Complex Numbers
Miscellaneous Exercise Solutions1. Evaluation & Simplification
Q1: Evaluate $[i^{18} + (\frac{1}{i})^{25}]^3$
Term 1: $i^{18} = (i^4)^4 \cdot i^2 = 1 \cdot (-1) = -1$
Term 2: $(\frac{1}{i})^{25} = \frac{1}{i^{25}} = \frac{1}{(i^4)^6 \cdot i} = \frac{1}{i}$
$\frac{1}{i} \times \frac{i}{i} = \frac{i}{-1} = -i$
$\frac{1}{i} \times \frac{i}{i} = \frac{i}{-1} = -i$
Expression becomes: $[-1 – i]^3 = – (1 + i)^3$
Expand $(1+i)^3 = 1^3 + 3(1)^2(i) + 3(1)(i^2) + i^3$
$= 1 + 3i – 3 – i = -2 + 2i$
Expand $(1+i)^3 = 1^3 + 3(1)^2(i) + 3(1)(i^2) + i^3$
$= 1 + 3i – 3 – i = -2 + 2i$
Multiply by initial negative sign: $-(-2 + 2i) = 2 – 2i$
Answer: $2 – 2i$
Q3: Reduce to standard form: $(\frac{1}{1-4i} – \frac{2}{1+i})(\frac{3-4i}{5+i})$
Simplifying First Bracket:
$= \frac{(1+i) – 2(1-4i)}{(1-4i)(1+i)} = \frac{1+i-2+8i}{1+i-4i-4i^2}$
$= \frac{-1+9i}{1-3i+4} = \frac{-1+9i}{5-3i}$
$= \frac{(1+i) – 2(1-4i)}{(1-4i)(1+i)} = \frac{1+i-2+8i}{1+i-4i-4i^2}$
$= \frac{-1+9i}{1-3i+4} = \frac{-1+9i}{5-3i}$
Multiply by Second Term:
$= \frac{-1+9i}{5-3i} \times \frac{3-4i}{5+i} = \frac{(-1+9i)(3-4i)}{(5-3i)(5+i)}$
$= \frac{-1+9i}{5-3i} \times \frac{3-4i}{5+i} = \frac{(-1+9i)(3-4i)}{(5-3i)(5+i)}$
Numerator: $-3 + 4i + 27i – 36i^2 = -3 + 31i + 36 = 33 + 31i$
Denominator: $25 + 5i – 15i – 3i^2 = 25 – 10i + 3 = 28 – 10i$
Denominator: $25 + 5i – 15i – 3i^2 = 25 – 10i + 3 = 28 – 10i$
Result: $\frac{33+31i}{28-10i}$. (Rationalize Denominator)
$\frac{(33+31i)(28+10i)}{(28-10i)(28+10i)} = \frac{924 + 330i + 868i – 310}{784 + 100}$
$= \frac{614 + 1198i}{884} = \frac{307 + 599i}{442}$
$\frac{(33+31i)(28+10i)}{(28-10i)(28+10i)} = \frac{924 + 330i + 868i – 310}{784 + 100}$
$= \frac{614 + 1198i}{884} = \frac{307 + 599i}{442}$
Answer: $\frac{307}{442} + i\frac{599}{442}$
2. Proofs & Identities
Q2: Prove Re($z_1 z_2$) = Re $z_1$ Re $z_2$ – Im $z_1$ Im $z_2$
Let $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$
$z_1 z_2 = (x_1 + iy_1)(x_2 + iy_2)$
$= x_1 x_2 + i x_1 y_2 + i y_1 x_2 + i^2 y_1 y_2$
$= (x_1 x_2 – y_1 y_2) + i(x_1 y_2 + y_1 x_2)$
$= x_1 x_2 + i x_1 y_2 + i y_1 x_2 + i^2 y_1 y_2$
$= (x_1 x_2 – y_1 y_2) + i(x_1 y_2 + y_1 x_2)$
Re($z_1 z_2$) = $x_1 x_2 – y_1 y_2$
Substitute back:
$x_1 = \text{Re } z_1, \quad x_2 = \text{Re } z_2$
$y_1 = \text{Im } z_1, \quad y_2 = \text{Im } z_2$
$x_1 = \text{Re } z_1, \quad x_2 = \text{Re } z_2$
$y_1 = \text{Im } z_1, \quad y_2 = \text{Im } z_2$
Re($z_1 z_2$) = Re $z_1$ Re $z_2$ – Im $z_1$ Im $z_2$ (Proved)
Q4: If $x – iy = \sqrt{\frac{a-ib}{c-id}}$, prove $(x^2+y^2)^2 = \frac{a^2+b^2}{c^2+d^2}$
Given: $x – iy = \sqrt{\frac{a-ib}{c-id}}$
Take modulus on both sides:
$|x – iy| = \left| \sqrt{\frac{a-ib}{c-id}} \right| = \sqrt{\left| \frac{a-ib}{c-id} \right|}$
Take modulus on both sides:
$|x – iy| = \left| \sqrt{\frac{a-ib}{c-id}} \right| = \sqrt{\left| \frac{a-ib}{c-id} \right|}$
$\sqrt{x^2 + (-y)^2} = \sqrt{\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}}$
Squaring both sides:
$x^2 + y^2 = \sqrt{\frac{a^2+b^2}{c^2+d^2}}$
$x^2 + y^2 = \sqrt{\frac{a^2+b^2}{c^2+d^2}}$
Squaring again:
$(x^2 + y^2)^2 = \frac{a^2+b^2}{c^2+d^2}$
$(x^2 + y^2)^2 = \frac{a^2+b^2}{c^2+d^2}$
Hence Proved
3. Complex Arithmetic & Modulus
Q9: Find modulus of $\frac{1+i}{1-i} – \frac{1-i}{1+i}$
Take LCM:
$\frac{(1+i)^2 – (1-i)^2}{(1-i)(1+i)}$
$\frac{(1+i)^2 – (1-i)^2}{(1-i)(1+i)}$
Expand Numerator:
$(1 + 2i + i^2) – (1 – 2i + i^2) = (1 + 2i – 1) – (1 – 2i – 1)$
$= 2i – (-2i) = 4i$
$(1 + 2i + i^2) – (1 – 2i + i^2) = (1 + 2i – 1) – (1 – 2i – 1)$
$= 2i – (-2i) = 4i$
Expand Denominator:
$1^2 – i^2 = 1 – (-1) = 2$
$1^2 – i^2 = 1 – (-1) = 2$
Simplify Fraction: $\frac{4i}{2} = 2i = 0 + 2i$
Modulus $|z| = \sqrt{0^2 + 2^2} = \sqrt{4} = 2$
Modulus = 2
4. Solving for Unknowns
Q12: Find number of non-zero integral solutions of $|1-i|^x = 2^x$
Calculate modulus $|1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$
Equation becomes: $(\sqrt{2})^x = 2^x$
$(2^{1/2})^x = 2^x$
$2^{x/2} = 2^x$
$(2^{1/2})^x = 2^x$
$2^{x/2} = 2^x$
Equating powers: $\frac{x}{2} = x$
$x = 2x \Rightarrow x = 0$
$x = 2x \Rightarrow x = 0$
Question asks for non-zero solutions.
Number of non-zero solutions = 0
Q14: Find least positive integer $m$ for $(\frac{1+i}{1-i})^m = 1$
Simplify base: $\frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{1^2-i^2}$
$= \frac{1+2i+i^2}{1-(-1)} = \frac{1+2i-1}{2} = \frac{2i}{2} = i$
Equation becomes: $i^m = 1$
We know $i^1=i, i^2=-1, i^3=-i, i^4=1$.
Least positive integer m = 4