RBSE Solutions Class 11 Maths Chapter 5 Complex Numbers Exercise 5.1

Question 1. Write the following in simplest form:
(i) \( i^{52} \)
(ii) \( \sqrt{-2} \sqrt{-3} \)
(iii) \( (1 + i)^5 (1 - i)^5 \)
Answer:
(i) To simplify \( i^{52} \):
We know that \( i^4 = 1 \).
So, \( i^{52} = (i^4)^{13} \).
This means \( i^{52} = (1)^{13} \).
Therefore, \( i^{52} = 1 \).
(ii) To simplify \( \sqrt{-2} \sqrt{-3} \):
We know that \( \sqrt{-1} = i \).
So, \( \sqrt{-2} = \sqrt{2}i \) and \( \sqrt{-3} = \sqrt{3}i \).
Multiply them: \( \sqrt{-2} \sqrt{-3} = (\sqrt{2}i)(\sqrt{3}i) \)
\( \implies \sqrt{2} \times \sqrt{3} \times i \times i \)
\( \implies \sqrt{6}i^2 \)
Since \( i^2 = -1 \), we get:
\( \implies \sqrt{6}(-1) \)
\( \implies -\sqrt{6} \).
Therefore, \( \sqrt{-2} \sqrt{-3} = -\sqrt{6} \).
(iii) To simplify \( (1 + i)^5 (1 - i)^5 \):
We can write this as \( [(1 + i)(1 - i)]^5 \).
Using the difference of squares formula, \( (a+b)(a-b) = a^2 - b^2 \):
\( \implies [1^2 - i^2]^5 \)
Since \( i^2 = -1 \), we substitute this:
\( \implies [1 - (-1)]^5 \)
\( \implies [1 + 1]^5 \)
\( \implies 2^5 \)
\( \implies 32 \).
Therefore, \( (1 + i)^5 (1 - i)^5 = 32 \).
In simple words: For powers of \( i \), remember that \( i^4 = 1 \). For square roots of negative numbers, use \( i \). For products of complex conjugates, use the difference of squares formula.

🎯 Exam Tip: Always remember the basic powers of \( i \) (\( i^1=i, i^2=-1, i^3=-i, i^4=1 \)) and how to simplify square roots of negative numbers using \( i \).

Question 2. Find the additive and multiplicative inverse of following numbers:
(i) \( 1 + 2i \)
(ii) \( \frac{1}{3 + 4i} \)
(iii) \( (3 + i)^2 \)
Answer:
(i) For \( z = 1 + 2i \):
The additive inverse is \( -z \).
\( -z = -(1 + 2i) \)
\( \implies -1 - 2i \).
The multiplicative inverse is \( \frac{1}{z} \).
\( \frac{1}{z} = \frac{1}{1 + 2i} \)
To rationalize, multiply the numerator and denominator by the conjugate of the denominator, which is \( 1 - 2i \).
\( \implies \frac{1}{1 + 2i} \times \frac{1 - 2i}{1 - 2i} \)
\( \implies \frac{1 - 2i}{1^2 - (2i)^2} \)
\( \implies \frac{1 - 2i}{1 - 4i^2} \)
Since \( i^2 = -1 \):
\( \implies \frac{1 - 2i}{1 - 4(-1)} \)
\( \implies \frac{1 - 2i}{1 + 4} \)
\( \implies \frac{1 - 2i}{5} \)
\( \implies \frac{1}{5} - \frac{2}{5}i \).
(ii) For \( z = \frac{1}{3 + 4i} \):
First, simplify \( z \). Multiply numerator and denominator by the conjugate of \( 3 + 4i \), which is \( 3 - 4i \).
\( z = \frac{1}{3 + 4i} \times \frac{3 - 4i}{3 - 4i} \)
\( \implies \frac{3 - 4i}{3^2 - (4i)^2} \)
\( \implies \frac{3 - 4i}{9 - 16i^2} \)
Since \( i^2 = -1 \):
\( \implies \frac{3 - 4i}{9 - 16(-1)} \)
\( \implies \frac{3 - 4i}{9 + 16} \)
\( \implies \frac{3 - 4i}{25} \)
\( \implies \frac{3}{25} - \frac{4}{25}i \).
Now, find the additive inverse \( -z \).
\( -z = -(\frac{3}{25} - \frac{4}{25}i) \)
\( \implies -\frac{3}{25} + \frac{4}{25}i \).
The multiplicative inverse is \( \frac{1}{z} \).
Since \( z = \frac{1}{3 + 4i} \), then \( \frac{1}{z} = 3 + 4i \).
(iii) For \( z = (3 + i)^2 \):
First, expand \( z \). Using the formula \( (a+b)^2 = a^2 + 2ab + b^2 \):
\( z = 3^2 + 2(3)(i) + i^2 \)
\( \implies 9 + 6i + (-1) \)
\( \implies 9 + 6i - 1 \)
\( \implies 8 + 6i \).
Now, find the additive inverse \( -z \).
\( -z = -(8 + 6i) \)
\( \implies -8 - 6i \).
The multiplicative inverse is \( \frac{1}{z} \).
\( \frac{1}{z} = \frac{1}{8 + 6i} \)
Multiply numerator and denominator by the conjugate of \( 8 + 6i \), which is \( 8 - 6i \).
\( \implies \frac{1}{8 + 6i} \times \frac{8 - 6i}{8 - 6i} \)
\( \implies \frac{8 - 6i}{8^2 - (6i)^2} \)
\( \implies \frac{8 - 6i}{64 - 36i^2} \)
Since \( i^2 = -1 \):
\( \implies \frac{8 - 6i}{64 - 36(-1)} \)
\( \implies \frac{8 - 6i}{64 + 36} \)
\( \implies \frac{8 - 6i}{100} \)
\( \implies \frac{8}{100} - \frac{6}{100}i \)
\( \implies \frac{2}{25} - \frac{3}{50}i \).
In simple words: The additive inverse of a complex number is found by changing the sign of both its real and imaginary parts. The multiplicative inverse is found by dividing 1 by the number, then simplifying by multiplying the numerator and denominator by the conjugate.

🎯 Exam Tip: Always simplify the complex number to the form \( a + bi \) first before finding its inverses. Remember that \( i^2 = -1 \) is key for these calculations.

Question 4. Find the modulus of the following:
(i) \( 4 + i \)
(ii) \( -2 - 3i \)
(iii) \( \frac{1}{3 - 2i} \)
Answer:
(i) For \( z = 4 + i \):
The modulus of a complex number \( z = a + bi \) is \( |z| = \sqrt{a^2 + b^2} \).
Here, \( a = 4 \) and \( b = 1 \).
\( |z| = \sqrt{4^2 + 1^2} \)
\( \implies \sqrt{16 + 1} \)
\( \implies \sqrt{17} \).
(ii) For \( z = -2 - 3i \):
Here, \( a = -2 \) and \( b = -3 \).
\( |z| = \sqrt{(-2)^2 + (-3)^2} \)
\( \implies \sqrt{4 + 9} \)
\( \implies \sqrt{13} \).
(iii) For \( z = \frac{1}{3 - 2i} \):
First, express \( z \) in the form \( a + bi \). Multiply the numerator and denominator by the conjugate of \( 3 - 2i \), which is \( 3 + 2i \).
\( z = \frac{1}{3 - 2i} \times \frac{3 + 2i}{3 + 2i} \)
\( \implies \frac{3 + 2i}{3^2 - (2i)^2} \)
\( \implies \frac{3 + 2i}{9 - 4i^2} \)
Since \( i^2 = -1 \):
\( \implies \frac{3 + 2i}{9 - 4(-1)} \)
\( \implies \frac{3 + 2i}{9 + 4} \)
\( \implies \frac{3 + 2i}{13} \)
\( \implies \frac{3}{13} + \frac{2}{13}i \).
Now, find the modulus of \( z = \frac{3}{13} + \frac{2}{13}i \).
Here, \( a = \frac{3}{13} \) and \( b = \frac{2}{13} \).
\( |z| = \sqrt{(\frac{3}{13})^2 + (\frac{2}{13})^2} \)
\( \implies \sqrt{\frac{9}{169} + \frac{4}{169}} \)
\( \implies \sqrt{\frac{13}{169}} \)
\( \implies \sqrt{\frac{1}{13}} \)
\( \implies \frac{1}{\sqrt{13}} \).
In simple words: The modulus of a complex number is like its length or distance from the origin on a graph. You calculate it using the Pythagorean theorem with its real and imaginary parts. Make sure to simplify any fractions first.

🎯 Exam Tip: When finding the modulus of a fraction, always convert the fraction into the \( a + bi \) form first, then apply the modulus formula. Do not take the modulus of the numerator and denominator separately.

Question 5. If \( a^2 + b^2 = 1 \) then find the value of \( \frac{1+b+ia}{1+b-ia} \).
Answer:
Let the given expression be \( E = \frac{1+b+ia}{1+b-ia} \).
To simplify this expression, multiply the numerator and denominator by the conjugate of the denominator. The denominator is \( (1+b) - ia \), so its conjugate is \( (1+b) + ia \).
\( E = \frac{(1+b+ia)}{(1+b-ia)} \times \frac{(1+b+ia)}{(1+b+ia)} \)
\( \implies E = \frac{((1+b)+ia)^2}{((1+b)-ia)((1+b)+ia)} \)
Use the formulas \( (X+Y)^2 = X^2 + Y^2 + 2XY \) and \( (X-Y)(X+Y) = X^2 - Y^2 \). Here, \( X = (1+b) \) and \( Y = ia \).
Numerator: \( (1+b)^2 + (ia)^2 + 2(1+b)(ia) \)
\( \implies (1 + 2b + b^2) + i^2a^2 + 2ia(1+b) \)
Since \( i^2 = -1 \):
\( \implies 1 + 2b + b^2 - a^2 + 2ia(1+b) \).
Denominator: \( (1+b)^2 - (ia)^2 \)
\( \implies (1 + 2b + b^2) - i^2a^2 \)
Since \( i^2 = -1 \):
\( \implies 1 + 2b + b^2 - (-1)a^2 \)
\( \implies 1 + 2b + b^2 + a^2 \).
Now substitute \( a^2 + b^2 = 1 \) into both numerator and denominator.
Numerator becomes: \( 1 + 2b + (b^2 - a^2) + 2ia(1+b) \).
We know \( b^2 - a^2 = b^2 - (1 - b^2) = 2b^2 - 1 \). So, \( 1 + 2b + 2b^2 - 1 + 2ia(1+b) = 2b + 2b^2 + 2ia(1+b) = 2b(1+b) + 2ia(1+b) \).
Denominator becomes: \( 1 + 2b + (a^2 + b^2) \)
\( \implies 1 + 2b + 1 \)
\( \implies 2 + 2b \)
\( \implies 2(1+b) \).
So, the expression \( E \) is:
\( E = \frac{2b(1+b) + 2ia(1+b)}{2(1+b)} \)
Now, divide both terms in the numerator by the denominator:
\( E = \frac{2b(1+b)}{2(1+b)} + \frac{2ia(1+b)}{2(1+b)} \)
Assuming \( 1+b \neq 0 \), we can cancel \( (1+b) \) and \( 2 \).
\( E = b + ia \).
In simple words: To simplify this complex fraction, we multiplied the top and bottom by the complex conjugate of the denominator. Then we expanded and used the given condition \( a^2 + b^2 = 1 \) to simplify the real parts, leading to the final simple form.

🎯 Exam Tip: In problems with conditions like \( a^2 + b^2 = 1 \), always remember to substitute this condition after expanding and simplifying to get the final answer. Rationalizing the denominator is often the first step for complex fractions.

Question 6. If \( a = \cos \theta + i \sin \theta \), then find the value of \( \frac{1+a}{1-a} \).
Answer:
Given \( a = \cos \theta + i \sin \theta \).
We need to find the value of \( \frac{1+a}{1-a} \).
Substitute the value of \( a \):
\( \frac{1+a}{1-a} = \frac{1 + (\cos \theta + i \sin \theta)}{1 - (\cos \theta + i \sin \theta)} \)
\( \implies \frac{(1 + \cos \theta) + i \sin \theta}{(1 - \cos \theta) - i \sin \theta} \).
To simplify, multiply the numerator and denominator by the conjugate of the denominator, which is \( (1 - \cos \theta) + i \sin \theta \).
Numerator: \( ((1 + \cos \theta) + i \sin \theta)((1 - \cos \theta) + i \sin \theta) \)
\( \implies (1 + \cos \theta)(1 - \cos \theta) + (1 + \cos \theta)(i \sin \theta) + (i \sin \theta)(1 - \cos \theta) + (i \sin \theta)(i \sin \theta) \)
\( \implies (1 - \cos^2 \theta) + i \sin \theta + i \cos \theta \sin \theta + i \sin \theta - i \cos \theta \sin \theta + i^2 \sin^2 \theta \)
\( \implies \sin^2 \theta + 2i \sin \theta - \sin^2 \theta \) (Since \( 1 - \cos^2 \theta = \sin^2 \theta \) and \( i^2 = -1 \)).
\( \implies 2i \sin \theta \).
Denominator: \( ((1 - \cos \theta) - i \sin \theta)((1 - \cos \theta) + i \sin \theta) \)
\( \implies (1 - \cos \theta)^2 - (i \sin \theta)^2 \)
\( \implies (1 - 2 \cos \theta + \cos^2 \theta) - i^2 \sin^2 \theta \)
\( \implies 1 - 2 \cos \theta + \cos^2 \theta + \sin^2 \theta \) (Since \( i^2 = -1 \)).
\( \implies 1 - 2 \cos \theta + 1 \) (Since \( \cos^2 \theta + \sin^2 \theta = 1 \)).
\( \implies 2 - 2 \cos \theta \)
\( \implies 2(1 - \cos \theta) \).
So, the expression becomes:
\( \frac{1+a}{1-a} = \frac{2i \sin \theta}{2(1 - \cos \theta)} \)
\( \implies \frac{i \sin \theta}{1 - \cos \theta} \).
Now, use half-angle identities: \( \sin \theta = 2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2}) \) and \( 1 - \cos \theta = 2 \sin^2(\frac{\theta}{2}) \).
\( \implies \frac{i (2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2}))}{2 \sin^2(\frac{\theta}{2})} \)
Cancel \( 2 \) and one \( \sin(\frac{\theta}{2}) \):
\( \implies i \frac{\cos(\frac{\theta}{2})}{\sin(\frac{\theta}{2})} \)
\( \implies i \cot(\frac{\theta}{2}) \).
In simple words: We put the complex number into the expression. To simplify the fraction, we multiplied by the conjugate of the bottom part. After simplifying, we used special trigonometric identities (half-angle formulas) to get a much shorter answer involving cotangent.

🎯 Exam Tip: Remember to use trigonometric identities, especially half-angle formulas for sine and cosine, when you see expressions like \( 1 + \cos \theta \) or \( 1 - \cos \theta \) in complex number problems. This often leads to significant simplification.

Question 7. Find the value of x and y which satisfy the equation \( \frac{(1+i)x-2i}{3+i} + \frac{(2-3i)y+i}{3-i} = i \).
Answer:
Given the equation: \( \frac{(1+i)x-2i}{3+i} + \frac{(2-3i)y+i}{3-i} = i \).
First, simplify the first term: \( T_1 = \frac{(1+i)x-2i}{3+i} = \frac{x+ix-2i}{3+i} \).
Multiply numerator and denominator by the conjugate of \( 3+i \), which is \( 3-i \).
\( T_1 = \frac{(x+ix-2i)(3-i)}{(3+i)(3-i)} \)
Denominator: \( 3^2 - i^2 = 9 - (-1) = 10 \).
Numerator: \( (x+ix-2i)(3-i) = 3x + 3ix - 6i - ix - i^2x + 2i^2 \)
\( \implies 3x + 3ix - 6i - ix + x - 2 \) (Since \( i^2 = -1 \)).
\( \implies (3x+x-2) + (3ix-ix-6i) \)
\( \implies (4x-2) + (2x-6)i \).
So, \( T_1 = \frac{(4x-2) + (2x-6)i}{10} \).
Next, simplify the second term: \( T_2 = \frac{(2-3i)y+i}{3-i} = \frac{2y-3iy+i}{3-i} \).
Multiply numerator and denominator by the conjugate of \( 3-i \), which is \( 3+i \).
\( T_2 = \frac{(2y-3iy+i)(3+i)}{(3-i)(3+i)} \)
Denominator: \( 3^2 - i^2 = 9 - (-1) = 10 \).
Numerator: \( (2y-3iy+i)(3+i) = 6y - 9iy + 3i + 2iy - 3i^2y + i^2 \)
\( \implies 6y - 9iy + 3i + 2iy + 3y - 1 \) (Since \( i^2 = -1 \)).
\( \implies (6y+3y-1) + (-9iy+2iy+3i) \)
\( \implies (9y-1) + (-7y+3)i \).
So, \( T_2 = \frac{(9y-1) + (-7y+3)i}{10} \).
Now, substitute \( T_1 \) and \( T_2 \) back into the original equation:
\( \frac{(4x-2) + (2x-6)i}{10} + \frac{(9y-1) + (-7y+3)i}{10} = i \)
Combine the fractions:
\( \frac{(4x-2 + 9y-1) + (2x-6-7y+3)i}{10} = i \)
\( \implies (4x+9y-3) + (2x-7y-3)i = 10i \).
Compare the real and imaginary parts on both sides of the equation.
Real part: \( 4x+9y-3 = 0 \)
\( \implies 4x+9y = 3 \) (Equation 1).
Imaginary part: \( 2x-7y-3 = 10 \)
\( \implies 2x-7y = 13 \) (Equation 2).
Now, solve the system of linear equations (1) and (2).
Multiply Equation 2 by 2:
\( 2(2x-7y) = 2(13) \)
\( \implies 4x-14y = 26 \) (Equation 3).
Subtract Equation 3 from Equation 1:
\( (4x+9y) - (4x-14y) = 3 - 26 \)
\( \implies 4x+9y-4x+14y = -23 \)
\( \implies 23y = -23 \)
\( \implies y = -1 \).
Substitute \( y = -1 \) into Equation 2:
\( 2x - 7(-1) = 13 \)
\( \implies 2x + 7 = 13 \)
\( \implies 2x = 13 - 7 \)
\( \implies 2x = 6 \)
\( \implies x = 3 \).
Thus, the values are \( x = 3 \) and \( y = -1 \).
In simple words: We first simplified each complex fraction by multiplying by the conjugate. Then we combined them and set the real and imaginary parts equal to the real and imaginary parts of \( i \) (which is \( 0 + 1i \)). This gave us two simple equations to solve for x and y.

🎯 Exam Tip: When solving equations with complex numbers, always separate the real and imaginary parts to form a system of real linear equations. This is a common and reliable method for finding unknown variables.

Question 8. If \( z_1 \) and \( z_2 \) are any two complex numbers, then prove that \( |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2 \).
Answer:
We know that for any complex number \( z \), \( |z|^2 = z \cdot \bar{z} \), where \( \bar{z} \) is the conjugate of \( z \). We also know that \( z + \bar{z} = 2 \text{Re}(z) \).
Let's expand the left-hand side (L.H.S.) of the equation.
First term: \( |z_1 + z_2|^2 \)
\( |z_1 + z_2|^2 = (z_1 + z_2)(\overline{z_1 + z_2}) \)
\( \implies (z_1 + z_2)(\bar{z_1} + \bar{z_2}) \)
\( \implies z_1 \bar{z_1} + z_1 \bar{z_2} + z_2 \bar{z_1} + z_2 \bar{z_2} \)
\( \implies |z_1|^2 + |z_2|^2 + (z_1 \bar{z_2} + \bar{z_1} z_2) \).
We know that \( z_1 \bar{z_2} + \bar{z_1} z_2 = 2 \text{Re}(z_1 \bar{z_2}) \).
So, \( |z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 + 2 \text{Re}(z_1 \bar{z_2}) \) (Equation A).
Second term: \( |z_1 - z_2|^2 \)
\( |z_1 - z_2|^2 = (z_1 - z_2)(\overline{z_1 - z_2}) \)
\( \implies (z_1 - z_2)(\bar{z_1} - \bar{z_2}) \)
\( \implies z_1 \bar{z_1} - z_1 \bar{z_2} - z_2 \bar{z_1} + z_2 \bar{z_2} \)
\( \implies |z_1|^2 + |z_2|^2 - (z_1 \bar{z_2} + \bar{z_1} z_2) \).
Again, using \( z_1 \bar{z_2} + \bar{z_1} z_2 = 2 \text{Re}(z_1 \bar{z_2}) \).
So, \( |z_1 - z_2|^2 = |z_1|^2 + |z_2|^2 - 2 \text{Re}(z_1 \bar{z_2}) \) (Equation B).
Now, add Equation A and Equation B:
\( |z_1 + z_2|^2 + |z_1 - z_2|^2 = (|z_1|^2 + |z_2|^2 + 2 \text{Re}(z_1 \bar{z_2})) + (|z_1|^2 + |z_2|^2 - 2 \text{Re}(z_1 \bar{z_2})) \)
\( \implies |z_1|^2 + |z_2|^2 + |z_1|^2 + |z_2|^2 \)
\( \implies 2|z_1|^2 + 2|z_2|^2 \).
This is the right-hand side (R.H.S.) of the given equation.
Hence, it is proven.
In simple words: This property shows that if you add the squares of the magnitudes of the sum and difference of two complex numbers, you get twice the sum of the squares of their individual magnitudes. It's similar to a parallelogram law in vector addition.

🎯 Exam Tip: Remember the fundamental property \( |z|^2 = z \bar{z} \) and \( z + \bar{z} = 2 \text{Re}(z) \) for complex numbers. These are crucial for proving identities involving moduli and conjugates.

Question 9. If \( a + ib = \frac{c+i}{c-i} \), where c is a real number, then prove that \( a^2 + b^2 = 1 \) and \( \frac{b}{a} = \frac{2c}{c^2-1} \).
Answer:
Given \( a + ib = \frac{c+i}{c-i} \).
First, simplify the right-hand side (RHS) by multiplying the numerator and denominator by the conjugate of the denominator, which is \( c+i \).
\( \frac{c+i}{c-i} = \frac{(c+i)(c+i)}{(c-i)(c+i)} \)
\( \implies \frac{c^2 + 2ci + i^2}{c^2 - i^2} \)
Since \( i^2 = -1 \):
\( \implies \frac{c^2 + 2ci - 1}{c^2 - (-1)} \)
\( \implies \frac{(c^2 - 1) + 2ci}{c^2 + 1} \)
\( \implies \frac{c^2 - 1}{c^2 + 1} + i \frac{2c}{c^2 + 1} \).
Now, we have \( a + ib = \frac{c^2 - 1}{c^2 + 1} + i \frac{2c}{c^2 + 1} \).
Comparing the real and imaginary parts, we get:
\( a = \frac{c^2 - 1}{c^2 + 1} \)
\( b = \frac{2c}{c^2 + 1} \).
**Part 1: Prove \( a^2 + b^2 = 1 \)**
Substitute the values of \( a \) and \( b \):
\( a^2 + b^2 = \left(\frac{c^2 - 1}{c^2 + 1}\right)^2 + \left(\frac{2c}{c^2 + 1}\right)^2 \)
\( \implies \frac{(c^2 - 1)^2}{(c^2 + 1)^2} + \frac{(2c)^2}{(c^2 + 1)^2} \)
\( \implies \frac{c^4 - 2c^2 + 1 + 4c^2}{(c^2 + 1)^2} \)
\( \implies \frac{c^4 + 2c^2 + 1}{(c^2 + 1)^2} \)
The numerator is a perfect square, \( (c^2 + 1)^2 \).
\( \implies \frac{(c^2 + 1)^2}{(c^2 + 1)^2} \)
\( \implies 1 \).
Thus, \( a^2 + b^2 = 1 \) is proven.
**Part 2: Prove \( \frac{b}{a} = \frac{2c}{c^2-1} \)**
Substitute the values of \( a \) and \( b \):
\( \frac{b}{a} = \frac{\frac{2c}{c^2 + 1}}{\frac{c^2 - 1}{c^2 + 1}} \)
Since the denominators are the same, they cancel out (assuming \( c^2+1 \neq 0 \), which is true for real c).
\( \implies \frac{2c}{c^2 - 1} \).
Thus, \( \frac{b}{a} = \frac{2c}{c^2-1} \) is proven.
In simple words: We first simplified the complex fraction to find the real part 'a' and imaginary part 'b'. Then, we plugged these values into the two equations we needed to prove. By doing the algebra, we showed that \( a^2+b^2 \) equals 1 and that the ratio \( b/a \) equals the other given expression.

🎯 Exam Tip: When proving identities involving complex numbers, always start by expressing the complex number in the standard \( a+bi \) form. This allows you to equate real and imaginary parts and use these relations for further proofs.

Question 10. If \( (x + iy)^{1/3} = a + ib \), then prove that \( \frac{x}{a} + \frac{y}{b} = 4(a^2-b^2) \).
Answer:
Given the equation \( (x + iy)^{1/3} = a + ib \).
Cube both sides of the equation:
\( ((x + iy)^{1/3})^3 = (a + ib)^3 \)
\( \implies x + iy = (a + ib)^3 \).
Expand \( (a + ib)^3 \) using the binomial expansion \( (A+B)^3 = A^3 + 3A^2B + 3AB^2 + B^3 \).
Here, \( A = a \) and \( B = ib \).
\( (a + ib)^3 = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3 \)
\( \implies a^3 + 3ia^2b + 3ai^2b^2 + i^3b^3 \).
Now, substitute the powers of \( i \): \( i^2 = -1 \) and \( i^3 = -i \).
\( \implies a^3 + 3ia^2b + 3a(-1)b^2 + (-i)b^3 \)
\( \implies a^3 + 3ia^2b - 3ab^2 - ib^3 \).
Group the real and imaginary parts:
\( x + iy = (a^3 - 3ab^2) + i(3a^2b - b^3) \).
Now, compare the real and imaginary parts on both sides of the equation.
Real part: \( x = a^3 - 3ab^2 \)
Take \( a \) common from the RHS:
\( x = a(a^2 - 3b^2) \)
Divide by \( a \) (assuming \( a \neq 0 \)):
\( \frac{x}{a} = a^2 - 3b^2 \) (Equation 1).
Imaginary part: \( y = 3a^2b - b^3 \)
Take \( b \) common from the RHS:
\( y = b(3a^2 - b^2) \)
Divide by \( b \) (assuming \( b \neq 0 \)):
\( \frac{y}{b} = 3a^2 - b^2 \) (Equation 2).
Now, we need to prove \( \frac{x}{a} + \frac{y}{b} = 4(a^2-b^2) \).
Add Equation 1 and Equation 2:
\( \frac{x}{a} + \frac{y}{b} = (a^2 - 3b^2) + (3a^2 - b^2) \)
\( \implies a^2 - 3b^2 + 3a^2 - b^2 \)
\( \implies (a^2 + 3a^2) + (-3b^2 - b^2) \)
\( \implies 4a^2 - 4b^2 \)
Take \( 4 \) common:
\( \implies 4(a^2 - b^2) \).
Thus, \( \frac{x}{a} + \frac{y}{b} = 4(a^2-b^2) \) is proven.
In simple words: We raised both sides of the equation to the power of 3 to remove the cube root. Then we expanded the complex cube and separated it into real and imaginary parts. By comparing these parts with \( x \) and \( y \), we found expressions for \( x/a \) and \( y/b \), which, when added, proved the given identity.

🎯 Exam Tip: When dealing with powers of complex numbers, expanding using the binomial theorem and then carefully grouping real and imaginary parts is a crucial step. Remember to use the correct values for powers of \( i \).

Question 11. If \( \frac{(x+i)^2}{3x+2} = a + ib \), then prove that \( \frac{(x^2+1)^2}{(3x+2)^2} = a^2 + b^2 \).
Answer:
Given the equation (1):
\( \frac{(x+i)^2}{3x+2} = a + ib \) (Equation 1).
Take the complex conjugate of both sides of Equation 1. The conjugate of \( a+ib \) is \( a-ib \). Since \( x \) is a real number, \( 3x+2 \) is also a real number, so its conjugate is itself. The conjugate of \( (x+i)^2 \) is \( (\overline{x+i})^2 = (x-i)^2 \).
So, the conjugate equation (2) is:
\( \frac{(x-i)^2}{3x+2} = a - ib \) (Equation 2).
Now, multiply Equation 1 by Equation 2:
\( \left(\frac{(x+i)^2}{3x+2}\right) \times \left(\frac{(x-i)^2}{3x+2}\right) = (a + ib)(a - ib) \).
On the left-hand side (LHS), combine the terms:
\( \frac{(x+i)^2 (x-i)^2}{(3x+2)(3x+2)} = \frac{((x+i)(x-i))^2}{(3x+2)^2} \).
Using the difference of squares formula \( (A+B)(A-B) = A^2 - B^2 \), we simplify \( (x+i)(x-i) \):
\( (x+i)(x-i) = x^2 - i^2 \).
Since \( i^2 = -1 \):
\( x^2 - (-1) = x^2 + 1 \).
So, the LHS becomes: \( \frac{(x^2+1)^2}{(3x+2)^2} \).
On the right-hand side (RHS), multiply \( (a+ib)(a-ib) \):
\( (a+ib)(a-ib) = a^2 - (ib)^2 \)
\( \implies a^2 - i^2b^2 \).
Since \( i^2 = -1 \):
\( \implies a^2 - (-1)b^2 \)
\( \implies a^2 + b^2 \).
Therefore, by equating the simplified LHS and RHS, we get:
\( \frac{(x^2+1)^2}{(3x+2)^2} = a^2 + b^2 \).
This proves the required identity.
In simple words: We took the original equation and its complex conjugate. By multiplying these two equations together, we used the property that \( z \cdot \bar{z} = |z|^2 \) (which is \( a^2+b^2 \) for \( a+ib \)) and simplified the complex parts on the other side, leading directly to the result.

🎯 Exam Tip: This type of proof often relies on taking the complex conjugate of both sides of a given equation and then multiplying the original equation by its conjugate. Remember that \( (a+bi)(a-bi) = a^2+b^2 \).