Solved Examples

How to Use This Section

Here are 32 fully worked problems covering the entire chapter — powers of i, arithmetic in all four operations, standard-form conversions, moduli and conjugates, inverses, Argand-plane geometry, and the miscellaneous-level heavyweights — arranged in a deliberate easy → medium → hard progression.

One suggestion that multiplies the value of every problem: attempt each yourself before reading the solution. In this chapter especially, decide your STRATEGY first — expand, or use modulus/conjugate properties? The property route is usually five times shorter. Problems 1-10 warm up on powers and arithmetic, 11-22 build the modulus-conjugate machinery, 23-32 are Board and JEE-level.

Solved Examples

Example 1: Standard form warm-up

Express $(5i)\left(-\frac{3}{5}i\right)$ in the form $a + ib$.

Solution:

Step 1 — Multiply coefficients and i's separately. $5 \times \left(-\frac{3}{5}\right) = -3$ and $i \cdot i = i^2 = -1$.

Step 2 — Combine. $(-3)(-1) = 3 = 3 + i0$.

Takeaway: Numbers with numbers, i's with i's.

Example 2: Power reduction

Evaluate $i^{-999}$.

Solution:

Step 1 — Add a free multiple of 4. $1000 = 4(250)$, so $i^{1000} = 1$ and $i^{-999} = i^{-999} \cdot i^{1000} = i^1$.

Step 2 — Read the wheel. $i^{-999} = i$.

Takeaway: 1000 is a multiple of 4, so $i^{1000} = 1$ costs nothing.

Example 3: A triple product

Express $(1 - i)(-3 + 6i)$ in the form $a + ib$, then multiply the result by $i$.

Solution:

Step 1 — Expand the first product. $-3 + 6i + 3i - 6i^2 = -3 + 9i + 6 = 3 + 9i$.

Step 2 — Multiply by i. $(3 + 9i)i = 3i + 9i^2 = -9 + 3i$.

Step 3 — Notice the pattern. $(a + bi) \cdot i = -b + ai$ — the parts swap with one sign flip.

Takeaway: Multiplying by i rotates the number: a preview of the rotation picture of multiplication.

Example 4: Collecting a long expression

Reduce $\left(\frac{1}{3} + i\frac{7}{3}\right) + \left(4 + i\frac{1}{3}\right) - \left(-\frac{4}{3} + i\right)$ to standard form.

Solution:

Step 1 — Collect the real parts. $\frac{1}{3} + 4 + \frac{4}{3} = \frac{1 + 12 + 4}{3} = \frac{17}{3}$.

Step 2 — Collect the imaginary parts. $\frac{7}{3} + \frac{1}{3} - 1 = \frac{7 + 1 - 3}{3} = \frac{5}{3}$.

Step 3 — Assemble. $\frac{17}{3} + \frac{5}{3}i$.

Takeaway: Line up the parts vertically and add like fractions — the i is just a column separator.

Example 5: The standard-form gauntlet

Reduce $\left(\frac{1}{1 - 4i} - \frac{2}{1 + i}\right)\left(\frac{3 - 4i}{5 + i}\right)$ to the standard form.

Solution:

Step 1 — Combine the first bracket over a common denominator. $\frac{(1 + i) - 2(1 - 4i)}{(1 - 4i)(1 + i)} = \frac{1 + i - 2 + 8i}{1 + i - 4i + 4} = \frac{-1 + 9i}{5 - 3i}$.

Step 2 — Multiply by the second factor. $\frac{(-1 + 9i)(3 - 4i)}{(5 - 3i)(5 + i)} = \frac{-3 + 4i + 27i + 36}{25 + 5i - 15i + 3} = \frac{33 + 31i}{28 - 10i}$.

Step 3 — Rationalise once, at the very end. $\frac{(33 + 31i)(28 + 10i)}{28^2 + 10^2} = \frac{924 + 330i + 868i - 310}{884} = \frac{614 + 1198i}{884}$.

Step 4 — Reduce the fractions. $\frac{614}{884} = \frac{307}{442}$ and $\frac{1198}{884} = \frac{599}{442}$: answer $\frac{307}{442} + \frac{599}{442}i$.

Takeaway: The longest computation in the chapter. Strategy: simplify each bracket fully BEFORE multiplying; rationalise once at the very end.

Example 6: Modulus evaluation

If $z_1 = 2 - i$ and $z_2 = 1 + i$, find $\left|\frac{z_1 + z_2 + 1}{z_1 - z_2 + 1}\right|$.

Solution:

Step 1 — Simplify the numerator. $z_1 + z_2 + 1 = (2 + 1 + 1) + (-1 + 1)i = 4$.

Step 2 — Simplify the denominator. $z_1 - z_2 + 1 = (2 - 1 + 1) + (-1 - 1)i = 2 - 2i$.

Step 3 — Take the modulus quotient. $\frac{|4|}{|2 - 2i|} = \frac{4}{\sqrt{4 + 4}} = \frac{4}{2\sqrt{2}} = \sqrt{2}$.

Takeaway: Simplify inside first; the final step is one modulus quotient.

Example 7: Real-imaginary extraction

Let $z_1 = 2 - i$, $z_2 = -2 + i$. Find (i) Re$\left(\frac{z_1 z_2}{\bar{z}_1}\right)$ (ii) Im$\left(\frac{1}{z_1\bar{z}_1}\right)$.

Solution:

Step 1 — (i) Compute the product. $z_1z_2 = (2 - i)(-2 + i) = -4 + 2i + 2i - i^2 = -3 + 4i$.

Step 2 — Divide by the conjugate. $\bar{z}_1 = 2 + i$: $\frac{-3 + 4i}{2 + i} \times \frac{2 - i}{2 - i} = \frac{-6 + 3i + 8i + 4}{5} = \frac{-2 + 11i}{5}$.

Step 3 — Read off. Re $= -\frac{2}{5}$.

Step 4 — (ii) Spot the shortcut. $z_1\bar{z}_1 = |z_1|^2 = 5$ is REAL, so $\frac{1}{5}$ is real: Im $= 0$ — no computation needed.

Takeaway: Part (ii) needs NO computation once you see $z\bar{z} = |z|^2$ is always real.

Example 8: u + iv from a cube

If $(x + iy)^3 = u + iv$, show that $\frac{u}{x} + \frac{v}{y} = 4(x^2 - y^2)$.

Solution:

Step 1 — Expand the cube. $(x + iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 = (x^3 - 3xy^2) + i(3x^2y - y^3)$.

Step 2 — Identify u and v. $u = x(x^2 - 3y^2)$ and $v = y(3x^2 - y^2)$.

Step 3 — Form the target combination. $\frac{u}{x} + \frac{v}{y} = (x^2 - 3y^2) + (3x^2 - y^2) = 4x^2 - 4y^2 = 4(x^2 - y^2)$. ∎

Takeaway: Expand, name u and v, and the target combination collapses.

Example 9: The a² + b² pattern

If $a + ib = \frac{(x + i)^2}{2x^2 + 1}$, prove that $a^2 + b^2 = \frac{(x^2 + 1)^2}{(2x^2 + 1)^2}$.

Solution:

Step 1 — Recognise $a^2 + b^2$ as a squared modulus. $a^2 + b^2 = |a + ib|^2 = \left|\frac{(x + i)^2}{2x^2 + 1}\right|^2 = \frac{|x + i|^4}{(2x^2 + 1)^2}$.

Step 2 — Compute the small modulus. $|x + i|^2 = x^2 + 1$, so $|x + i|^4 = (x^2 + 1)^2$.

Step 3 — Conclude. $a^2 + b^2 = \frac{(x^2 + 1)^2}{(2x^2 + 1)^2}$. ∎

Takeaway: Every "prove $a^2 + b^2 = \ldots$" is a modulus-squared statement; never expand.

Example 10: Conjugate equation

Find all complex z with $z^2 = \bar{z}$.

Solution:

Step 1 — Take moduli of both sides. $|z|^2 = |\bar{z}| = |z|$, so $|z|(|z| - 1) = 0$: $|z| = 0$ or $|z| = 1$.

Step 2 — Case |z| = 0. z = 0, which satisfies $0 = \bar{0}$ ✓.

Step 3 — Case |z| = 1: multiply the equation by z. $z^3 = z\bar{z} = |z|^2 = 1$ — z is a CUBE ROOT of unity: $z = 1, \omega, \omega^2$.

Step 4 — Collect. Solution set $\{0, 1, \omega, \omega^2\}$ — four solutions.

Takeaway: Taking moduli first splits the problem into clean cases; multiplying by z converted the conjugate into $|z|^2$. (Meet $\omega$ properly in the JEE Corner.)

Example 11: A modulus equation

Find all complex z with $|z| = z + 1 + 2i$.

Solution:

Step 1 — Compare imaginary parts. The left side |z| is REAL, so the right side must be too: with z = x + iy, $y + 2 = 0$ gives $y = -2$.

Step 2 — Write the real-part equation. $\sqrt{x^2 + 4} = x + 1$ — which also demands $x + 1 \geq 0$.

Step 3 — Square. $x^2 + 4 = x^2 + 2x + 1$ gives $2x = 3$, $x = \frac{3}{2}$ (and $\frac{3}{2} + 1 > 0$ ✓).

Step 4 — Assemble. $z = \frac{3}{2} - 2i$.

Takeaway: When a modulus equals a complex expression, the expression's imaginary part must vanish — harvest that condition first.

Example 12: Real quotient condition

Find real $\lambda$ so that $\frac{2 + \lambda i}{1 - i}$ is (i) purely real (ii) purely imaginary.

Solution:

Step 1 — Rationalise. $\frac{(2 + \lambda i)(1 + i)}{(1 - i)(1 + i)} = \frac{(2 - \lambda) + (2 + \lambda)i}{2}$.

Step 2 — (i) Purely real: kill the imaginary part. $2 + \lambda = 0$: $\lambda = -2$.

Step 3 — (ii) Purely imaginary: kill the real part. $2 - \lambda = 0$: $\lambda = 2$.

Takeaway: Rationalise, then kill whichever part the question demands.

Example 13: Modulus of a power

Find $\left|(1 + i)^{10}\right|$ without expanding.

Solution:

Step 1 — Apply $|z^n| = |z|^n$. $|(1+i)^{10}| = |1 + i|^{10}$.

Step 2 — Evaluate. $|1 + i| = \sqrt{2}$, so the answer is $(\sqrt{2})^{10} = 2^5 = 32$.

Takeaway: $|z^n| = |z|^n$ — ten-fold expansion avoided.

Example 14: Real and imaginary parts of 1/z

For $z = 4 - 3i$, find Re$\left(\frac{1}{z}\right)$ and Im$\left(\frac{1}{z}\right)$.

Solution:

Step 1 — Use the inverse formula. $\frac{1}{z} = \frac{\bar{z}}{|z|^2} = \frac{4 + 3i}{16 + 9} = \frac{4 + 3i}{25}$.

Step 2 — Read off the parts. Re $= \frac{4}{25}$, Im $= \frac{3}{25}$.

Takeaway: One formula, both parts — the inverse re-read as parts.

Example 15: An i-quotient chain

Simplify $\frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}} - 1$.

Solution:

Step 1 — Factor out the smallest power upstairs and downstairs. Numerator: $i^{584}(i^8 + i^6 + i^4 + i^2 + 1)$; denominator: $i^{574}(i^8 + i^6 + i^4 + i^2 + 1)$.

Step 2 — Cancel the identical brackets. Quotient $= i^{584 - 574} = i^{10}$.

Step 3 — Reduce. $10 = 4(2) + 2$: $i^{10} = -1$.

Step 4 — Subtract 1. $-1 - 1 = -2$.

Takeaway: Factor out the smallest power; matching brackets cancel no matter how ugly.

Example 16: Equality with a quotient

Find real x, y if $\frac{x + iy}{2 + 3i} = 1 - i$.

Solution:

Step 1 — Clear the denominator. $x + iy = (1 - i)(2 + 3i)$.

Step 2 — Expand. $2 + 3i - 2i - 3i^2 = 2 + i + 3 = 5 + i$.

Step 3 — Match parts. x = 5, y = 1.

Takeaway: Clearing the denominator first avoids rationalisation entirely.

Example 17: Conjugate of a power

Find $\overline{(1 + 2i)^2}$ two ways.

Solution:

Step 1 — Way 1: expand, then conjugate. $(1 + 2i)^2 = 1 + 4i + 4i^2 = -3 + 4i$; conjugate: $-3 - 4i$.

Step 2 — Way 2: conjugate, then square. $\overline{(1 + 2i)} = 1 - 2i$, and $(1 - 2i)^2 = 1 - 4i - 4 = -3 - 4i$.

Step 3 — Compare. Same answer ✓ — the bar passes through powers.

Takeaway: $\overline{z^n} = (\bar{z})^n$ — use whichever order is lighter.

Example 18: A symmetric system

If $z_1 + z_2 = 4 + 2i$ and $z_1 - z_2 = 2 + 4i$, find $z_1$, $z_2$ and compare $|z_1|$ with $|z_2|$.

Solution:

Step 1 — Add the equations. $2z_1 = 6 + 6i$: $z_1 = 3 + 3i$.

Step 2 — Subtract them. $2z_2 = (4 - 2) + (2 - 4)i = 2 - 2i$: $z_2 = 1 - i$.

Step 3 — Compare moduli. $|z_1| = \sqrt{9 + 9} = 3\sqrt{2}$ and $|z_2| = \sqrt{2}$ — NOT equal ($|z_1| = 3|z_2|$).

Takeaway: Complex linear systems solve exactly like real ones — add and subtract.

Example 19: The (1 + i)ⁿ real test

Find the smallest positive integer n for which $(1 + i)^{2n} = (1 - i)^{2n}$.

Solution:

Step 1 — Divide both sides. $\left(\frac{1 + i}{1 - i}\right)^{2n} = 1$, and the base is the standard quotient i: $i^{2n} = 1$.

Step 2 — Apply the cycle. $i^m = 1$ exactly when m is a multiple of 4, so 2n must be a multiple of 4 — n must be even.

Step 3 — Take the smallest. n = 2. Check: $(1+i)^4 = (2i)^2 = -4$ and $(1-i)^4 = (-2i)^2 = -4$ ✓.

Takeaway: A classic JEE one-liner — reduce to a power of i and quote the cycle.

Example 20: Modulus locus algebra

Show that $|z + 1| = |z - 1|$ forces z to be purely imaginary; then describe the locus $|z - 1| = |z - i|$.

Solution:

Step 1 — First locus: read as distances. Equidistant from $-1$ and 1 → the perpendicular bisector of the segment → the imaginary axis → Re z = 0: z is purely imaginary (or 0).

Step 2 — Second locus: same tool. Equidistant from 1 (point (1, 0)) and i (point (0, 1)) → the perpendicular bisector of THAT segment.

Step 3 — Identify the line. The midpoint is $\left(\frac{1}{2}, \frac{1}{2}\right)$ and the segment has slope $-1$, so the bisector is the line y = x: Re z = Im z.

Takeaway: Equal-modulus conditions are bisector statements; no algebra required once the picture is drawn.

Example 21: Inverse of a product

Find the multiplicative inverse of $(3 + 4i)(2 - i)$ without expanding the product first.

Solution:

Step 1 — Invert factor by factor. $\left[(3+4i)(2-i)\right]^{-1} = (3+4i)^{-1}(2-i)^{-1} = \frac{3 - 4i}{25} \cdot \frac{2 + i}{5}$.

Step 2 — Expand the light numerator. $(3 - 4i)(2 + i) = 6 + 3i - 8i - 4i^2 = 10 - 5i$.

Step 3 — Divide. $\frac{10 - 5i}{125} = \frac{2}{25} - \frac{1}{25}i$.

Takeaway: Properties first, arithmetic second — the light expansion happened in the numerator only.

Example 22: A purely imaginary product condition

For what real a is $(1 + ai)^2$ purely imaginary?

Solution:

Step 1 — Expand. $(1 + ai)^2 = 1 + 2ai + a^2i^2 = (1 - a^2) + 2ai$.

Step 2 — Impose both conditions. Purely imaginary needs $1 - a^2 = 0$ AND $2a \neq 0$.

Step 3 — Solve. $a = \pm 1$ (both give $2a \neq 0$ ✓).

Step 4 — Check. $(1 + i)^2 = 2i$ ✓ and $(1 - i)^2 = -2i$ ✓.

Takeaway: The check confirms the beloved shortcut $(1 \pm i)^2 = \pm 2i$ from the other side.

Example 23: The least-m power question

Find the least positive integer m for which $\left(\frac{1 + i}{1 - i}\right)^m = 1$.

Solution:

Step 1 — Collapse the base. $\frac{1+i}{1-i} = i$ (multiply by the conjugate: $\frac{(1+i)^2}{2} = i$).

Step 2 — Apply the cycle. $i^m = 1$ requires m to be a multiple of 4.

Step 3 — Take the least. m = 4.

Takeaway: After the base collapses to i, the cycle answers instantly.

Example 24: Sum with conjugate weights

If $z = 3 + 4i$, evaluate $z\bar{z} + (z + \bar{z}) - |z|$.

Solution:

Step 1 — Compute each piece by formula. $z\bar{z} = |z|^2 = 9 + 16 = 25$; $z + \bar{z} = 2\,\mathrm{Re}\,z = 6$; $|z| = 5$.

Step 2 — Combine. $25 + 6 - 5 = 26$.

Takeaway: All three quantities are REAL — computed from formulas, never from expansion.

Example 25: A two-variable matching

Solve for real x, y: $\frac{(1+i)x - 2i}{3 + i} + \frac{(2 - 3i)y + i}{3 - i} = i$.

Solution:

Step 1 — Clear both denominators at once. Multiply through by $(3+i)(3-i) = 10$: $[(1+i)x - 2i](3 - i) + [(2-3i)y + i](3 + i) = 10i$.

Step 2 — Expand the first bracket. $(1+i)(3-i) = 3 - i + 3i + 1 = 4 + 2i$, so it contributes $4x + 2xi - 6i - 2$ (using $-2i(3 - i) = -6i + 2i^2$).

Step 3 — Expand the second. $(2-3i)(3+i) = 6 + 2i - 9i + 3 = 9 - 7i$, so it contributes $9y - 7yi + 3i - 1$ (using $i(3 + i) = 3i - 1$).

Step 4 — Collect real and imaginary totals. Real: $4x + 9y - 3$; imaginary: $2x - 7y - 3$. Matching with $0 + 10i$: $4x + 9y = 3$ and $2x - 7y = 13$.

Step 5 — Solve the linear system. Doubling the second and subtracting from the first: $9y + 14y = 3 - 26$, $23y = -23$, y = $-1$; then $4x = 3 + 9 = 12$, x = 3.

Takeaway: The heaviest matching problem of the family — clear denominators once, expand ONCE, then it is a linear system.

Example 26: Modulus inequality preview

Show that $|z_1 + z_2| \leq |z_1| + |z_2|$ for $z_1 = 3 + 4i$, $z_2 = -4 + 3i$, and identify when equality would hold.

Solution:

Step 1 — Compute the left side. $z_1 + z_2 = -1 + 7i$, so $|z_1 + z_2| = \sqrt{1 + 49} = 5\sqrt{2} \approx 7.07$.

Step 2 — Compute the right side. $|z_1| + |z_2| = 5 + 5 = 10$.

Step 3 — Compare. $5\sqrt{2} < 10$ ✓ — the inequality holds strictly here.

Step 4 — When is it equality? Only when $z_1$ and $z_2$ point the same way (one is a positive real multiple of the other) — not the case for these two.

Takeaway: The triangle inequality in action; the JEE Corner states it in general.

Example 27: Argand geometry of multiplication by i

Plot z = $2 + i$, then iz, $i^2z$, $i^3z$. What figure do the four points form?

Solution:

Step 1 — Compute the four numbers. $iz = i(2 + i) = -1 + 2i$; $i^2z = -z = -2 - i$; $i^3z = -iz = 1 - 2i$.

Step 2 — Compare moduli and positions. All four have modulus $\sqrt{5}$, and each is the previous one turned 90° anticlockwise about the origin.

Step 3 — Name the figure. Four points, equal distances from O, successive quarter-turns: a SQUARE centred at the origin.

Takeaway: Multiplication by i = quarter-turn — the geometric soul of $i^2 = -1$ (two quarter-turns = reversal).

Example 28: A conjugate-heavy proof

Prove that for any z, the number $\frac{z - \bar{z}}{2i}$ is real, and equals Im z.

Solution:

Step 1 — Write z = x + iy and subtract. $z - \bar{z} = (x + iy) - (x - iy) = 2iy$.

Step 2 — Divide by 2i. $\frac{2iy}{2i} = y$ — a real number, exactly Im z. ∎

Takeaway: Companion to Re z = $\frac{z + \bar{z}}{2}$; the pair extracts coordinates from z algebraically.

Example 29: Square roots by matching parts

Find the square roots of $-8 + 6i$: solve $(x + iy)^2 = -8 + 6i$ for real x, y.

Solution:

Step 1 — Match real and imaginary parts. $(x + iy)^2 = (x^2 - y^2) + 2xyi$: so $x^2 - y^2 = -8$ and $2xy = 6$.

Step 2 — Add the modulus equation. $x^2 + y^2 = |-8 + 6i| = \sqrt{64 + 36} = 10$.

Step 3 — Solve the sum-difference pair. Adding and halving: $x^2 = 1$; subtracting: $y^2 = 9$.

Step 4 — Fix the signs. $xy = 3 > 0$: x and y share a sign. Roots: $\pm(1 + 3i)$. Check: $(1 + 3i)^2 = 1 + 6i - 9 = -8 + 6i$ ✓.

Takeaway: The three-equation method (real part, imaginary part, modulus) finds square roots of any complex number — formalised in the JEE Corner.

Example 30: Powers of a unit-modulus number

If $z = \frac{\sqrt{3}}{2} + \frac{1}{2}i$, compute $|z|$, and show $z^6 = -1$.

Solution:

Step 1 — Modulus. $|z| = \sqrt{\frac{3}{4} + \frac{1}{4}} = 1$ — z lies on the unit circle.

Step 2 — Square. $z^2 = \frac{3}{4} - \frac{1}{4} + 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2}i = \frac{1}{2} + \frac{\sqrt{3}}{2}i$.

Step 3 — Square again. $(z^2)^2 = \frac{1}{4} - \frac{3}{4} + 2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2}i = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$.

Step 4 — One more multiplication. $z^6 = z^4 \cdot z^2 = \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = -\frac{1}{4} - \frac{\sqrt{3}}{4}i + \frac{\sqrt{3}}{4}i - \frac{3}{4} = -1$. ∎

Takeaway: This z sits at 30° on the unit circle; six steps of 30° reach 180° — the polar form (JEE Corner) sees this instantly.

Example 31: A collected drill

Express in the form $a + ib$: (i) $\frac{3 + 2i}{-2 - i}$ (ii) $(3 - 2i)(3 + 2i)(1 + i)$.

Solution:

Step 1 — (i) Rationalise. $\frac{(3 + 2i)(-2 + i)}{(-2 - i)(-2 + i)} = \frac{-6 + 3i - 4i + 2i^2}{4 + 1} = \frac{-8 - i}{5}$.

Step 2 — Standard form. $-\frac{8}{5} - \frac{1}{5}i$.

Step 3 — (ii) Spot the conjugate pair FIRST. $(3 - 2i)(3 + 2i) = 9 + 4 = 13$ — real, instantly.

Step 4 — Finish. $13(1 + i) = 13 + 13i$.

Takeaway: Spot conjugate pairs before expanding — they collapse to $|z|^2$ and halve the work.

Example 32: The grand mixed finale

Given $z_1 = 1 + i$ and $z_2 = 2 - 3i$, compute: (i) $z_1z_2$ (ii) $\frac{z_1}{z_2}$ (iii) $|z_1^4 z_2^2|$ (iv) $\overline{z_1 z_2} - \bar{z}_1\bar{z}_2$.

Solution:

Step 1 — (i) Expand. $(1 + i)(2 - 3i) = 2 - 3i + 2i - 3i^2 = 5 - i$.

Step 2 — (ii) Rationalise. $\frac{(1 + i)(2 + 3i)}{4 + 9} = \frac{2 + 3i + 2i - 3}{13} = \frac{-1 + 5i}{13}$.

Step 3 — (iii) Use modulus rules, no expansion. $|z_1^4z_2^2| = |z_1|^4|z_2|^2 = (\sqrt{2})^4(\sqrt{13})^2 = 4 \times 13 = 52$.

Step 4 — (iv) Use the conjugate-product rule. $\overline{z_1z_2} = \bar{z}_1\bar{z}_2$ identically, so the difference is 0 — no computation.

Takeaway: Four questions, four different tools — expansion, rationalisation, modulus rules, conjugate rules. Choosing the right tool is the whole exam skill for this chapter.