2 \).
Answer:
Let \( z = (6 + 5i)^2 \).
First, expand the square using \( (a+b)^2 = a^2 + 2ab + b^2 \):
\( z = 6^2 + 2(6)(5i) + (5i)^2 \)
\( z = 36 + 60i + 25i^2 \)
Since \( i^2 = -1 \):
\( z = 36 + 60i - 25 \)
\( z = 11 + 60i \)
Now, find the conjugate of z. The conjugate of \( a+bi \) is \( a-bi \).
\( \bar{z} = \overline{11 + 60i} = 11 - 60i \)
In simple words: First, you square the complex number just like a regular algebra expression, remembering that \( i^2 \) becomes -1. Once you have a simple complex number (like 'a + bi'), you find its conjugate by changing the sign of the 'i' part.

🎯 Exam Tip: Always fully simplify the complex number into the form \( a+bi \) before finding its conjugate or performing other operations. This reduces the chance of error.

Question 7. Write the additive inverse of the following
(i) – 2 + 3i
(ii) 3 - 4i
Answer:
(i) For \( z = -2 + 3i \):
The additive inverse of a complex number \( z \) is \( -z \).
\( -z = -(-2 + 3i) = 2 - 3i \)

(ii) For \( z = 3 - 4i \):
The additive inverse of a complex number \( z \) is \( -z \).
\( -z = -(3 - 4i) = -3 + 4i \)
In simple words: To find the additive inverse of any complex number, you just change the sign of both its real part and its imaginary part. It's like multiplying the entire number by -1.

🎯 Exam Tip: The additive inverse of a complex number \( a+bi \) is \( -a-bi \). It's the number that, when added to the original number, results in zero.

Question 8. Determine the multiplicative inverse of each of the following complex numbers when it exists.
(i) 2 + 2i
(ii) – 7 + 0i
(iii) 0 + 0i
(iv) – 16
(v) \( \frac{i}{1+i} \)
(vi) \( (1 + i)² \)
(vii) \( \frac{3+4 i}{4-5 i} \)
(viii) \( (6 + 5i)² \)
(ix) \( \frac{(2+3 i)(3+2 i) i}{5+i} \)
Answer:
(i) For \( z = 2 + 2i \):
The multiplicative inverse is \( \frac{1}{z} \).
\( \frac{1}{2+2i} = \frac{1 \cdot (2-2i)}{(2+2i)(2-2i)} = \frac{2-2i}{2^2-(2i)^2} = \frac{2-2i}{4-(-4)} = \frac{2-2i}{8} = \frac{2}{8} - \frac{2}{8}i = \frac{1}{4} - \frac{1}{4}i \)

(ii) For \( z = -7 + 0i = -7 \):
The multiplicative inverse is \( \frac{1}{z} \).
\( \frac{1}{-7} = -\frac{1}{7} = -\frac{1}{7} + 0i \)

(iii) For \( z = 0 + 0i = 0 \):
The multiplicative inverse does not exist, because division by zero is not possible.

(iv) For \( z = -16 \):
The multiplicative inverse is \( \frac{1}{z} \).
\( \frac{1}{-16} = -\frac{1}{16} = -\frac{1}{16} + 0i \)

(v) For \( z = \frac{i}{1+i} \):
First, simplify \( z \):
\( z = \frac{i}{1+i} = \frac{i(1-i)}{(1+i)(1-i)} = \frac{i-i^2}{1^2-i^2} = \frac{i-(-1)}{1-(-1)} = \frac{1+i}{2} \)
Now, find the multiplicative inverse \( \frac{1}{z} \):
\( \frac{1}{z} = \frac{1}{\frac{1+i}{2}} = \frac{2}{1+i} = \frac{2(1-i)}{(1+i)(1-i)} = \frac{2(1-i)}{1^2-i^2} = \frac{2(1-i)}{1-(-1)} = \frac{2(1-i)}{2} = 1-i \)

(vi) For \( z = (1 + i)^2 \):
First, simplify \( z \):
\( z = (1+i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i \)
Now, find the multiplicative inverse \( \frac{1}{z} \):
\( \frac{1}{z} = \frac{1}{2i} = \frac{1 \cdot i}{2i \cdot i} = \frac{i}{2i^2} = \frac{i}{2(-1)} = -\frac{i}{2} = 0 - \frac{1}{2}i \)

(vii) For \( z = \frac{3+4 i}{4-5 i} \):
The multiplicative inverse is \( \frac{1}{z} \), which means we swap the numerator and denominator:
\( \frac{1}{z} = \frac{4-5i}{3+4i} \)
Now, rationalize the denominator:
\( \frac{4-5i}{3+4i} = \frac{(4-5i)(3-4i)}{(3+4i)(3-4i)} = \frac{12-16i-15i+20i^2}{3^2-(4i)^2} = \frac{12-31i-20}{9-(-16)} = \frac{-8-31i}{25} = -\frac{8}{25} - \frac{31}{25}i \)

(viii) For \( z = (6 + 5i)^2 \):
First, simplify \( z \):
\( z = (6+5i)^2 = 6^2 + 2(6)(5i) + (5i)^2 = 36 + 60i + 25i^2 = 36 + 60i - 25 = 11 + 60i \)
Now, find the multiplicative inverse \( \frac{1}{z} \):
\( \frac{1}{z} = \frac{1}{11+60i} = \frac{1 \cdot (11-60i)}{(11+60i)(11-60i)} = \frac{11-60i}{11^2-(60i)^2} = \frac{11-60i}{121-(-3600)} = \frac{11-60i}{121+3600} = \frac{11-60i}{3721} = \frac{11}{3721} - \frac{60}{3721}i \)

(ix) For \( z = \frac{(2+3 i)(3+2 i) i}{5+i} \):
First, simplify the numerator of \( z \):
\( (2+3i)(3+2i)i = (6+4i+9i+6i^2)i = (6+13i-6)i = (13i)i = 13i^2 = -13 \)
So, \( z = \frac{-13}{5+i} \)
Now, find the multiplicative inverse \( \frac{1}{z} \):
\( \frac{1}{z} = \frac{5+i}{-13} = -\frac{5}{13} - \frac{1}{13}i \)
In simple words: The multiplicative inverse of a complex number is like its "flip" or reciprocal, which you get by dividing 1 by the number. You then multiply the top and bottom of the fraction by the conjugate of the bottom number to make sure there's no 'i' in the denominator. Remember that zero has no multiplicative inverse.

🎯 Exam Tip: Always check if a complex number can be simplified first (e.g., \( (1+i)^2 \) simplifies to \( 2i \)) before finding its multiplicative inverse. This often makes the calculation much easier.

Question 9. Simplify:
(i) \( (1 + i)⁻¹ \)
(ii) \( \sqrt{-\frac{49}{25}} \sqrt{-\frac{1}{9}} \)
(iii) \( \sqrt{-64} \cdot(3+\sqrt{-361}) \)
(iv) \( (3 – 7i)² \)
(v) \( \left(-\frac{1}{2}-\frac{\sqrt{3}}{2} i\right)^2 \)
(vi) \( \frac{(1-i)^3}{\left(1-i^3\right)} \)
(vii) \( \left(\frac{1+i}{1-i}\right)^{4 n+1} \) (n is a positive integer)
(viii) \( \frac{\sqrt{(5+12 i)}+\sqrt{(5-12 i)}}{\sqrt{(5+12 i)}-\sqrt{(5-12 i)}} \)
Answer:
(i) To simplify \( (1 + i)^{-1} \):
\( (1+i)^{-1} = \frac{1}{1+i} = \frac{1 \cdot (1-i)}{(1+i)(1-i)} = \frac{1-i}{1^2-i^2} = \frac{1-i}{1-(-1)} = \frac{1-i}{2} = \frac{1}{2} - \frac{1}{2}i \)

(ii) To simplify \( \sqrt{-\frac{49}{25}} \sqrt{-\frac{1}{9}} \):
\( = \sqrt{\frac{49}{25}i^2} \sqrt{\frac{1}{9}i^2} \)
\( = \left(\frac{7}{5}i\right) \left(\frac{1}{3}i\right) \)
\( = \frac{7}{15}i^2 \)
Since \( i^2 = -1 \):
\( = -\frac{7}{15} \)

(iii) To simplify \( \sqrt{-64} \cdot(3+\sqrt{-361}) \):
\( = \sqrt{64i^2} \cdot (3+\sqrt{361i^2}) \)
\( = (8i) \cdot (3+19i) \)
\( = 8i \cdot 3 + 8i \cdot 19i \)
\( = 24i + 152i^2 \)
Since \( i^2 = -1 \):
\( = 24i - 152 \)

(iv) To simplify \( (3 – 7i)^2 \):
Using \( (a-b)^2 = a^2 - 2ab + b^2 \):
\( = 3^2 - 2(3)(7i) + (7i)^2 \)
\( = 9 - 42i + 49i^2 \)
Since \( i^2 = -1 \):
\( = 9 - 42i - 49 \)
\( = -40 - 42i \)

(v) To simplify \( \left(-\frac{1}{2}-\frac{\sqrt{3}}{2} i\right)^2 \):
This is the same as Question 4(ii).
\( = \left(-\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)\right)^2 = \left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^2 \)
\( = \left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2} i\right) + \left(\frac{\sqrt{3}}{2} i\right)^2 \)
\( = \frac{1}{4} + \frac{\sqrt{3}}{2} i + \frac{3}{4}i^2 \)
\( = \frac{1}{4} + \frac{\sqrt{3}}{2} i - \frac{3}{4} \)
\( = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \)

(vi) To simplify \( \frac{(1-i)^3}{\left(1-i^3\right)} \):
First, simplify the numerator \( (1-i)^3 \):
\( (1-i)^3 = 1^3 - 3(1)^2(i) + 3(1)(i)^2 - i^3 \)
\( = 1 - 3i + 3i^2 - i^3 \)
Since \( i^2 = -1 \) and \( i^3 = -i \):
\( = 1 - 3i + 3(-1) - (-i) \)
\( = 1 - 3i - 3 + i \)
\( = -2 - 2i \)
Next, simplify the denominator \( (1-i^3) \):
\( 1 - i^3 = 1 - (-i) = 1 + i \)
Now, divide the simplified numerator by the simplified denominator:
\( \frac{-2-2i}{1+i} = \frac{-2(1+i)}{1+i} = -2 \)

(vii) To simplify \( \left(\frac{1+i}{1-i}\right)^{4 n+1} \):
First, simplify the fraction \( \frac{1+i}{1-i} \):
\( \frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{(1+i)^2}{1^2-i^2} = \frac{1+2i+i^2}{1-(-1)} = \frac{1+2i-1}{2} = \frac{2i}{2} = i \)
Now, the expression becomes \( i^{4n+1} \).
Since n is a positive integer, \( i^{4n} = (i^4)^n = (1)^n = 1 \).
So, \( i^{4n+1} = i^{4n} \cdot i^1 = 1 \cdot i = i \)

(viii) To simplify \( \frac{\sqrt{(5+12 i)}+\sqrt{(5-12 i)}}{\sqrt{(5+12 i)}-\sqrt{(5-12 i)}} \):
We know that \( (3+2i)^2 = 9+12i+4i^2 = 5+12i \)
And \( (3-2i)^2 = 9-12i+4i^2 = 5-12i \)
So, \( \sqrt{5+12i} = 3+2i \) and \( \sqrt{5-12i} = 3-2i \).
Substitute these into the expression:
\( = \frac{(3+2i) + (3-2i)}{(3+2i) - (3-2i)} \)
\( = \frac{3+2i+3-2i}{3+2i-3+2i} \)
\( = \frac{6}{4i} \)
Now, rationalize the denominator:
\( = \frac{6 \cdot i}{4i \cdot i} = \frac{6i}{4i^2} = \frac{6i}{-4} = -\frac{3}{2}i \)
In simple words: To simplify these complex expressions, we break them down into smaller steps. This involves using properties of 'i' like \( i^2 = -1 \), expanding brackets, and simplifying fractions by using the conjugate. Recognizing patterns like perfect squares or differences of squares can make some problems much quicker to solve.

🎯 Exam Tip: When simplifying powers of 'i', remember the cycle: \( i^1=i, i^2=-1, i^3=-i, i^4=1 \). Any power of 'i' can be reduced to one of these four values. For complex fractions, always aim to rationalize the denominator.

Question 10. Prove that \( \left[\left(\frac{3+2 i}{2-5 i}\right)+\left(\frac{3-2 i}{2+5 i}\right)\right] \) is rational.
Answer:
Let the given expression be Z.
\( Z = \left(\frac{3+2 i}{2-5 i}\right)+\left(\frac{3-2 i}{2+5 i}\right) \)
Notice that the second term is the conjugate of the first term. Let \( w = \frac{3+2i}{2-5i} \). Then the expression is \( w + \bar{w} \).
We know that the sum of a complex number and its conjugate is always twice its real part, which is a real number. We need to show it's rational.
First, simplify the first term \( w \):
\( w = \frac{3+2i}{2-5i} = \frac{(3+2i)(2+5i)}{(2-5i)(2+5i)} = \frac{6+15i+4i+10i^2}{2^2-(5i)^2} = \frac{6+19i-10}{4-(-25)} = \frac{-4+19i}{29} \)
The second term is the conjugate of \( w \):
\( \bar{w} = \frac{3-2i}{2+5i} = \frac{(3-2i)(2-5i)}{(2+5i)(2-5i)} = \frac{6-15i-4i+10i^2}{2^2-(5i)^2} = \frac{6-19i-10}{4-(-25)} = \frac{-4-19i}{29} \)
Now, add the two terms:
\( Z = w + \bar{w} = \frac{-4+19i}{29} + \frac{-4-19i}{29} \)
\( = \frac{-4+19i-4-19i}{29} \)
\( = \frac{-8}{29} \)
Since \( -\frac{8}{29} \) is a fraction of two integers, it is a rational number.
Thus, the given expression is rational.
In simple words: We are adding a complex number to its conjugate. When you add a number to its conjugate, the imaginary parts always cancel out, leaving only a real number. In this case, that real number turned out to be a fraction of two whole numbers, which means it is a rational number.

🎯 Exam Tip: Remember the property that \( z + \bar{z} = 2 \text{Re}(z) \). If \( \text{Re}(z) \) is rational, then \( z + \bar{z} \) will also be rational. This can simplify proofs related to sums of conjugates.

Question 11. Show that \( \frac{1+2 i}{3+4 i} \times \frac{1-2 i}{3-4 i} \) is real.
Answer:
Let the given expression be Z.
\( Z = \frac{1+2 i}{3+4 i} \times \frac{1-2 i}{3-4 i} \)
We can rearrange the terms and multiply the numerators together and the denominators together:
\( Z = \frac{(1+2i)(1-2i)}{(3+4i)(3-4i)} \)
Now, use the difference of squares formula, \( (a+b)(a-b) = a^2 - b^2 \), for both the numerator and the denominator.
Numerator: \( (1+2i)(1-2i) = 1^2 - (2i)^2 = 1 - 4i^2 \)
Since \( i^2 = -1 \):
\( = 1 - 4(-1) = 1 + 4 = 5 \)
Denominator: \( (3+4i)(3-4i) = 3^2 - (4i)^2 = 9 - 16i^2 \)
Since \( i^2 = -1 \):
\( = 9 - 16(-1) = 9 + 16 = 25 \)
So, substitute these values back into the expression for Z:
\( Z = \frac{5}{25} \)
\( Z = \frac{1}{5} \)
Since \( \frac{1}{5} \) is a real number (it has no imaginary part), the expression is real.
In simple words: We multiply the top parts and bottom parts of the fractions. Using a special math rule, we see that the 'i' parts cancel out in both the top and the bottom. What's left is a simple fraction of whole numbers, which is a real number, meaning it does not have any 'i' component.

🎯 Exam Tip: Recognizing pairs of complex numbers like \( (a+bi)(a-bi) \) is key, as their product \( a^2+b^2 \) is always a real number. This property is very useful for showing expressions are real or rational.

Question 12. Perform the indicated operation and give your answer in the form x + yi, where x and y are real numbers and i = \( \sqrt{-1} \).
(i) \( (3 + 4i)^{-1} \)
(ii) \( \frac{2-\sqrt{-25}}{1-\sqrt{-16}} \)
(iii) \( \frac{5+2i}{-1+\sqrt{3}i} \)
(iv) \( \frac{5-3i}{6+i} \)
(v) \( (\sqrt{5}-7i)(\sqrt{5}-7i)^2 + (-2+7i)^2 \)
Answer:
(i) We need to find the inverse of \( 3+4i \). We can do this by multiplying the complex number by its conjugate and dividing by the square of its modulus.
\( (3+4i)^{-1} = \frac{1}{3+4i} \times \frac{3-4i}{3-4i} \)
\( \implies = \frac{3-4i}{3^2-(4i)^2} \)
\( \implies = \frac{3-4i}{9-16i^2} \)
\( \implies = \frac{3-4i}{9+16} \)
\( \implies = \frac{3-4i}{25} \)
\( \implies = \frac{3}{25} - \frac{4}{25}i \) This result is in the form \( x+yi \).
(ii) First, simplify the square roots with negative numbers. We know \( \sqrt{-a} = i\sqrt{a} \).
\( \sqrt{-25} = 5i \) and \( \sqrt{-16} = 4i \).
So, the expression becomes \( \frac{2-5i}{1-4i} \).
To simplify, we multiply the numerator and denominator by the conjugate of the denominator.
\( \frac{2-5i}{1-4i} \times \frac{1+4i}{1+4i} \)
\( \implies = \frac{(2-5i)(1+4i)}{1^2-(4i)^2} \)
\( \implies = \frac{2(1)+2(4i)-5i(1)-5i(4i)}{1-16i^2} \)
\( \implies = \frac{2+8i-5i-20i^2}{1+16} \)
\( \implies = \frac{2+3i+20}{17} \)
\( \implies = \frac{22+3i}{17} \)
\( \implies = \frac{22}{17} + \frac{3}{17}i \) This is the simplified form \( x+yi \).
(iii) To simplify this fraction, we multiply the numerator and denominator by the conjugate of the denominator, which is \( -1-\sqrt{3}i \).
\( \frac{5+2i}{-1+\sqrt{3}i} \times \frac{-1-\sqrt{3}i}{-1-\sqrt{3}i} \)
\( \implies = \frac{(5+2i)(-1-\sqrt{3}i)}{(-1)^2-(\sqrt{3}i)^2} \)
\( \implies = \frac{5(-1)+5(-\sqrt{3}i)+2i(-1)+2i(-\sqrt{3}i)}{1-3i^2} \)
\( \implies = \frac{-5-5\sqrt{3}i-2i-2\sqrt{3}i^2}{1+3} \)
\( \implies = \frac{-5-5\sqrt{3}i-2i+2\sqrt{3}}{4} \)
\( \implies = \frac{(-5+2\sqrt{3}) + (-5\sqrt{3}-2)i}{4} \)
\( \implies = \frac{-5+2\sqrt{3}}{4} + \frac{-5\sqrt{3}-2}{4}i \) This is the simplified form \( x+yi \).
(iv) To simplify the fraction, we multiply the numerator and denominator by the conjugate of the denominator, which is \( 6-i \).
\( \frac{5-3i}{6+i} \times \frac{6-i}{6-i} \)
\( \implies = \frac{(5-3i)(6-i)}{6^2-i^2} \)
\( \implies = \frac{5(6)+5(-i)-3i(6)-3i(-i)}{36-(-1)} \)
\( \implies = \frac{30-5i-18i+3i^2}{36+1} \)
\( \implies = \frac{30-23i-3}{37} \)
\( \implies = \frac{27-23i}{37} \)
\( \implies = \frac{27}{37} - \frac{23}{37}i \) This is the simplified form \( x+yi \).
(v) First, we simplify the terms by expanding the powers.
The expression is \( (\sqrt{5}-7i)(\sqrt{5}-7i)^2 + (-2+7i)^2 \).
This can be written as \( (\sqrt{5}-7i)^3 + (-2+7i)^2 \).
Let's expand \( (\sqrt{5}-7i)^2 \):
\( (\sqrt{5}-7i)^2 = (\sqrt{5})^2 - 2(\sqrt{5})(7i) + (7i)^2 = 5 - 14\sqrt{5}i - 49 = -44 - 14\sqrt{5}i \).
Now, the first term: \( (\sqrt{5}-7i)(-44-14\sqrt{5}i) \)
\( = \sqrt{5}(-44) + \sqrt{5}(-14\sqrt{5}i) -7i(-44) -7i(-14\sqrt{5}i) \)
\( = -44\sqrt{5} - 14(5)i + 308i + 98\sqrt{5}i^2 \)
\( = -44\sqrt{5} - 70i + 308i - 98\sqrt{5} \)
\( = (-44\sqrt{5} - 98\sqrt{5}) + (-70i + 308i) \)
\( = -142\sqrt{5} + 238i \).
Next, expand the second term: \( (-2+7i)^2 \)
\( = (-2)^2 + 2(-2)(7i) + (7i)^2 = 4 - 28i - 49 = -45 - 28i \).
Finally, add the two simplified terms:
\( (-142\sqrt{5} + 238i) + (-45 - 28i) \)
\( = (-142\sqrt{5} - 45) + (238 - 28)i \)
\( = (-142\sqrt{5} - 45) + 210i \) This is the final form \( x+yi \).
In simple words: To simplify complex fractions, always multiply the top and bottom by the conjugate of the bottom part. For powers, expand them carefully, remembering that \( i^2 \) is -1. Combining like terms (real parts with real parts, imaginary parts with imaginary parts) at each step helps avoid mistakes.

🎯 Exam Tip: When dealing with complex fractions, rationalizing the denominator by multiplying by its conjugate is crucial. Remember that \( i^2 = -1 \), and this substitution is key to simplifying expressions and getting the answer in the \( x+yi \) form.

Question 13. If \( x+yi = \frac{u+vi}{u-vi} \) prove that \( x^2+y^2=1 \).
Answer: We are given the equation \( x+yi = \frac{u+vi}{u-vi} \).
To prove \( x^2+y^2=1 \), we can use the property of moduli of complex numbers. The modulus of a complex number \( z=x+yi \) is \( |z|=\sqrt{x^2+y^2} \), and \( |z^n|=|z|^n \). Also, \( \left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} \).
Take the modulus on both sides of the given equation:
\( |x+yi| = \left|\frac{u+vi}{u-vi}\right| \)
\( \implies \sqrt{x^2+y^2} = \frac{|u+vi|}{|u-vi|} \)
The modulus of \( u+vi \) is \( \sqrt{u^2+v^2} \).
The modulus of \( u-vi \) is \( \sqrt{u^2+(-v)^2} = \sqrt{u^2+v^2} \).
\( \implies \sqrt{x^2+y^2} = \frac{\sqrt{u^2+v^2}}{\sqrt{u^2+v^2}} \)
\( \implies \sqrt{x^2+y^2} = 1 \). When you divide a number by itself, you get 1.
Now, square both sides of the equation:
\( (\sqrt{x^2+y^2})^2 = 1^2 \)
\( \implies x^2+y^2 = 1 \). This completes the proof.
In simple words: We take the "size" (modulus) of both sides of the given equation. The size of \( x+yi \) is \( \sqrt{x^2+y^2} \). The size of a fraction is the size of the top divided by the size of the bottom. Since the top and bottom parts of the fraction \( \frac{u+vi}{u-vi} \) have the same size (because \( u-vi \) is the conjugate of \( u+vi \)), their ratio is 1. So, \( \sqrt{x^2+y^2} = 1 \). When we square both sides, we get \( x^2+y^2=1 \).

🎯 Exam Tip: This type of proof often relies on the properties of complex moduli. Remembering that \( |z|=|\bar{z}| \) and \( \left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} \) are key concepts to apply here.

Question 14. Prove that: \( [4+3\sqrt{-20}]^{1/2}+[4-3\sqrt{-20}]^{1/2} = 6 \).
Answer: We need to show that the sum of these two complex square roots is 6.
Let's simplify \( \sqrt{-20} \): \( \sqrt{-20} = \sqrt{4 \times (-5)} = 2\sqrt{-5} = 2\sqrt{5}i \).
So the expression becomes: \( [4+3(2\sqrt{5}i)]^{1/2} + [4-3(2\sqrt{5}i)]^{1/2} \)
\( \implies [4+6\sqrt{5}i]^{1/2} + [4-6\sqrt{5}i]^{1/2} \).
Let's look for a complex number whose square is \( 4+6\sqrt{5}i \). We can test values. Try \( (3+\sqrt{5}i)^2 \):
\( (3+\sqrt{5}i)^2 = 3^2 + (\sqrt{5}i)^2 + 2(3)(\sqrt{5}i) \)
\( \implies = 9 + 5i^2 + 6\sqrt{5}i \)
\( \implies = 9 - 5 + 6\sqrt{5}i \)
\( \implies = 4 + 6\sqrt{5}i \). This is exactly the first term inside the square root. This means \( [4+6\sqrt{5}i]^{1/2} = 3+\sqrt{5}i \).
Similarly, for the second term, try \( (3-\sqrt{5}i)^2 \):
\( (3-\sqrt{5}i)^2 = 3^2 + (\sqrt{5}i)^2 - 2(3)(\sqrt{5}i) \)
\( \implies = 9 - 5 - 6\sqrt{5}i \)
\( \implies = 4 - 6\sqrt{5}i \). This is the second term inside the square root. So \( [4-6\sqrt{5}i]^{1/2} = 3-\sqrt{5}i \).
Now, substitute these back into the original sum:
\( (3+\sqrt{5}i) + (3-\sqrt{5}i) \)
\( \implies = 3+3 + \sqrt{5}i - \sqrt{5}i \)
\( \implies = 6 + 0i \)
\( \implies = 6 \). This proves the statement. Finding the square root of complex numbers often involves recognizing perfect squares or using algebraic methods.
In simple words: First, change the square root of negative numbers into imaginary numbers. Then, we look for two complex numbers that, when squared, give us the terms inside the square roots. We find that \( (3+\sqrt{5}i)^2 \) equals the first term and \( (3-\sqrt{5}i)^2 \) equals the second. Adding these two complex numbers \( (3+\sqrt{5}i) \) and \( (3-\sqrt{5}i) \) gives us 6, because the imaginary parts cancel each other out.

🎯 Exam Tip: When asked to find square roots of complex numbers, try to recognize them as perfect squares \( (a \pm bi)^2 \). If not immediately obvious, use the formula for square roots of complex numbers, which can be found by setting \( \sqrt{x+yi} = a+bi \) and solving for a and b.

Question 15. Express the following in the form a + bi :
(i) \( \sqrt{\frac{5(2+i)}{2-i}} \)
(ii) \( \frac{(3-i)^2}{2+i} \)
(iii) \( (1+i)^{-3} \)
(iv) \( \left(\frac{4i^3-i}{2i+1}\right)^2 \)
(v) \( \frac{i-1}{i+1} \)
(vi) \( \frac{2+i}{(3-i)(1+2i)} \)
(vii) \( \frac{5}{2i-7i^2} \)
Answer:
(i) First, we rationalize the denominator inside the square root by multiplying by its conjugate.
\( \sqrt{\frac{5(2+i)}{2-i}} = \sqrt{\frac{5(2+i)}{2-i} \times \frac{2+i}{2+i}} \)
\( \implies = \sqrt{\frac{5(2+i)^2}{2^2-i^2}} \)
\( \implies = \sqrt{\frac{5(2+i)^2}{4-(-1)}} \)
\( \implies = \sqrt{\frac{5(2+i)^2}{5}} \)
\( \implies = \sqrt{(2+i)^2} \)
\( \implies = 2+i \). This is already in the form \( a+bi \), where \( a=2 \) and \( b=1 \).
(ii) First, expand the numerator \( (3-i)^2 \).
\( (3-i)^2 = 3^2 - 2(3)(i) + i^2 = 9 - 6i - 1 = 8 - 6i \).
So the expression becomes \( \frac{8-6i}{2+i} \).
Now, rationalize the denominator by multiplying by its conjugate, \( 2-i \).
\( \frac{8-6i}{2+i} \times \frac{2-i}{2-i} \)
\( \implies = \frac{(8-6i)(2-i)}{2^2-i^2} \)
\( \implies = \frac{8(2)+8(-i)-6i(2)-6i(-i)}{4-(-1)} \)
\( \implies = \frac{16-8i-12i+6i^2}{4+1} \)
\( \implies = \frac{16-20i-6}{5} \)
\( \implies = \frac{10-20i}{5} \)
\( \implies = \frac{10}{5} - \frac{20}{5}i \)
\( \implies = 2-4i \). This is in the form \( a+bi \).
(iii) The expression \( (1+i)^{-3} \) means \( \frac{1}{(1+i)^3} \).
First, expand the denominator \( (1+i)^3 \). We use \( (a+b)^3 = a^3+3a^2b+3ab^2+b^3 \).
\( (1+i)^3 = 1^3 + 3(1)^2(i) + 3(1)(i)^2 + i^3 \)
\( \implies = 1 + 3i + 3(-1) + (-i) \)
\( \implies = 1 + 3i - 3 - i \)
\( \implies = -2 + 2i \).
So the expression becomes \( \frac{1}{-2+2i} \).
Now, rationalize the denominator by multiplying by its conjugate, \( -2-2i \).
\( \frac{1}{-2+2i} \times \frac{-2-2i}{-2-2i} \)
\( \implies = \frac{-2-2i}{(-2)^2-(2i)^2} \)
\( \implies = \frac{-2-2i}{4-4i^2} \)
\( \implies = \frac{-2-2i}{4+4} \)
\( \implies = \frac{-2-2i}{8} \)
\( \implies = -\frac{2}{8} - \frac{2}{8}i \)
\( \implies = -\frac{1}{4} - \frac{1}{4}i \). This is in the form \( a+bi \).
(iv) First, simplify the numerator \( 4i^3-i \). Remember \( i^3 = i^2 \times i = -1 \times i = -i \).
\( 4i^3-i = 4(-i)-i = -4i-i = -5i \).
So the expression inside the parenthesis becomes \( \frac{-5i}{2i+1} \).
Now, we need to square this expression.
\( \left(\frac{-5i}{2i+1}\right)^2 = \frac{(-5i)^2}{(2i+1)^2} \)
\( \implies = \frac{25i^2}{(2i)^2+2(2i)(1)+1^2} \)
\( \implies = \frac{-25}{4i^2+4i+1} \)
\( \implies = \frac{-25}{-4+4i+1} \)
\( \implies = \frac{-25}{-3+4i} \).
Now, rationalize the denominator by multiplying by its conjugate, \( -3-4i \).
\( \frac{-25}{-3+4i} \times \frac{-3-4i}{-3-4i} \)
\( \implies = \frac{-25(-3-4i)}{(-3)^2-(4i)^2} \)
\( \implies = \frac{75+100i}{9-16i^2} \)
\( \implies = \frac{75+100i}{9+16} \)
\( \implies = \frac{75+100i}{25} \)
\( \implies = \frac{75}{25} + \frac{100}{25}i \)
\( \implies = 3+4i \). This is in the form \( a+bi \).
(v) To simplify this fraction, we multiply the numerator and denominator by the conjugate of the denominator, which is \( i-1 \).
\( \frac{i-1}{i+1} \times \frac{i-1}{i-1} \)
\( \implies = \frac{(i-1)^2}{i^2-1^2} \)
\( \implies = \frac{i^2-2i+1}{-1-1} \)
\( \implies = \frac{-1-2i+1}{-2} \)
\( \implies = \frac{-2i}{-2} \)
\( \implies = i \). This can be written as \( 0+1i \), which is in the form \( a+bi \).
(vi) First, simplify the denominator by multiplying the two complex numbers.
\( (3-i)(1+2i) = 3(1)+3(2i)-i(1)-i(2i) \)
\( \implies = 3+6i-i-2i^2 \)
\( \implies = 3+5i+2 \)
\( \implies = 5+5i \).
So the expression becomes \( \frac{2+i}{5+5i} \).
Now, rationalize the denominator by multiplying by its conjugate, \( 5-5i \). We can also factor out 5 from the denominator and multiply by \( 1-i \).
\( \frac{2+i}{5(1+i)} = \frac{2+i}{5(1+i)} \times \frac{1-i}{1-i} \)
\( \implies = \frac{(2+i)(1-i)}{5(1^2-i^2)} \)
\( \implies = \frac{2(1)+2(-i)+i(1)+i(-i)}{5(1-(-1))} \)
\( \implies = \frac{2-2i+i-i^2}{5(2)} \)
\( \implies = \frac{2-i+1}{10} \)
\( \implies = \frac{3-i}{10} \)
\( \implies = \frac{3}{10} - \frac{1}{10}i \). This is in the form \( a+bi \).
(vii) First, simplify the denominator \( 2i-7i^2 \). Remember \( i^2 = -1 \).
\( 2i-7i^2 = 2i-7(-1) = 2i+7 = 7+2i \).
So the expression becomes \( \frac{5}{7+2i} \).
Now, rationalize the denominator by multiplying by its conjugate, \( 7-2i \).
\( \frac{5}{7+2i} \times \frac{7-2i}{7-2i} \)
\( \implies = \frac{5(7-2i)}{7^2-(2i)^2} \)
\( \implies = \frac{35-10i}{49-4i^2} \)
\( \implies = \frac{35-10i}{49+4} \)
\( \implies = \frac{35-10i}{53} \)
\( \implies = \frac{35}{53} - \frac{10}{53}i \). This is in the form \( a+bi \). Complex numbers can be simplified into a standard form using algebraic operations.
In simple words: For each problem, the goal is to get the complex number into the simple form \( a+bi \), where \( a \) is the real part and \( b \) is the imaginary part. We use rules like \( i^2 = -1 \) to remove powers of \( i \). When a complex number is in the denominator, we multiply the top and bottom by its conjugate to make the denominator a real number, which makes it easy to separate \( a \) and \( b \).

🎯 Exam Tip: Always remember that \( i^2 = -1 \), \( i^3 = -i \), and \( i^4 = 1 \). When simplifying fractions with complex numbers, multiply the numerator and denominator by the conjugate of the denominator to remove \( i \) from the bottom.

Question 16. Prove that \( \left(\frac{-1+i \sqrt{3}}{2}\right)^3 \) is a positive integer.
Answer: We need to calculate the cube of the given complex number and show that the result is a positive integer.
Let \( z = \frac{-1+i \sqrt{3}}{2} \). This complex number is one of the cube roots of unity, often denoted as \( \omega \). We know that \( \omega^3 = 1 \).
Let's expand \( z^3 \) using the formula \( (a+b)^3 = a^3+3a^2b+3ab^2+b^3 \), where \( a=-1 \) and \( b=i\sqrt{3} \).
\( z^3 = \left(\frac{-1+i \sqrt{3}}{2}\right)^3 = \frac{1}{2^3}(-1+i \sqrt{3})^3 \)
\( \implies = \frac{1}{8} [(-1)^3 + 3(-1)^2(i \sqrt{3}) + 3(-1)(i \sqrt{3})^2 + (i \sqrt{3})^3] \)
\( \implies = \frac{1}{8} [-1 + 3(1)(i \sqrt{3}) - 3(1)(i^2 \cdot 3) + i^3 (\sqrt{3})^3] \)
\( \implies = \frac{1}{8} [-1 + 3i \sqrt{3} - 3(-1 \cdot 3) + (-i) (3\sqrt{3})] \)
\( \implies = \frac{1}{8} [-1 + 3i \sqrt{3} + 9 - 3i \sqrt{3}] \)
\( \implies = \frac{1}{8} [(-1+9) + (3\sqrt{3}-3\sqrt{3})i] \)
\( \implies = \frac{1}{8} [8 + 0i] \)
\( \implies = \frac{8}{8} \)
\( \implies = 1 \). Since 1 is a positive integer, the proof is complete. This complex number is a special case in mathematics.
In simple words: We need to multiply the given complex number by itself three times. When we do this, using the rules for multiplying complex numbers and knowing that \( i^2 = -1 \) and \( i^3 = -i \), all the imaginary parts cancel out. The final answer is 1, which is a positive whole number.

🎯 Exam Tip: Recognizing special complex numbers like the cube roots of unity \( \left(\omega = \frac{-1 \pm i\sqrt{3}}{2}\right) \) can simplify calculations, as their powers follow a cyclical pattern (e.g., \( \omega^3=1 \)). If you don't recognize it, careful expansion using the binomial theorem is essential.

Question 17. If one of the values of x of the equation \( 2x^2-6x+k=0 \) be \( \frac{1}{2}(a+5i) \), find the values of a and k.
Answer: We are given a quadratic equation \( 2x^2-6x+k=0 \) and one of its roots, \( x = \frac{1}{2}(a+5i) \).
We can find the roots of the quadratic equation using the quadratic formula: \( x = \frac{-B \pm \sqrt{B^2-4AC}}{2A} \).
For \( 2x^2-6x+k=0 \), we have \( A=2 \), \( B=-6 \), \( C=k \).
\( x = \frac{-(-6) \pm \sqrt{(-6)^2-4(2)(k)}}{2(2)} \)
\( \implies x = \frac{6 \pm \sqrt{36-8k}}{4} \)
\( \implies x = \frac{6 \pm \sqrt{4(9-2k)}}{4} \)
\( \implies x = \frac{6 \pm 2\sqrt{9-2k}}{4} \)
\( \implies x = \frac{3 \pm \sqrt{9-2k}}{2} \). This gives the general form of the roots.
We are given that one of the values of x is \( \frac{1}{2}(a+5i) \), which can be written as \( \frac{a+5i}{2} \).
Comparing this with the form we found, \( \frac{3 \pm \sqrt{9-2k}}{2} \):
By comparing the real parts, we see that \( a=3 \).
By comparing the imaginary parts, we have \( 5i = \pm \sqrt{9-2k} \).
To find k, we square both sides of the imaginary part equality:
\( (5i)^2 = (\pm \sqrt{9-2k})^2 \)
\( \implies 25i^2 = 9-2k \)
\( \implies 25(-1) = 9-2k \)
\( \implies -25 = 9-2k \)
\( \implies 2k = 9+25 \)
\( \implies 2k = 34 \)
\( \implies k = 17 \).
So, the values are \( a=3 \) and \( k=17 \). Quadratic equations can have complex roots, and these roots always come in conjugate pairs.
In simple words: We find the general way to solve the given quadratic equation. Then we compare this general solution with the specific root we were given. By matching the real and imaginary parts of these solutions, we can figure out that \( a \) must be 3. Then, by solving for the imaginary part, we find that \( k \) is 17.

🎯 Exam Tip: When a quadratic equation has complex roots, they always appear in conjugate pairs. If one root is \( \alpha + \beta i \), the other must be \( \alpha - \beta i \). This property is key for questions involving complex roots and unknown coefficients.

Question 18. Define conjugate complex numbers and show that their sum and product are real numbers.
Answer:
Definition of Conjugate Complex Numbers: For any complex number \( z = x+iy \), where \( x \) and \( y \) are real numbers and \( i = \sqrt{-1} \), its conjugate is denoted by \( \bar{z} \) and is defined as \( \bar{z} = x-iy \). The conjugate is formed by changing the sign of the imaginary part of the complex number.
Sum of Conjugate Complex Numbers:
Let \( z = x+iy \). Then its conjugate is \( \bar{z} = x-iy \).
Their sum is: \( z + \bar{z} = (x+iy) + (x-iy) \)
\( \implies = x+iy+x-iy \)
\( \implies = (x+x) + (iy-iy) \)
\( \implies = 2x \).
Since \( x \) is a real number, \( 2x \) is also a real number. Therefore, the sum of a complex number and its conjugate is always a real number.
Product of Conjugate Complex Numbers:
Let \( z = x+iy \). Then its conjugate is \( \bar{z} = x-iy \).
Their product is: \( z \bar{z} = (x+iy)(x-iy) \)
This is in the form \( (a+b)(a-b) = a^2-b^2 \).
\( \implies = x^2 - (iy)^2 \)
\( \implies = x^2 - i^2y^2 \)
\( \implies = x^2 - (-1)y^2 \)
\( \implies = x^2+y^2 \).
Since \( x \) and \( y \) are real numbers, \( x^2 \) and \( y^2 \) are also real numbers. The sum of two real numbers, \( x^2+y^2 \), is always a real number. Therefore, the product of a complex number and its conjugate is always a real number. This property is useful for rationalizing denominators.
In simple words: A conjugate complex number is found by just flipping the sign of the imaginary part (the part with 'i'). For example, if you have \( 3+4i \), its conjugate is \( 3-4i \). When you add a complex number and its conjugate, the 'i' parts cancel out, leaving only a real number. When you multiply them, the 'i' parts again disappear because \( i^2 \) turns into -1, leaving only a real number.

🎯 Exam Tip: Remember the definitions and properties of complex conjugates. The sum \( z+\bar{z}=2Re(z) \) and the product \( z\bar{z}=|z|^2 \) are fundamental and frequently used in complex number problems.

Question 19. If \( \bar{z} = -z \neq 0 \), show that z is necessarily a purely imaginary number.
Answer: We need to prove that if the conjugate of a complex number \( z \) is equal to its negative (and \( z \) is not zero), then \( z \) must be a purely imaginary number.
Let \( z \) be a complex number. We can write \( z \) in its standard form as \( z = x+iy \), where \( x \) and \( y \) are real numbers.
The conjugate of \( z \) is \( \bar{z} = x-iy \).
The negative of \( z \) is \( -z = -(x+iy) = -x-iy \).
We are given the condition \( \bar{z} = -z \).
Substitute the expressions for \( \bar{z} \) and \( -z \):
\( x-iy = -x-iy \).
Now, we can add \( iy \) to both sides of the equation:
\( x = -x \).
Add \( x \) to both sides:
\( 2x = 0 \)
\( \implies x = 0 \).
Since \( x=0 \), the complex number \( z \) becomes \( z = 0+iy = iy \).
This means that \( z \) has no real part and only an imaginary part. Therefore, \( z \) is a purely imaginary number. The condition \( z \neq 0 \) implies that \( y \neq 0 \).
In simple words: We start with a complex number that has a real part and an imaginary part. If its "conjugate" (which means flipping the sign of its imaginary part) is the same as its "negative" (which means flipping the signs of both its real and imaginary parts), then the only way this can be true is if the real part of the number is zero. If the real part is zero, the number is only made of its imaginary part, meaning it is purely imaginary.

🎯 Exam Tip: A complex number \( z=x+iy \) is purely real if \( Im(z)=0 \) (i.e., \( y=0 \)) and purely imaginary if \( Re(z)=0 \) (i.e., \( x=0 \)). Use this definition when analyzing properties related to real and imaginary components.

Question 20. z and z' are complex numbers such that their product \( zz' = 3-4i \). Given that \( z' = 5+3i \), express z in the form \( a+bi \) where a and b are rational numbers.
Answer: We are given the product of two complex numbers \( zz' = 3-4i \) and the value of one of the numbers, \( z' = 5+3i \). We need to find \( z \) in the form \( a+bi \).
Substitute the value of \( z' \) into the product equation:
\( z(5+3i) = 3-4i \).
To find \( z \), we divide \( 3-4i \) by \( 5+3i \):
\( z = \frac{3-4i}{5+3i} \).
To express \( z \) in the form \( a+bi \), we need to rationalize the denominator. Multiply the numerator and denominator by the conjugate of \( 5+3i \), which is \( 5-3i \).
\( z = \frac{3-4i}{5+3i} \times \frac{5-3i}{5-3i} \)
\( \implies = \frac{(3-4i)(5-3i)}{5^2-(3i)^2} \)
\( \implies = \frac{3(5)+3(-3i)-4i(5)-4i(-3i)}{25-9i^2} \)
\( \implies = \frac{15-9i-20i+12i^2}{25-(-9)} \)
\( \implies = \frac{15-29i-12}{25+9} \)
\( \implies = \frac{3-29i}{34} \)
\( \implies = \frac{3}{34} - \frac{29}{34}i \). This is the value of \( z \) in the required \( a+bi \) form. Rational numbers are numbers that can be expressed as a fraction of two integers.
In simple words: We are given an equation where two complex numbers multiply to give another complex number. We know one of the numbers, so we divide to find the other. To divide complex numbers, we multiply the top and bottom of the fraction by the conjugate of the bottom number. This removes the 'i' from the bottom, letting us write the answer in the simple \( a+bi \) form.

🎯 Exam Tip: Always remember that the conjugate of \( A+Bi \) is \( A-Bi \). Multiplying a complex number by its conjugate eliminates the imaginary part from the denominator, making it a real number, which is essential for expressing the result in \( a+bi \) form.

Question 21. If \( a+bi = \frac{(x+i)^2}{2x^2+1} \), prove that \( a^2+b^2 = \frac{(x^2+1)^2}{(2x^2+1)^2} \).
Answer: We are given the complex number \( a+bi = \frac{(x+i)^2}{2x^2+1} \). We need to prove the relationship between \( a^2+b^2 \) and \( x \).
We know that for any complex number \( z = a+bi \), its modulus is \( |z| = \sqrt{a^2+b^2} \), so \( |z|^2 = a^2+b^2 \).
Also, we use the properties of moduli: \( |z^n|=|z|^n \) and \( \left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} \).
Let's take the modulus on both sides of the given equation:
\( |a+bi| = \left|\frac{(x+i)^2}{2x^2+1}\right| \)
\( \implies \sqrt{a^2+b^2} = \frac{|(x+i)^2|}{|2x^2+1|} \).
Now, let's find the modulus of the numerator and denominator separately:
For the numerator, \( |(x+i)^2| = |x+i|^2 \).
\( |x+i| = \sqrt{x^2+1^2} = \sqrt{x^2+1} \).
So, \( |x+i|^2 = (\sqrt{x^2+1})^2 = x^2+1 \).
For the denominator, \( 2x^2+1 \): Since \( x \) is a real number, \( x^2 \ge 0 \), so \( 2x^2+1 \) is always a positive real number. Thus, \( |2x^2+1| = 2x^2+1 \).
Substitute these back into the modulus equation:
\( \sqrt{a^2+b^2} = \frac{x^2+1}{2x^2+1} \). This is an intermediate step in the proof.
Finally, square both sides to get \( a^2+b^2 \):
\( (\sqrt{a^2+b^2})^2 = \left(\frac{x^2+1}{2x^2+1}\right)^2 \)
\( \implies a^2+b^2 = \frac{(x^2+1)^2}{(2x^2+1)^2} \). This proves the given relationship. Modulus provides a simple way to deal with the magnitude of complex numbers.
In simple words: We use the idea of "modulus" (which is like the length or size) of a complex number. The square of the modulus of \( a+bi \) is \( a^2+b^2 \). We take the modulus of both sides of the given equation. The modulus of a fraction is the modulus of the top divided by the modulus of the bottom. After calculating these moduli and squaring the result, we find that \( a^2+b^2 \) equals the desired expression.

🎯 Exam Tip: The property \( |z_1z_2| = |z_1||z_2| \) and \( |z^n|=|z|^n \) are very useful for proving identities involving the squares of real and imaginary parts. Always check if a term in the denominator is purely real or can be simplified before calculating the modulus.

Question 22. Let \( z_1 = 2-i \), \( z_2 = -2+i \), find
(i) \( Re\left(\frac{z_1 z_2}{\bar{z_1}}\right) \)
(ii) \( Im\left(\frac{1}{z_1 \bar{z_2}}\right) \)
Answer:
(i) We are asked to find the real part of \( \frac{z_1 z_2}{\bar{z_1}} \).
First, find \( \bar{z_1} \): The conjugate of \( z_1 = 2-i \) is \( \bar{z_1} = 2+i \).
Next, find the product \( z_1 z_2 \):
\( z_1 z_2 = (2-i)(-2+i) = 2(-2)+2(i)-i(-2)-i(i) \)
\( \implies = -4+2i+2i-i^2 \)
\( \implies = -4+4i-(-1) \)
\( \implies = -4+4i+1 \)
\( \implies = -3+4i \).
Now, we form the fraction \( \frac{z_1 z_2}{\bar{z_1}} \):
\( \frac{-3+4i}{2+i} \).
To find its real part, we rationalize the denominator by multiplying by \( 2-i \).
\( \frac{-3+4i}{2+i} \times \frac{2-i}{2-i} \)
\( \implies = \frac{(-3)(2)+(-3)(-i)+4i(2)+4i(-i)}{2^2-i^2} \)
\( \implies = \frac{-6+3i+8i-4i^2}{4-(-1)} \)
\( \implies = \frac{-6+11i+4}{4+1} \)
\( \implies = \frac{-2+11i}{5} \)
\( \implies = -\frac{2}{5} + \frac{11}{5}i \).
The real part of this complex number is \( Re\left(\frac{z_1 z_2}{\bar{z_1}}\right) = -\frac{2}{5} \).
(ii) We are asked to find the imaginary part of \( \frac{1}{z_1 \bar{z_2}} \).
First, find \( \bar{z_2} \): The conjugate of \( z_2 = -2+i \) is \( \bar{z_2} = -2-i \).
Next, find the product \( z_1 \bar{z_2} \):
\( z_1 \bar{z_2} = (2-i)(-2-i) = 2(-2)+2(-i)-i(-2)-i(-i) \)
\( \implies = -4-2i+2i+i^2 \)
\( \implies = -4-1 \)
\( \implies = -5 \).
Now, we form the fraction \( \frac{1}{z_1 \bar{z_2}} \):
\( \frac{1}{-5} = -\frac{1}{5} \).
This is a purely real number. A purely real number can be written as \( -\frac{1}{5} + 0i \).
The imaginary part of this complex number is \( Im\left(\frac{1}{z_1 \bar{z_2}}\right) = 0 \). Calculating real and imaginary parts is fundamental in complex number operations.
In simple words: For the first part, we first multiply \( z_1 \) and \( z_2 \), then we divide by the conjugate of \( z_1 \). After simplifying the result into the \( a+bi \) form, we pick out the 'a' part as the real part. For the second part, we multiply \( z_1 \) by the conjugate of \( z_2 \). Then we find the reciprocal of this product. If the result has no 'i' term, its imaginary part is 0.

🎯 Exam Tip: Always remember that \( Re(z) \) is the real part and \( Im(z) \) is the coefficient of \( i \) in the complex number \( z=x+iy \). Be careful with signs when calculating conjugates and multiplying complex numbers.

Question 23. If \( z_1 = 3+5i \) and \( z_2 = 2-3i \), then verify that \( \overline{\left(\frac{z_1}{z_2}\right)} = \frac{\bar{z_1}}{\bar{z_2}} \).
Answer: We need to verify a property of complex conjugates related to division. This property states that the conjugate of a quotient of two complex numbers is equal to the quotient of their conjugates.
Let's calculate the Left Hand Side (LHS) first: \( \overline{\left(\frac{z_1}{z_2}\right)} \).
First, calculate \( \frac{z_1}{z_2} \):
\( \frac{z_1}{z_2} = \frac{3+5i}{2-3i} \).
Rationalize the denominator by multiplying by its conjugate, \( 2+3i \):
\( \frac{3+5i}{2-3i} \times \frac{2+3i}{2+3i} \)
\( \implies = \frac{(3+5i)(2+3i)}{2^2-(3i)^2} \)
\( \implies = \frac{3(2)+3(3i)+5i(2)+5i(3i)}{4-9i^2} \)
\( \implies = \frac{6+9i+10i+15i^2}{4-(-9)} \)
\( \implies = \frac{6+19i-15}{4+9} \)
\( \implies = \frac{-9+19i}{13} \)
\( \implies = -\frac{9}{13} + \frac{19}{13}i \).
Now, find the conjugate of this result for the LHS:
\( \overline{\left(\frac{z_1}{z_2}\right)} = \overline{-\frac{9}{13} + \frac{19}{13}i} = -\frac{9}{13} - \frac{19}{13}i \).
Next, let's calculate the Right Hand Side (RHS): \( \frac{\bar{z_1}}{\bar{z_2}} \).
First, find the conjugates of \( z_1 \) and \( z_2 \):
\( \bar{z_1} = \overline{3+5i} = 3-5i \).
\( \bar{z_2} = \overline{2-3i} = 2+3i \).
Now, form the fraction \( \frac{\bar{z_1}}{\bar{z_2}} \):
\( \frac{3-5i}{2+3i} \).
Rationalize the denominator by multiplying by its conjugate, \( 2-3i \):
\( \frac{3-5i}{2+3i} \times \frac{2-3i}{2-3i} \)
\( \implies = \frac{(3-5i)(2-3i)}{2^2-(3i)^2} \)
\( \implies = \frac{3(2)+3(-3i)-5i(2)-5i(-3i)}{4-9i^2} \)
\( \implies = \frac{6-9i-10i+15i^2}{4-(-9)} \)
\( \implies = \frac{6-19i-15}{4+9} \)
\( \implies = \frac{-9-19i}{13} \)
\( \implies = -\frac{9}{13} - \frac{19}{13}i \).
Comparing the LHS and RHS results, we see that \( -\frac{9}{13} - \frac{19}{13}i = -\frac{9}{13} - \frac{19}{13}i \). This confirms the property for these specific complex numbers. This shows that the order of operations (division then conjugate, or conjugate then division) doesn't change the outcome.
In simple words: We want to check if taking the conjugate of a complex division is the same as dividing the conjugates of the complex numbers. We calculate both sides separately. For the left side, we first divide \( z_1 \) by \( z_2 \) and then take the conjugate of the answer. For the right side, we first find the conjugates of \( z_1 \) and \( z_2 \) and then divide them. If both results are the same, the property is proven.

🎯 Exam Tip: This problem verifies a fundamental property of complex conjugates. There are similar properties for addition \( \overline{z_1+z_2} = \bar{z_1}+\bar{z_2} \), subtraction \( \overline{z_1-z_2} = \bar{z_1}-\bar{z_2} \), and multiplication \( \overline{z_1z_2} = \bar{z_1}\bar{z_2} \). Knowing these properties can save time in complex calculations.

Question 24. If \( x = -2-\sqrt{3}i \), find that value of \( 2x^4+5x^3+7x^2-x+41 \).
Answer: We are given a value for \( x \) and need to find the value of a polynomial expression. This often involves finding a simpler quadratic equation that \( x \) satisfies.
Given \( x = -2-\sqrt{3}i \).
Move the real part to the left side:
\( x+2 = -\sqrt{3}i \).
Square both sides to eliminate the imaginary part:
\( (x+2)^2 = (-\sqrt{3}i)^2 \)
\( \implies x^2+4x+4 = (\sqrt{3})^2 i^2 \)
\( \implies x^2+4x+4 = 3(-1) \)
\( \implies x^2+4x+4 = -3 \).
Move the constant to the left side to get a quadratic equation:
\( x^2+4x+7 = 0 \). This is a crucial relationship.
Now, we use this quadratic equation to simplify the given polynomial \( P(x) = 2x^4+5x^3+7x^2-x+41 \). We can express \( P(x) \) in terms of \( (x^2+4x+7) \).
\( P(x) = 2x^2(x^2+4x+7) - 8x^3 - 14x^2 + 5x^3+7x^2-x+41 \) (Subtracting terms added by \( 2x^2(4x+7) \))
\( \implies = 2x^2(x^2+4x+7) -3x^3-7x^2-x+41 \). (Combine like terms)
Since \( x^2+4x+7=0 \), the first term \( 2x^2(x^2+4x+7) \) becomes \( 2x^2(0) = 0 \).
So, \( P(x) = -3x^3-7x^2-x+41 \).
Now, we continue factoring out \( (x^2+4x+7) \):
\( P(x) = -3x(x^2+4x+7) + 12x^2+21x -7x^2-x+41 \) (Subtracting terms added by \( -3x(4x+7) \))
\( \implies = -3x(x^2+4x+7) + 5x^2+20x+41 \). (Combine like terms)
Again, \( -3x(x^2+4x+7) \) becomes \( -3x(0) = 0 \).
So, \( P(x) = 5x^2+20x+41 \).
Finally, factor out \( (x^2+4x+7) \) one more time:
\( P(x) = 5(x^2+4x+7) - 20x - 35 + 20x + 41 \) (Subtracting terms added by \( 5(4x+7) \))
\( \implies = 5(x^2+4x+7) + 6 \).
Since \( x^2+4x+7=0 \), the term \( 5(x^2+4x+7) \) becomes \( 5(0) = 0 \).
Therefore, \( P(x) = 6 \). The evaluation of polynomials at complex roots can be greatly simplified by finding a minimal polynomial for the root.
In simple words: First, we change the given value of \( x \) into a simple equation without \( i \), which turns out to be \( x^2+4x+7=0 \). Then, we repeatedly use this simple equation to break down the big polynomial expression. Each time we find a part that is a multiple of \( x^2+4x+7 \), we replace it with zero. After removing all the zero parts, we are left with the final numerical value.

🎯 Exam Tip: When given a complex root to evaluate a high-degree polynomial, first derive a quadratic equation with real coefficients that the root satisfies. Then, use polynomial division (or repeated factorization as shown here) to reduce the degree of the original polynomial and simplify the evaluation.

Question 25. If \( z = -3+\sqrt{-2} \), then prove that \( z^4+5z^3+8z^2+7z+4 = -29 \).
Answer: We need to prove that a polynomial expression equals a specific value for the given complex number \( z \). Similar to the previous problem, we start by finding a simple quadratic equation that \( z \) satisfies.
Given \( z = -3+\sqrt{-2} \).
Simplify the square root: \( \sqrt{-2} = \sqrt{2}i \).
So, \( z = -3+\sqrt{2}i \).
Move the real part to the left side:
\( z+3 = \sqrt{2}i \).
Square both sides to eliminate the imaginary part:
\( (z+3)^2 = (\sqrt{2}i)^2 \)
\( \implies z^2+6z+9 = (\sqrt{2})^2 i^2 \)
\( \implies z^2+6z+9 = 2(-1) \)
\( \implies z^2+6z+9 = -2 \).
Move the constant to the left side:
\( z^2+6z+11 = 0 \). This is our key relationship.
Now, we use this to simplify the polynomial \( P(z) = z^4+5z^3+8z^2+7z+4 \). We will express the polynomial in terms of \( (z^2+6z+11) \).
\( P(z) = z^2(z^2+6z+11) - z^3 - 3z^2 + 7z + 4 \) (We started with \( z^4+5z^3+8z^2 \) and created \( z^2(z^2+6z+11) = z^4+6z^3+11z^2 \). To keep the original terms, we subtract \( (6-5)z^3+(11-8)z^2 = z^3+3z^2 \), so we actually have \( -z^3-3z^2 \), thus the OCR line is correct for the transformation).
Since \( z^2+6z+11=0 \), the first term \( z^2(z^2+6z+11) \) becomes \( z^2(0) = 0 \).
So, \( P(z) = -z^3 - 3z^2 + 7z + 4 \).
Now, let's factor out \( -z \) from an expression similar to \( (z^2+6z+11) \):
\( P(z) = -z(z^2+6z+11) + 3z^2+18z+4 \) (Here, \( -z(z^2+6z+11) = -z^3-6z^2-11z \). To get back to \( -z^3-3z^2+7z+4 \), we need to add \( (-3 - (-6))z^2 + (7 - (-11))z + 4 = 3z^2+18z+4 \). The OCR line correctly reflects this).
Again, \( -z(z^2+6z+11) \) becomes \( -z(0) = 0 \).
So, \( P(z) = 3z^2+18z+4 \).
Finally, factor out \( 3 \):
\( P(z) = 3(z^2+6z+11) - 33 + 4 \) (Here, \( 3(z^2+6z+11) = 3z^2+18z+33 \). To get back to \( 3z^2+18z+4 \), we need to subtract \( 33-4 = 29 \), which is \( -29 \) from the final 4).
The last line in OCR `3 (z^2 + 6z + 11) – 29` is correct. This is \( 3z^2+18z+33-29 = 3z^2+18z+4 \).
Since \( z^2+6z+11=0 \), the term \( 3(z^2+6z+11) \) becomes \( 3(0) = 0 \).
Therefore, \( P(z) = -29 \). This completes the proof. Finding specific polynomial values often requires using properties of the roots.
In simple words: We first change the given complex number \( z \) into a simple equation without \( i \), which is \( z^2+6z+11=0 \). Then, we cleverly rewrite the long polynomial expression by taking out parts that are multiples of \( (z^2+6z+11) \). Since \( z^2+6z+11 \) is zero, these parts become zero. We keep doing this until we are left with only a constant number, which turns out to be -29.

🎯 Exam Tip: For complex numbers that are roots of a polynomial, forming a quadratic (or higher-degree) equation with real coefficients that the complex number satisfies is the first and most crucial step. This equation then acts as a 'reducer' for higher powers of \( z \) in the main polynomial.