Maharashtra Board Class 11 Maths Part 1 Chapter 3 Complex Numbers 3.1 Solutions

Question 1. Write the conjugates of the following complex numbers:
(i) \( 3 + i \)
(ii) \( 3 - i \)
(iii) \( -\sqrt{5} - \sqrt{7}i \)
(iv) \( -\sqrt{-5} \)
(v) \( 5i \)
(vi) \( \sqrt{5} - i \)
(vii) \( \sqrt{2} + \sqrt{3}i \)
Answer:
(i) The conjugate of \( 3 + i \) is \( 3 - i \).
(ii) The conjugate of \( 3 - i \) is \( 3 + i \).
(iii) The conjugate of \( -\sqrt{5} - \sqrt{7}i \) is \( -\sqrt{5} + \sqrt{7}i \).
(iv) Since \( -\sqrt{-5} = -\sqrt{5}i \), its conjugate is \( \sqrt{5}i \) (or \( \sqrt{-5} \)).
(v) The conjugate of \( 5i \) is \( -5i \).
(vi) The conjugate of \( \sqrt{5} - i \) is \( \sqrt{5} + i \).
(vii) The conjugate of \( \sqrt{2} + \sqrt{3}i \) is \( \sqrt{2} - \sqrt{3}i \). Finding the conjugate simply involves changing the sign of the imaginary part of the complex number.
In simple words: To find the conjugate of a complex number, you just change the sign in front of the imaginary part (the part with 'i') from plus to minus, or minus to plus.

🎯 Exam Tip: Remember that the real part remains completely unchanged when finding a conjugate; only change the sign of the imaginary term containing \( i \).

Question 1. Find the conjugate of the following complex numbers:
Answer:
(i) Conjugate of \( (3 + i) \) is \( (3 - i) \)
(ii) Conjugate of \( (3 - i) \) is \( (3 + i) \)
(iii) Conjugate of \( (-\sqrt{5} - \sqrt{7}i) \) is \( (-\sqrt{5} + \sqrt{7}i) \)
(iv) \( -\sqrt{-5} = -\sqrt{5} \times \sqrt{-1} = -\sqrt{5}i \)
Conjugate of \( -\sqrt{-5} \) is \( \sqrt{5}i \)
(v) Conjugate of \( 5i \) is \( -5i \)
(vi) Conjugate of \( \sqrt{5} - i \) is \( \sqrt{5} + i \)
(vii) Conjugate of \( \sqrt{2} + \sqrt{3}i \) is \( \sqrt{2} - \sqrt{3}i \)
In simple words: To find the conjugate of a complex number, we simply change the sign of its imaginary part (the part with \( i \)).

🎯 Exam Tip: Remember that the conjugate only changes the sign of the imaginary part; the real part remains exactly the same.

Question 2. Express the following in the form of \( a + ib \), \( a, b \in \mathbb{R} \), \( i = \sqrt{-1} \). State the values of \( a \) and \( b \):
(i) \( (1 + 2i)(-2 + i) \)
(ii) \( \frac{i(4+3i)}{1-i} \)
(iii) \( \frac{2+i}{(3-i)(1+2i)} \)
(iv) \( \frac{3+2i}{2-5i} + \frac{3-2i}{2+5i} \)
(v) \( \frac{2+\sqrt{-3}}{4+\sqrt{-3}} \)
(vi) \( (2 + 3i)(2 - 3i) \)
(vii) \( \frac{4i^8 - 3i^9 + 3}{3i^{11} - 4i^{10} - 2} \)
Answer:
(i) \( (1 + 2i)(-2 + i) = -2 + i - 4i + 2i^2 = -2 - 3i - 2 = -4 - 3i \)
Here, \( a = -4 \) and \( b = -3 \).

(ii) \( \frac{i(4+3i)}{1-i} = \frac{4i-3}{1-i} = \frac{(-3+4i)(1+i)}{(1-i)(1+i)} = \frac{-3-3i+4i-4}{1+1} = \frac{-7+i}{2} = -\frac{7}{2} + \frac{1}{2}i \)
Here, \( a = -\frac{7}{2} \) and \( b = \frac{1}{2} \).

(iii) \( \frac{2+i}{(3-i)(1+2i)} = \frac{2+i}{3+6i-i+2} = \frac{2+i}{5+5i} = \frac{(2+i)(1-i)}{5(1+i)(1-i)} = \frac{2-2i+i+1}{5(2)} = \frac{3-i}{10} = \frac{3}{10} - \frac{1}{10}i \)
Here, \( a = \frac{3}{10} \) and \( b = -\frac{1}{10} \).

(iv) \( \frac{3+2i}{2-5i} + \frac{3-2i}{2+5i} = \frac{(3+2i)(2+5i) + (3-2i)(2-5i)}{(2-5i)(2+5i)} = \frac{(6+15i+4i-10) + (6-15i-4i-10)}{4+25} = \frac{-4+19i-4-19i}{29} = -\frac{8}{29} + 0i \)
Here, \( a = -\frac{8}{29} \) and \( b = 0 \).

(v) \( \frac{2+\sqrt{-3}}{4+\sqrt{-3}} = \frac{2+\sqrt{3}i}{4+\sqrt{3}i} = \frac{(2+\sqrt{3}i)(4-\sqrt{3}i)}{(4+\sqrt{3}i)(4-\sqrt{3}i)} = \frac{8-2\sqrt{3}i+4\sqrt{3}i+3}{16+3} = \frac{11+2\sqrt{3}i}{19} = \frac{11}{19} + \frac{2\sqrt{3}}{19}i \)
Here, \( a = \frac{11}{19} \) and \( b = \frac{2\sqrt{3}}{19} \).

(vi) \( (2 + 3i)(2 - 3i) = 2^2 - (3i)^2 = 4 - 9(-1) = 13 = 13 + 0i \)
Here, \( a = 13 \) and \( b = 0 \).

(vii) \( \frac{4i^8 - 3i^9 + 3}{3i^{11} - 4i^{10} - 2} = \frac{4(1) - 3(i) + 3}{3(-i) - 4(-1) - 2} = \frac{7-3i}{2-3i} = \frac{(7-3i)(2+3i)}{(2-3i)(2+3i)} = \frac{14+21i-6i+9}{4+9} = \frac{23+15i}{13} = \frac{23}{13} + \frac{15}{13}i \)
Here, \( a = \frac{23}{13} \) and \( b = \frac{15}{13} \).
In simple words: To write any complex fraction in standard form, we multiply the top and bottom by the conjugate of the bottom part to get rid of the imaginary unit \( i \) in the denominator.

🎯 Exam Tip: Always simplify powers of \( i \) first (like \( i^8 = 1 \), \( i^9 = i \)) to make the calculations much simpler and avoid arithmetic errors.

Question 1. Simplify the following and find the values of \( a \) and \( b \):
(i) \( (1 + 2i)(-2 + i) \)
(ii) \( \frac{i(4 - 3i)}{1 - i} \)
(iii) \( \frac{2 + i}{(3 - i)(1 + 2i)} \)
Answer:
(i)
\( (1 + 2i)(-2 + i) = -2 + i - 4i + 2i^2 \)
\( = -2 - 3i + 2(-1) \) ... \( [\because i^2 = -1] \)
\( \therefore (1 + 2i)(-2 + i) = -4 - 3i \)
Comparing this with the standard form \( a + ib \), we can easily determine the real and imaginary parts.
\( \therefore a = -4 \) and \( b = -3 \)

(ii)
\( \frac{i(4 - 3i)}{1 - i} = \frac{4i + 3i^2}{1 - i} \)
\( = \frac{-3 + 4i}{1 - i} \) ... \( [\because i^2 = -1] \)
Multiplying the numerator and denominator by the conjugate of the denominator, we get:
\( = \frac{(-3 + 4i)(1 + i)}{(1 - i)(1 + i)} \)
\( = \frac{-3 - 3i + 4i + 4i^2}{1 - i^2} \)
\( = \frac{-3 + i + 4(-1)}{1 - (-1)} \) ... \( [\because i^2 = -1] \)
\( = \frac{-7 + i}{2} \)
\( \therefore \frac{i(4 - 3i)}{1 - i} = \frac{-7}{2} + \frac{1}{2}i \)
\( \therefore a = \frac{-7}{2} \) and \( b = \frac{1}{2} \)

(iii)
\( \frac{2 + i}{(3 - i)(1 + 2i)} = \frac{2 + i}{3 + 6i - i - 2i^2} \)
\( = \frac{2 + i}{3 + 5i - 2(-1)} \) ... \( [\because i^2 = -1] \)
\( = \frac{2 + i}{5 + 5i} \)
\( = \frac{2 + i}{5(1 + i)} = \frac{(2 + i)(1 - i)}{5(1 + i)(1 - i)} \)
\( = \frac{2 - 2i + i - i^2}{5(1 - i^2)} \)
\( = \frac{2 - i - (-1)}{5[1 - (-1)]} \) ... \( [\because i^2 = -1] \)
\( = \frac{3 - i}{10} \)
\( = \frac{3}{10} - \frac{1}{10}i \)
\( \therefore a = \frac{3}{10} \) and \( b = -\frac{1}{10} \)
In simple words: To simplify complex fractions, we multiply the top and bottom by the conjugate of the bottom part to remove the imaginary unit 'i' from the denominator. Then, we substitute \( i^2 = -1 \) and separate the real and imaginary parts to find \( a \) and \( b \).

🎯 Exam Tip: Always remember to substitute \( i^2 = -1 \) carefully and double-check your signs when multiplying complex conjugates in the denominator to avoid simple calculation errors.

Question 1. Express the following in the form \( a + ib \) and find the values of \( a \) and \( b \):
(iv) \( \frac{3+2i}{2-5i} + \frac{3-2i}{2+5i} \)
(v) \( \frac{2+\sqrt{-3}}{4+\sqrt{-3}} \)
Answer:
(iv) \( \frac{3+2i}{2-5i} + \frac{3-2i}{2+5i} \)
\( = \frac{(3+2i)(2+5i) + (2-5i)(3-2i)}{(2-5i)(2+5i)} \)
\( = \frac{6 + 15i + 4i + 10i^2 + 6 - 4i - 15i + 10i^2}{4 - 25i^2} \)
\( = \frac{12 + 20i^2}{4 - 25i^2} \)
\( = \frac{12 + 20(-1)}{4 - 25(-1)} \) ... [\( \because i^2 = -1 \)]
\( = \frac{-8}{29} \)
\( \therefore \frac{3+2i}{2-5i} + \frac{3-2i}{2+5i} = -\frac{8}{29} + 0i \)
\( \dots a = -\frac{8}{29} \) and \( b = 0 \)

(v) \( \frac{2+\sqrt{-3}}{4+\sqrt{-3}} = \frac{2+\sqrt{3}i}{4+\sqrt{3}i} \)
\( = \frac{(2+\sqrt{3}i)(4-\sqrt{3}i)}{(4+\sqrt{3}i)(4-\sqrt{3}i)} \)
\( = \frac{8 - 2\sqrt{3}i + 4\sqrt{3}i - 3i^2}{16 - 3i^2} \)
\( = \frac{8 + 2\sqrt{3}i - 3(-1)}{16 - 3(-1)} \) ... [\( \because i^2 = -1 \)]
\( = \frac{11 + 2\sqrt{3}i}{19} \)
\( \therefore \frac{2+\sqrt{-3}}{4+\sqrt{-3}} = \frac{11}{19} + \frac{2\sqrt{3}}{19}i \)
\( \dots a = \frac{11}{19} \) and \( b = \frac{2\sqrt{3}}{19} \)
In simple words: To simplify these complex fractions, we cross-multiply or multiply the numerator and denominator by the conjugate of the denominator. This removes the imaginary unit 'i' from the bottom, allowing us to separate the real part (a) and the imaginary part (b) clearly.

🎯 Exam Tip: Always remember that \( i^2 = -1 \) and double-check your signs when expanding brackets, as a simple sign error is the most common mistake in complex number simplification.

Question 2(vi). Express the following in the form of a + ib, find the values of a and b: \( (2 + 3i)(2 - 3i) \)
Answer:
\( (2 + 3i)(2 - 3i) = 4 - 9i^2 \)
\( = 4 - 9(-1) \quad [\because i^2 = -1] \)
\( = 4 + 9 = 13 \)
\( \therefore (2 + 3i)(2 - 3i) = 13 + 0i \)
\( \therefore a = 13 \) and \( b = 0 \)
In simple words: When we multiply a complex number by its conjugate, the imaginary parts cancel out, leaving us with a purely real number.

🎯 Exam Tip: Remember that \( (x + yi)(x - yi) = x^2 + y^2 \). This shortcut helps you quickly find the product of complex conjugates without expanding every term.

Question 2(vii). Express the following in the form of a + ib, find the values of a and b: \( \frac{4i^8 - 3i^9 + 3}{3i^{11} - 4i^{10} - 2} \)
Answer:
\( \frac{4i^8 - 3i^9 + 3}{3i^{11} - 4i^{10} - 2} = \frac{4(i^4)^2 - 3(i^4)^2 \cdot i + 3}{3(i^4)^2 \cdot i^3 - 4(i^4)^2 \cdot i^2 - 2} \)
Since, \( i^2 = -1, i^3 = -i \) and \( i^4 = 1 \)
\( \therefore \frac{4i^8 - 3i^9 + 3}{3i^{11} - 4i^{10} - 2} = \frac{4(1)^2 - 3(1)^2 \cdot i + 3}{3(1)^2(-i) - 4(1)^2(-1) - 2} \)
\( = \frac{4 - 3i + 3}{-3i + 4 - 2} \)
\( = \frac{7 - 3i}{2 - 3i} \)
\( = \frac{(7 - 3i)(2 + 3i)}{(2 - 3i)(2 + 3i)} \)
\( = \frac{14 + 21i - 6i - 9i^2}{4 - 9i^2} \)
\( = \frac{14 + 15i - 9(-1)}{4 - 9(-1)} \)
\( = \frac{23 + 15i}{13} \)
\( \therefore \frac{4i^8 - 3i^9 + 3}{3i^{11} - 4i^{10} - 2} = \frac{23}{13} + \frac{15}{13}i \)
\( \therefore a = \frac{23}{13} \) and \( b = \frac{15}{13} \)
In simple words: First, simplify the high powers of \( i \) by dividing the exponents by 4. Then, multiply the top and bottom by the conjugate of the bottom to remove \( i \) from the denominator.

🎯 Exam Tip: Always reduce higher powers of \( i \) to \( \pm 1 \) or \( \pm i \) first to make the fraction much simpler to rationalize.

Question 3. Show that \( (-1 + \sqrt{3}i)^3 \) is a real number.
Answer:
\( (-1 + \sqrt{3}i)^3 \)
\( = (-1)^3 + 3(-1)^2(\sqrt{3}i) + 3(-1)(\sqrt{3}i)^2 + (\sqrt{3}i)^3 \quad [\because (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3] \)
\( = -1 + 3\sqrt{3}i - 3(3i^2) + 3\sqrt{3}i^3 \)
\( = -1 + 3\sqrt{3}i - 3(-3) - 3\sqrt{3}i \quad [\because i^2 = -1, i^3 = -i] \)
\( = -1 + 9 \)
\( = 8 \), which is a real number.
In simple words: We expand the expression using the cubic formula. After substituting the values of \( i^2 \) and \( i^3 \), the imaginary terms cancel out completely, leaving us with the real number 8.

🎯 Exam Tip: Clearly state the algebraic identity used for expansion and show the step-by-step substitution of \( i^2 = -1 \) and \( i^3 = -i \) to secure full marks.

Question 4. Evaluate the following:
(i) \( i^{35} \)
Answer:
\( i^{35} = i^{32} \cdot i^3 \)
\( = (i^4)^8 \cdot i^3 \)
\( = (1)^8 \cdot (-i) \quad [\text{since } i^4 = 1 \text{ and } i^3 = -i] \)
\( = -i \)
In simple words: To find the value of a high power of \( i \), divide the power by 4 and find the remainder. Here, 35 divided by 4 leaves a remainder of 3, so \( i^{35} \) is the same as \( i^3 \text{, which is } -i \).

🎯 Exam Tip: For any large power \( i^n \), divide \( n \) by 4. The remainder \( r \) will give you the answer as \( i^r \), where \( i^0=1, i^1=i, i^2=-1, i^3=-i \).

Question 4. Evaluate:
(i) \( i^{35} \)
(ii) \( i^{888} \)
(iii) \( i^{93} \)
(iv) \( i^{116} \)
(v) \( i^{403} \)
(vi) \( \frac{1}{i^{58}} \)
(vii) \( i^{30} + i^{40} + i^{50} + i^{60} \)
Answer:
We know that, \( i^2 = -1 \), \( i^3 = -i \), \( i^4 = 1 \)
(i) \( i^{35} = (i^4)^8 (i^2) i = (1)^8 (-1) i = -i \)
(ii) \( i^{888} = (i^4)^{222} = (1)^{222} = 1 \)
(iii) \( i^{93} = (i^4)^{23} \cdot i = (1)^{23} \cdot i = i \)
(iv) \( i^{116} = (i^4)^{29} = (1)^{29} = 1 \)
(v) \( i^{403} = (i^4)^{100} (i^2) i = (1)^{100} (-1) i = -i \)
(vi) \( \frac{1}{i^{58}} = \frac{1}{(i^4)^{14} \cdot i^2} = \frac{1}{(1)^{14}(-1)} = -1 \)
(vii) \( i^{30} + i^{40} + i^{50} + i^{60} = (i^4)^7 i^2 + (i^4)^{10} + (i^4)^{12} i^2 + (i^4)^{15} = (1)^7 (-1) + (1)^{10} + (1)^{12} (-1) + (1)^{15} = -1 + 1 - 1 + 1 = 0 \). These calculations demonstrate how higher powers of the imaginary unit always simplify to one of four basic values.
In simple words: To find the value of any high power of \( i \), we divide the power by 4. Since \( i^4 = 1 \), we can easily simplify the expression using the remainder.

🎯 Exam Tip: Remember that any power of \( i \) can be simplified by dividing the exponent by 4 and looking at the remainder: \( i^0=1, i^1=i, i^2=-1, i^3=-i \).

Question 5. Show that \( 1 + i^{10} + i^{20} + i^{30} \) is a real number.
Answer:
\( 1 + i^{10} + i^{20} + i^{30} \)
\( = 1 + (i^4)^2 \cdot i^2 + (i^4)^5 + (i^4)^7 \cdot i^2 \)
\( = 1 + (1)^2 (-1) + (1)^5 + (1)^7 (-1) \) [\( \because i^4 = 1, i^2 = -1 \)]
\( = 1 - 1 + 1 - 1 \)
\( = 0 \), which is a real number. Since zero has no imaginary part, it is classified as a purely real number.
In simple words: We simplify each term by writing the powers of \( i \) in terms of \( i^4 \). When we add them up, the imaginary parts cancel out completely, leaving us with 0, which is a real number.

🎯 Exam Tip: When asked to show a number is real, simplify the expression until all terms containing \( i \) disappear or cancel out to leave a constant.

Question 6. Find the value of
(i) \( i^{49} + i^{68} + i^{89} + i^{110} \)
(ii) \( i + i^2 + i^3 + i^4 \)
Answer:
(i) \( i^{49} + i^{68} + i^{89} + i^{110} \)
\( = (i^4)^{12} \cdot i + (i^4)^{17} + (i^4)^{22} \cdot i + (i^4)^{27} \cdot i^2 \)
\( = (1)^{12} \cdot i + (1)^{17} + (1)^{22} \cdot i + (1)^{27} \cdot (-1) \)
\( = i + 1 + i - 1 \)
\( = 2i \)

(ii) \( i + i^2 + i^3 + i^4 \)
\( = i + (-1) + (-i) + 1 \)
\( = 0 \). This shows that the sum of any four consecutive powers of \( i \) is always equal to zero.
In simple words: We break down each power of \( i \) using the rule \( i^4 = 1 \). After simplifying and adding the terms together, we get the final simplified values.

🎯 Exam Tip: Keep in mind that the sum of any four consecutive powers of \( i \) (like \( i^1 + i^2 + i^3 + i^4 \)) is always 0, which is a very useful shortcut for multiple-choice questions.

Question 7. Find the value of \( 1 + i^2 + i^4 + i^6 + i^8 + \dots + i^{20} \).
Answer:
Let the given expression be:
\( S = 1 + i^2 + i^4 + i^6 + i^8 + \dots + i^{20} \)
We can group the terms in pairs of consecutive powers to simplify the calculation:
\( S = 1 + (i^2 + i^4) + (i^6 + i^8) + (i^{10} + i^{12}) + (i^{14} + i^{16}) + (i^{18} + i^{20}) \)
Since \( i^2 = -1 \), we can express higher powers of \( i \) in terms of \( i^2 \):
\( S = 1 + [i^2 + (i^2)^2] + [(i^2)^3 + (i^2)^4] + [(i^2)^5 + (i^2)^6] + [(i^2)^7 + (i^2)^8] + [(i^2)^9 + (i^2)^{10}] \)
Substituting \( i^2 = -1 \):
\( S = 1 + [-1 + (-1)^2] + [(-1)^3 + (-1)^4] + [(-1)^5 + (-1)^6] + [(-1)^7 + (-1)^8] + [(-1)^9 + (-1)^{10}] \)
\( S = 1 + [-1 + 1] + [-1 + 1] + [-1 + 1] + [-1 + 1] + [-1 + 1] \)
\( S = 1 + 0 + 0 + 0 + 0 + 0 \)
\( S = 1 \)
Thus, the simplified value of the expression is 1.
In simple words: Since \( i^2 = -1 \), the terms cancel each other out in pairs (like \( -1 + 1 = 0 \)), leaving only the very first number, which is 1.

🎯 Exam Tip: Remember that the sum of any two consecutive even powers of \( i \) (like \( i^{2k} + i^{2k+2} \)) is always zero, which simplifies long series instantly.

Question 8. Find the values of x and y which satisfy the following equations (\( x, y \in \mathbb{R} \)):
(i) \( (x + 2y) + (2x - 3y)i + 4i = 5 \)
(ii) \( \frac{x+1}{1+i} + \frac{y-1}{1-i} = i \)
Answer:
(i) \( (x + 2y) + (2x - 3y)i + 4i = 5 \)
Given equation:
\( (x + 2y) + (2x - 3y)i + 4i = 5 \)

\( \implies (x + 2y) + (2x - 3y)i = 5 - 4i \)
Equating the real and imaginary parts on both sides, we get:
\( x + 2y = 5 \) ——— (1)
\( 2x - 3y = -4 \) ——— (2)
Multiplying equation (1) by 2, we get:
\( 2x + 4y = 10 \) ——— (3)
Subtracting equation (2) from equation (3):
\( (2x + 4y) - (2x - 3y) = 10 - (-4) \)

\( \implies 7y = 14 \)

\( \implies y = 2 \)
Substituting \( y = 2 \) in equation (1):
\( x + 2(2) = 5 \)

\( \implies x + 4 = 5 \)

\( \implies x = 1 \)
Therefore, \( x = 1 \) and \( y = 2 \).

(ii) \( \frac{x+1}{1+i} + \frac{y-1}{1-i} = i \)
Given equation:
\( \frac{x+1}{1+i} + \frac{y-1}{1-i} = i \)
Taking the common denominator on the left-hand side to combine the fractions:
\( \frac{(x+1)(1-i) + (y-1)(1+i)}{(1+i)(1-i)} = i \)
Since \( (1+i)(1-i) = 1^2 - i^2 = 1 - (-1) = 2 \), we have:
\( \frac{(x - xi + 1 - i) + (y + yi - 1 - i)}{2} = i \)

\( \implies (x + y) + (-x + y - 2)i = 2i \)
Equating the real and imaginary parts on both sides, we get:
\( x + y = 0 \) ——— (1)
\( -x + y - 2 = 2 \)

\( \implies -x + y = 4 \) ——— (2)
Adding equation (1) and equation (2):
\( (x + y) + (-x + y) = 0 + 4 \)

\( \implies 2y = 4 \)

\( \implies y = 2 \)
Substituting \( y = 2 \) in equation (1):
\( x + 2 = 0 \)

\( \implies x = -2 \)
Therefore, \( x = -2 \) and \( y = 2 \).
In simple words: To solve these equations, we group the real numbers together and the imaginary numbers (with \( i \)) together, then set them equal to the corresponding parts on the other side of the equals sign.

🎯 Exam Tip: When equating complex numbers, always ensure both sides are in the standard form \( a + bi \) before comparing the real and imaginary parts.

Question 8. Find the values of x and y which satisfy the following equation:
(ii) \( \frac{x+1}{1+i} + \frac{y-1}{1-i} = i \)
Answer: Given equation:
\( \frac{x+1}{1+i} + \frac{y-1}{1-i} = i \)
\( \implies \frac{(x+1)(1-i) + (y-1)(1+i)}{(1+i)(1-i)} = i \)
\( \implies \frac{x - xi + 1 - i + y + yi - 1 - i}{1 - i^2} = i \)
\( \implies \frac{(x+y) + (y-x-2)i}{1 - (-1)} = i \) ... [\( \because i^2 = -1 \)]
\( \implies \frac{(x+y) + (y-x-2)i}{2} = i \)
\( \implies (x+y) + (y-x-2)i = 2i \)
\( \implies (x+y) + (y-x-2)i = 0 + 2i \)
Equating real and imaginary parts, we get:
\( x + y = 0 \) and \( y - x - 2 = 2 \)
\( \implies x + y = 0 \) ......(i)
and \( -x + y = 4 \) ......(ii)
Adding (i) and (ii), we get:
\( 2y = 4 \)
\( \implies y = 2 \)
Putting \( y = 2 \) in (i), we get:
\( x + 2 = 0 \)
\( \implies x = -2 \)
\( \implies x = -2 \) and \( y = 2 \)
In simple words: We simplify the complex fractions by finding a common denominator, then we separate the real and imaginary parts to form two simple equations and solve for x and y.

🎯 Exam Tip: Always multiply the numerator and denominator by the conjugate of the complex denominator to simplify fractions easily.

Question 9. Find the value of:
(i) \( x^3 - x^2 + x + 46 \), if \( x = 2 + 3i \)
(ii) \( 2x^3 - 11x^2 + 44x + 27 \), if \( x = \frac{25}{3-4i} \)
Answer:
(i) Given: \( x = 2 + 3i \)
\( \implies x - 2 = 3i \)
Squaring both sides, we get:
\( (x - 2)^2 = (3i)^2 \)
\( \implies x^2 - 4x + 4 = -9 \)
\( \implies x^2 - 4x + 13 = 0 \)
Now, we divide \( x^3 - x^2 + x + 46 \) by \( x^2 - 4x + 13 \):
\( x^3 - x^2 + x + 46 = x(x^2 - 4x + 13) + 3(x^2 - 4x + 13) + 7 \)
\( \implies x^3 - x^2 + x + 46 = (x^2 - 4x + 13)(x + 3) + 7 \)
Substituting \( x^2 - 4x + 13 = 0 \):
\( \implies x^3 - x^2 + x + 46 = (0)(x + 3) + 7 = 7 \)
Thus, the value of \( x^3 - x^2 + x + 46 \) is 7.

(ii) Given: \( x = \frac{25}{3-4i} \)
First, simplify \( x \) by multiplying the numerator and denominator by the conjugate \( 3 + 4i \):
\( x = \frac{25(3+4i)}{(3-4i)(3+4i)} \)
\( \implies x = \frac{25(3+4i)}{3^2 - (4i)^2} \)
\( \implies x = \frac{25(3+4i)}{9 - 16(-1)} \)
\( \implies x = \frac{25(3+4i)}{25} \)
\( \implies x = 3 + 4i \)
\( \implies x - 3 = 4i \)
Squaring both sides, we get:
\( (x - 3)^2 = (4i)^2 \)
\( \implies x^2 - 6x + 9 = -16 \)
\( \implies x^2 - 6x + 25 = 0 \)
Now, we divide \( 2x^3 - 11x^2 + 44x + 27 \) by \( x^2 - 6x + 25 \):
\( 2x^3 - 11x^2 + 44x + 27 = 2x(x^2 - 6x + 25) + 1(x^2 - 6x + 25) + 2 \)
\( \implies 2x^3 - 11x^2 + 44x + 27 = (x^2 - 6x + 25)(2x + 1) + 2 \)
Substituting \( x^2 - 6x + 25 = 0 \):
\( \implies 2x^3 - 11x^2 + 44x + 27 = (0)(2x + 1) + 2 = 2 \)
Thus, the value of \( 2x^3 - 11x^2 + 44x + 27 \) is 2.
In simple words: To find the value of these long expressions, we first rearrange the given value of x to form a quadratic equation equal to zero, and then we divide the main expression by this quadratic equation to find the remaining constant value.

🎯 Exam Tip: Expressing the polynomial in terms of the quadratic factor that equals zero is the fastest way to find its value without tedious direct substitution.

Question 1. Solve the following:
(i) Find the value of \( x^3 - x^2 + x + 46 \) if \( x = 2 + 3i \).
(ii) Find the value of \( x^2 - 6x + 25 \) if \( x = \frac{25}{3-4i} \).
Answer:
(i) Given: \( x = 2 + 3i \)
\( \implies x - 2 = 3i \)
Squaring both sides:
\( \implies (x - 2)^2 = (3i)^2 \)
\( \implies x^2 - 4x + 4 = 9i^2 \)
\( \implies x^2 - 4x + 4 = 9(-1) \quad [\because i^2 = -1] \)
\( \implies x^2 - 4x + 13 = 0 \) ———(i)
Now, dividing \( x^3 - x^2 + x + 46 \) by \( x^2 - 4x + 13 \), we get:
\( x^3 - x^2 + x + 46 = (x^2 - 4x + 13)(x + 3) + 7 \)
\( \implies x^3 - x^2 + x + 46 = 0(x + 3) + 7 \quad [\text{From (i)}] \)
\( \implies x^3 - x^2 + x + 46 = 7 \)

(ii) Given: \( x = \frac{25}{3-4i} \)
Multiplying the numerator and denominator by the conjugate \( 3 + 4i \):
\( \implies x = \frac{25(3+4i)}{(3-4i)(3+4i)} \)
\( \implies x = \frac{25(3+4i)}{9 - 16i^2} \)
\( \implies x = \frac{25(3+4i)}{9 - 16(-1)} \quad [\because i^2 = -1] \)
\( \implies x = \frac{25(3+4i)}{25} \)
\( \implies x = 3 + 4i \)
\( \implies x - 3 = 4i \)
Squaring both sides:
\( \implies (x - 3)^2 = (4i)^2 \)
\( \implies x^2 - 6x + 9 = 16i^2 \)
\( \implies x^2 - 6x + 9 = 16(-1) \quad [\because i^2 = -1] \)
\( \implies x^2 - 6x + 25 = 0 \) ———(i)
In simple words: To find the value of these expressions, we first rearrange the given complex number to form a quadratic equation equal to zero. Then, we use this zero-value equation to simplify and find the final numerical value of the larger expression.

🎯 Exam Tip: Always rationalize the denominator first when dealing with complex fractions, and remember that \( i^2 = -1 \) to simplify quadratic terms quickly.

Question. Find the value of \( 2x^3 - 11x^2 + 44x + 27 \) given that \( x^2 - 6x + 25 = 0 \).
Answer: First, we perform polynomial long division to divide \( 2x^3 - 11x^2 + 44x + 27 \) by \( x^2 - 6x + 25 \): \[ \begin{array}{rll} & 2x + 1 & \\ \cline{2-2} x^2 - 6x + 25 & \smash{\big)} 2x^3 - 11x^2 + 44x + 27 & \\ & \underline{2x^3 - 12x^2 + 50x} & \\ & \phantom{2x^3 - {}} x^2 - 6x + 27 & \\ & \phantom{2x^3 - {}} \underline{x^2 - 6x + 25} & \\ & \phantom{2x^3 - 12x^2 + 50x} 2 & \end{array} \] By the division algorithm:
\( \text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder} \)
\( \therefore 2x^3 - 11x^2 + 44x + 27 = (x^2 - 6x + 25)(2x + 1) + 2 \)
Substituting \( x^2 - 6x + 25 = 0 \) from the given condition:
\( = 0 \cdot (2x + 1) + 2 \)
\( = 0 + 2 \)
\( = 2 \)
Thus, the value of the expression is 2.
In simple words: To find the value of a large expression easily, we divide it by the smaller expression that equals zero. Since the divisor is zero, the entire part multiplied by it becomes zero, leaving only the remainder as the final answer.

🎯 Exam Tip: When solving polynomial evaluation questions, always use long division to express the higher-degree polynomial in terms of the quadratic equation that equals zero. This simplifies the calculation to just finding the remainder.