9709 P3 §3.9 — Complex Numbers Cartesian Form Comprehensive Review

🧠 Rapid Recall

60-second brain dump

Quick check — pick the correct option in each of these six quick-fire questions.

Review 1 — For $z = 3 - 4\mathrm{i}$, the value of $|z|$ is:

$1$

$7$

$5$

$25$

$|z| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$, and $\arg(z) = -\arctan(4/3) \approx -0.927$ rad (in Q4).

Review 2 — Compute $(2 + \mathrm{i})(2 - \mathrm{i})$. The result is:

$3$

$4 + \mathrm{i}$

$2 + 2\mathrm{i}$

$5$

$(2+\mathrm{i})(2-\mathrm{i}) = 4 - \mathrm{i}^2 = 4 + 1 = 5 = |2+\mathrm{i}|^2$. This is the conjugate trick: $z \cdot z^* = |z|^2$.

Review 3 — Simplify $\dfrac{1 + \mathrm{i}}{1 - \mathrm{i}}$ to the form $x + \mathrm{i}y$. The result is:

$- \mathrm{i}$

$1$

$\mathrm{i}$

$2\mathrm{i}$

$\dfrac{1+\mathrm{i}}{1-\mathrm{i}} \cdot \dfrac{1+\mathrm{i}}{1+\mathrm{i}} = \dfrac{(1+\mathrm{i})^2}{2} = \dfrac{2\mathrm{i}}{2} = \mathrm{i}$.

Review 4 — If $z = 5 + 2\mathrm{i}$ is a root of a real polynomial, another root is:

$-5 + 2\mathrm{i}$

$5 - 2\mathrm{i}$

$-5 - 2\mathrm{i}$

$2 + 5\mathrm{i}$

By the conjugate root theorem for real-coefficient polynomials, the conjugate $z^* = 5 - 2\mathrm{i}$ is also a root.

Review 5 — $|z| = 4$ and $\arg(z) = \dfrac{\pi}{3}$. In Cartesian form, $z$ is:

$2\sqrt{3} + 2\mathrm{i}$

$2 + 2\sqrt{3}\,\mathrm{i}$

$4 + 4\mathrm{i}$

$2 - 2\sqrt{3}\,\mathrm{i}$

$z = 4\left(\cos\dfrac{\pi}{3} + \mathrm{i}\sin\dfrac{\pi}{3}\right) = 4\left(\dfrac{1}{2} + \mathrm{i}\dfrac{\sqrt{3}}{2}\right) = 2 + 2\sqrt{3}\,\mathrm{i}$.

Review 6 — Multiplying a complex number by $\mathrm{i}$ has the geometric effect of:

a reflection in the real axis

a rotation by $+\dfrac{\pi}{2}$ (90° anticlockwise) about the origin

a rotation by $\pi$ about the origin

a scaling by a factor of 2

Multiplication by $\mathrm{i}$ rotates the point by $+\dfrac{\pi}{2}$ (90° anticlockwise) about the origin, since $|\mathrm{i}| = 1$ and $\arg(\mathrm{i}) = \dfrac{\pi}{2}$.

⚠️ Misconception Trap: Three Classic Errors

A student is asked to simplify $z \cdot z^*$ for $z = 3 + 4\mathrm{i}$, divide $\dfrac{6 + 2\mathrm{i}}{1 + \mathrm{i}}$, and expand $(2 + \mathrm{i})^2$. Their working is shown below.

Error 1: $z \cdot z^* = |z| = 5$

Error 2: $\dfrac{6 + 2\mathrm{i}}{1 + \mathrm{i}} = 6 + 2\mathrm{i} - 1 - \mathrm{i} = 5 + \mathrm{i}$ (dividing term by term)

Error 3: $(2 + \mathrm{i})^2 = 4 + 4\mathrm{i} + \mathrm{i}^2 = 4 + 4\mathrm{i} + 1 = 5 + 4\mathrm{i}$

Can you spot all three mistakes?

👁 What went wrong?

Error 1: The correct identity is $z \cdot z^* = |z|^2$, not $|z|$. Here $z \cdot z^* = (3+4\mathrm{i})(3-4\mathrm{i}) = 9 + 16 = 25 = |z|^2$. The value $|z| = 5$ is the square root of $z \cdot z^*$, not $z \cdot z^*$ itself.

Error 2: You cannot divide complex numbers term by term. You must multiply numerator and denominator by the conjugate of the denominator: $\dfrac{(6+2\mathrm{i})(1-\mathrm{i})}{(1+\mathrm{i})(1-\mathrm{i})} = \dfrac{6 - 6\mathrm{i} + 2\mathrm{i} + 2}{2} = \dfrac{8 - 4\mathrm{i}}{2} = 4 - 2\mathrm{i}$.

Error 3: $\mathrm{i}^2 = -1$, not $+1$. The correct expansion is $(2+\mathrm{i})^2 = 4 + 4\mathrm{i} + \mathrm{i}^2 = 4 + 4\mathrm{i} - 1 = 3 + 4\mathrm{i}$.

📋 Review Points: All Week 10 Skills

1 Cartesian Form: $z = x + \mathrm{i}y$

A complex number $z = x + \mathrm{i}y$ has real part $\operatorname{Re}(z) = x$, imaginary part $\operatorname{Im}(z) = y$, modulus $|z| = \sqrt{x^2 + y^2}$, principal argument $\arg(z) = \arctan(y/x)$ (adjusted for quadrant), and conjugate $z^* = x - \mathrm{i}y$.

Key identities: $\operatorname{Re}(z) = \dfrac{z + z^*}{2}$, $\operatorname{Im}(z) = \dfrac{z - z^*}{2\mathrm{i}}$, $(z^*)^* = z$.

2 Operations: Addition, Subtraction, Multiplication, Division

Add/Subtract: combine real and imaginary parts separately.

Multiply: $(a + b\mathrm{i})(c + d\mathrm{i}) = (ac - bd) + (ad + bc)\mathrm{i}$.

Divide: multiply numerator and denominator by the conjugate of the denominator.

Power patterns: $z^2 = (a^2 - b^2) + 2ab\,\mathrm{i}$, $z^3 = (a^3 - 3ab^2) + (3a^2b - b^3)\mathrm{i}$.

3 Conjugate: $z \cdot z^* = |z|^2$

The product of a complex number and its conjugate is always real and non-negative: $z \cdot z^* = (x + \mathrm{i}y)(x - \mathrm{i}y) = x^2 + y^2 = |z|^2$. This is the master key for clearing complex denominators.

Division: $\dfrac{w}{z} = \dfrac{w \cdot z^*}{z \cdot z^*} = \dfrac{w \cdot z^*}{|z|^2}$.

4 Argand Diagram: Geometric Representation

A complex number $z = x + \mathrm{i}y$ is plotted as the point $(x, y)$ in the complex plane. The real axis is horizontal, the imaginary axis is vertical. The modulus $|z|$ is the distance from the origin, and $\arg(z)$ is the angle from the positive real axis.

Loci: $|z - z_0| = r$ is a circle centred at $z_0$ with radius $r$. $|z - z_1| = |z - z_2|$ is the perpendicular bisector of the segment joining $z_1$ and $z_2$.

5 Geometric Effects of Operations

Conjugation $z \mapsto z^*$: reflection in the real axis.

Addition $z \mapsto z + w$: translation by the vector representing $w$ (parallelogram law).

Multiplication $z \mapsto zw$: rotation by $\arg(w)$ and scaling by $|w|$. In particular, multiplying by $\mathrm{i}$ rotates by $+\dfrac{\pi}{2}$.

Modulus rules: $|zw| = |z| \cdot |w|$, $\left|\dfrac{z}{w}\right| = \dfrac{|z|}{|w|}$, $\arg(zw) = \arg(z) + \arg(w)$.

6 Polynomial Roots: Conjugate Pair Theorem

If a polynomial has real coefficients and $z = a + b\mathrm{i}$ (with $b \neq 0$) is a root, then $z^* = a - b\mathrm{i}$ is also a root. The conjugate pair forms a real quadratic factor: $(z - z_0)(z - z_0^*) = z^2 - 2\operatorname{Re}(z_0)z + |z_0|^2$.

Strategy: given one complex root, immediately write down its conjugate, form the quadratic factor, then use polynomial division to find the remaining factor(s).

Group · Ch18 Activity 1: Group Learning & Peer Tutoring — Comprehensive Review Challenge

In groups of 3–4, students tackle a multi-step review problem that draws on at least three of the six W10 skills. Each group member takes a role: algebra specialist, geometry specialist, checker, and presenter. Groups that finish early send their presenter to tutor a group that is stuck.

Think-First (90 seconds of silence): Pose the challenge and wait. “Given that $w = \dfrac{2 + 4\mathrm{i}}{1 + 3\mathrm{i}}$, find $|w^2|$ and $\arg(w^2)$, then determine whether $w$ lies inside, on, or outside the circle $|z| = 2$.” Students attempt it individually first — no discussion, no calculators.

Step 1 (Group work, 4 minutes): Compare individual attempts. Identify which skills are needed: (a) division using the conjugate, (b) modulus and argument, (c) modulus of a power, (d) geometric interpretation. Work together to produce a clean solution.

Step 2 (Peer tutoring, 2 minutes): Groups that have a verified answer send their presenter to a neighbouring group to explain their method. The receiving group asks one clarifying question.

Step 3 (Class share, 2 minutes): The teacher collects two different approaches on the board (e.g., direct computation vs. modulus-argument shortcut). The class votes on which is more efficient.

👁 Teacher Answer Key

$w = \dfrac{(2+4\mathrm{i})(1-3\mathrm{i})}{(1+3\mathrm{i})(1-3\mathrm{i})} = \dfrac{2 - 6\mathrm{i} + 4\mathrm{i} + 12}{10} = \dfrac{14 - 2\mathrm{i}}{10} = \dfrac{7}{5} - \dfrac{1}{5}\mathrm{i}$.

$|w| = \sqrt{\dfrac{49}{25} + \dfrac{1}{25}} = \sqrt{\dfrac{50}{25}} = \sqrt{2}$, so $|w^2| = |w|^2 = 2$.

$\arg(w) = \arctan\!\left(\dfrac{-1/5}{7/5}\right) = \arctan\!\left(-\dfrac{1}{7}\right) \approx -0.142$ rad, so $\arg(w^2) = 2\arg(w) \approx -0.284$ rad.

Since $|w| = \sqrt{2} \approx 1.414 < 2$, the point $w$ lies inside the circle $|z| = 2$.

Individual · Ch24 Activity 2: Confident Questioning — Vote & Defend

Every student commits to an answer before the solution is revealed. This ensures 100% participation and surfaces lingering misconceptions. After voting, students discuss in pairs, then the teacher guides a class debate before revealing the answer.

Question: “The complex number $z = -1 - \sqrt{3}\,\mathrm{i}$ is a root of $z^2 + (2\sqrt{3}\,\mathrm{i})z - 4 = 0$. What is the other root?”

Step 1 (Vote, 30 seconds): Four options are displayed. Every student silently commits to A, B, C, or D using fingers (1=A, 2=B, 3=C, 4=D) held to their chest so only the teacher can see.

A. $1 + \sqrt{3}\,\mathrm{i}$ (conjugate of the given root)

B. $1 - \sqrt{3}\,\mathrm{i}$ (the “other” root from the quadratic formula)

C. $-1 + \sqrt{3}\,\mathrm{i}$ (negate the imaginary part only)

D. $1 + \sqrt{3}\,\mathrm{i}$ and $-1 + \sqrt{3}\,\mathrm{i}$ are both roots (both conjugates)

Step 2 (Pair discussion, 90 seconds): Students compare with a neighbour and justify their choice. If they disagree, they must argue it out. The teacher listens for the key misconception: confusing “the other root of the quadratic” with “the conjugate of the given root”.

Step 3 (Class debate & reveal, 90 seconds): Collect the vote distribution. Ask a student who chose B to explain. The coefficients are NOT all real (the $z$ coefficient is $2\sqrt{3}\,\mathrm{i}$), so the conjugate root theorem does NOT apply. The correct answer is B: using the quadratic formula, $z = \dfrac{-2\sqrt{3}\,\mathrm{i} \pm \sqrt{(2\sqrt{3}\,\mathrm{i})^2 + 16}}{2} = \dfrac{-2\sqrt{3}\,\mathrm{i} \pm 2}{2} = -\sqrt{3}\,\mathrm{i} \pm 1$, giving $z = 1 - \sqrt{3}\,\mathrm{i}$ and $z = -1 - \sqrt{3}\,\mathrm{i}$.

Group · Guided Worked Example 1: Modulus, Argument & Loci (Past Paper) Exercise 11G Q8 · 9709/33 Nov 2015 Q9

(i) It is given that $(1 + 3\mathrm{i})w = 2 + 4\mathrm{i}$. Showing all necessary working, prove that the exact value of $|w^2|$ is 2 and find $\arg(w^2)$ correct to 3 significant figures.

(ii) On a single Argand diagram sketch the loci $|z| = 5$ and $|z - 5| = |z|$. Hence determine the complex numbers represented by points common to both loci, giving each answer in the form $re^{\mathrm{i}\theta}$.

👁 Reveal Solution

Step 1: Find $w$ by division

$w = \dfrac{2 + 4\mathrm{i}}{1 + 3\mathrm{i}} = \dfrac{(2 + 4\mathrm{i})(1 - 3\mathrm{i})}{(1 + 3\mathrm{i})(1 - 3\mathrm{i})} = \dfrac{2 - 6\mathrm{i} + 4\mathrm{i} - 12\mathrm{i}^2}{1 + 9} = \dfrac{2 - 2\mathrm{i} + 12}{10} = \dfrac{14 - 2\mathrm{i}}{10} = \dfrac{7}{5} - \dfrac{1}{5}\mathrm{i}$.

Step 2: Compute $|w|$ and $|w^2|$

$|w| = \sqrt{\left(\dfrac{7}{5}\right)^2 + \left(-\dfrac{1}{5}\right)^2} = \sqrt{\dfrac{49}{25} + \dfrac{1}{25}} = \sqrt{\dfrac{50}{25}} = \sqrt{2}$.

Therefore $|w^2| = |w|^2 = (\sqrt{2})^2 = \boxed{2}$. $\checkmark$

Step 3: Compute $\arg(w)$ and $\arg(w^2)$

$\arg(w) = \arctan\!\left(\dfrac{-1/5}{7/5}\right) = \arctan\!\left(-\dfrac{1}{7}\right) \approx -0.1419$ rad (Q4: real part positive, imaginary part negative).

$\arg(w^2) = 2\arg(w) \approx 2 \times (-0.1419) \approx \boxed{-0.284}$ rad (3 s.f.).

Step 4: Sketch the loci

$|z| = 5$: circle of radius 5 centred at the origin.

$|z - 5| = |z|$: the set of points equidistant from $5$ and $0$, which is the perpendicular bisector of the segment $[0, 5]$. This is the vertical line $\operatorname{Re}(z) = \dfrac{5}{2}$.

Step 5: Find the intersection points

Substitute $x = \dfrac{5}{2}$ into $x^2 + y^2 = 25$:

$\dfrac{25}{4} + y^2 = 25 \Rightarrow y^2 = 25 - \dfrac{25}{4} = \dfrac{75}{4} \Rightarrow y = \pm\dfrac{5\sqrt{3}}{2}$.

So the two intersection points are $z = \dfrac{5}{2} \pm \dfrac{5\sqrt{3}}{2}\mathrm{i}$.

Step 6: Convert to $re^{\mathrm{i}\theta}$ form

Both points have $|z| = 5$ (they lie on $|z| = 5$).

For $z = \dfrac{5}{2} + \dfrac{5\sqrt{3}}{2}\mathrm{i}$: $\theta = \arctan(\sqrt{3}) = \dfrac{\pi}{3}$.

For $z = \dfrac{5}{2} - \dfrac{5\sqrt{3}}{2}\mathrm{i}$: $\theta = -\arctan(\sqrt{3}) = -\dfrac{\pi}{3}$.

$$\boxed{z = 5e^{\mathrm{i}\pi/3} \quad \text{and} \quad z = 5e^{-\mathrm{i}\pi/3}}$$

Individual · Guided Worked Example 2: Quadratic with Complex Coefficient & Equilateral Triangle Exercise 11F Q5

(i) Solve the equation $z^2 + (2\sqrt{3}\mathrm{i})z - 4 = 0$, giving your answers in the form $x + \mathrm{i}y$, where $x$ and $y$ are real.

(ii) Sketch an Argand diagram showing the points representing the roots.

(iii) Find the modulus and argument of each root.

(iv) Show that the origin and the points representing the roots are the vertices of an equilateral triangle.

👁 Reveal Solution

Step 1: Apply the quadratic formula

$z = \dfrac{-(2\sqrt{3}\,\mathrm{i}) \pm \sqrt{(2\sqrt{3}\,\mathrm{i})^2 - 4(1)(-4)}}{2} = \dfrac{-2\sqrt{3}\,\mathrm{i} \pm \sqrt{-12 + 16}}{2} = \dfrac{-2\sqrt{3}\,\mathrm{i} \pm \sqrt{4}}{2} = \dfrac{-2\sqrt{3}\,\mathrm{i} \pm 2}{2}$.

Step 2: Write the two roots

$z_1 = \dfrac{2 - 2\sqrt{3}\,\mathrm{i}}{2} = \boxed{1 - \sqrt{3}\,\mathrm{i}}$.

$z_2 = \dfrac{-2 - 2\sqrt{3}\,\mathrm{i}}{2} = \boxed{-1 - \sqrt{3}\,\mathrm{i}}$.

Note: these are NOT conjugates — the coefficients are not all real.

Step 3: Sketch the Argand diagram

$z_1 = 1 - \sqrt{3}\,\mathrm{i}$ is at the point $(1, -\sqrt{3})$ — Q4.

$z_2 = -1 - \sqrt{3}\,\mathrm{i}$ is at the point $(-1, -\sqrt{3})$ — Q3.

Both points are in the lower half-plane, symmetric about the imaginary axis.

Step 4: Modulus and argument of each root

$z_1 = 1 - \sqrt{3}\,\mathrm{i}$: $|z_1| = \sqrt{1 + 3} = 2$, $\arg(z_1) = -\arctan(\sqrt{3}) = \boxed{-\dfrac{\pi}{3}}$ (Q4).

$z_2 = -1 - \sqrt{3}\,\mathrm{i}$: $|z_2| = \sqrt{1 + 3} = 2$, $\arg(z_2) = -\pi + \arctan(\sqrt{3}) = -\pi + \dfrac{\pi}{3} = \boxed{-\dfrac{2\pi}{3}}$ (Q3).

Step 5: Verify the equilateral triangle

The three vertices are: $O = (0, 0)$, $z_1 = (1, -\sqrt{3})$, $z_2 = (-1, -\sqrt{3})$.

Side 1: $|z_1| = |z_1 - 0| = 2$ (distance from $O$ to $z_1$).

Side 2: $|z_2| = |z_2 - 0| = 2$ (distance from $O$ to $z_2$).

Side 3: $|z_1 - z_2| = |(1 - (-1)) + (-\sqrt{3} - (-\sqrt{3}))\mathrm{i}| = |2| = 2$ (distance from $z_1$ to $z_2$).

All three sides equal 2, so the triangle is equilateral. $\boxed{\checkmark}$

✏️ Concept Quiz

Score: 0 / 4

1. For $z = 3 - 5\mathrm{i}$, what is $\operatorname{Re}(z^*)$?

$3$

$-3$

$5$

$-5$

$z^* = 3 + 5\mathrm{i}$ (conjugate flips the sign of the imaginary part only). So $\operatorname{Re}(z^*) = 3$. The real part is unchanged by conjugation. The trap $-3$ wrongly negates the real part. The traps $5$ and $-5$ confuse the real and imaginary parts.

2. Simplify $(2 + \mathrm{i})(2 - \mathrm{i})$.

$3$

$5$

$4 + 4\mathrm{i}$

$4 - 4\mathrm{i}$

$(2+\mathrm{i})(2-\mathrm{i}) = 4 - \mathrm{i}^2 = 4 - (-1) = 4 + 1 = 5$. This is the conjugate trick: $z \cdot z^* = |z|^2 = 2^2 + 1^2 = 5$. The trap $3$ computes $4 - 1$ (treating $\mathrm{i}^2 = +1$). The traps $4 \pm 4\mathrm{i}$ expand incorrectly as $4 \pm 4\mathrm{i}$.

3. What is the geometric effect of multiplying a complex number by $\mathrm{i}$?

Reflection in the real axis

Rotation by $+\dfrac{\pi}{2}$ (anticlockwise) about the origin

Rotation by $\pi$ about the origin

Scaling by a factor of 2

Multiplication by $\mathrm{i}$ scales by $|\mathrm{i}| = 1$ (no scaling) and rotates by $\arg(\mathrm{i}) = \dfrac{\pi}{2}$. So the effect is a pure $90°$ anticlockwise rotation about the origin. Reflection in the real axis is the effect of conjugation ($z \mapsto z^*$). Rotation by $\pi$ is the effect of multiplying by $-1$. Scaling by 2 is the effect of multiplying by 2.

4. If $z = 2 + \mathrm{i}$ is a root of a polynomial with real coefficients, which of the following must also be a root?

$-2 + \mathrm{i}$

$-2 - \mathrm{i}$

$2 - \mathrm{i}$

$1 + 2\mathrm{i}$

By the conjugate root theorem, if a real-coefficient polynomial has root $z = a + b\mathrm{i}$, then $z^* = a - b\mathrm{i}$ is also a root. Here $z = 2 + \mathrm{i}$, so $z^* = 2 - \mathrm{i}$ must also be a root. The trap $-2+\mathrm{i}$ negates the real part. The trap $-2-\mathrm{i}$ negates both parts. The trap $1+2\mathrm{i}$ swaps the real and imaginary parts.

🎉

You scored 0 / 4

✏️ More Practice

Exercise 11A Q4 [adapted] Standard

Given that $z = 2 + 3\mathrm{i}$ and $w = 6 - 4\mathrm{i}$, find the following. [Attempt parts (i) to (vi) first.]

(i) $\operatorname{Re}(z)$ (ii) $\operatorname{Im}(w)$ (iii) $z^*$ (iv) $w^*$ (v) $z^* + w^*$ (vi) $z^* - w^*$

💡 Hint

Recall: $\operatorname{Re}(x + \mathrm{i}y) = x$, $\operatorname{Im}(x + \mathrm{i}y) = y$, and the conjugate of $x + \mathrm{i}y$ is $x - \mathrm{i}y$.

For addition/subtraction of conjugates: add/subtract the real parts and imaginary parts separately.

👁 Reveal Solution

Parts (i)–(iv): Basic properties

(i) $\operatorname{Re}(z) = \operatorname{Re}(2 + 3\mathrm{i}) = \boxed{2}$.

(ii) $\operatorname{Im}(w) = \operatorname{Im}(6 - 4\mathrm{i}) = \boxed{-4}$.

(iii) $z^* = (2 + 3\mathrm{i})^* = \boxed{2 - 3\mathrm{i}}$.

(iv) $w^* = (6 - 4\mathrm{i})^* = \boxed{6 + 4\mathrm{i}}$.

Parts (v)–(vi): Conjugate arithmetic

(v) $z^* + w^* = (2 - 3\mathrm{i}) + (6 + 4\mathrm{i}) = \boxed{8 + \mathrm{i}}$.

(vi) $z^* - w^* = (2 - 3\mathrm{i}) - (6 + 4\mathrm{i}) = (2 - 6) + (-3 - 4)\mathrm{i} = \boxed{-4 - 7\mathrm{i}}$.

Exercise 11B Q5 Medium

Find all the complex numbers $z$ for which $z^2 = 2z^*$

💡 Hint 1

Let $z = x + \mathrm{i}y$ where $x, y$ are real. Then $z^* = x - \mathrm{i}y$.

Compute $z^2 = (x + \mathrm{i}y)^2 = (x^2 - y^2) + 2xy\,\mathrm{i}$.

Set $z^2 = 2z^*$ and equate real and imaginary parts to get two equations in $x$ and $y$.

💡 Hint 2

Equating: $x^2 - y^2 = 2x$ (real parts) and $2xy = -2y$ (imaginary parts).

From the second equation: $2y(x + 1) = 0$, so either $y = 0$ or $x = -1$.

Consider each case separately and substitute back into the first equation.

👁 Reveal Solution

Step 1: Set up equations

Let $z = x + \mathrm{i}y$. Then $z^2 = (x^2 - y^2) + 2xy\,\mathrm{i}$ and $2z^* = 2x - 2y\,\mathrm{i}$.

Equating real and imaginary parts: $x^2 - y^2 = 2x$ … (1), $2xy = -2y$ … (2).

Step 2: Case 1 — $y = 0$

From (1): $x^2 = 2x \Rightarrow x(x - 2) = 0 \Rightarrow x = 0$ or $x = 2$.

Solutions: $z = 0$ and $z = 2$.

Step 3: Case 2 — $x = -1$

From (1): $1 - y^2 = -2 \Rightarrow y^2 = 3 \Rightarrow y = \pm\sqrt{3}$.

Solutions: $z = -1 + \sqrt{3}\,\mathrm{i}$ and $z = -1 - \sqrt{3}\,\mathrm{i}$.

Final Answer

$$\boxed{z = 0, \quad z = 2, \quad z = -1 + \sqrt{3}\,\mathrm{i}, \quad z = -1 - \sqrt{3}\,\mathrm{i}}$$

Verification for $z = -1 + \sqrt{3}\,\mathrm{i}$: $z^2 = 1 - 2\sqrt{3}\,\mathrm{i} - 3 = -2 - 2\sqrt{3}\,\mathrm{i}$ and $2z^* = 2(-1 - \sqrt{3}\,\mathrm{i}) = -2 - 2\sqrt{3}\,\mathrm{i}$. $\checkmark$

Exercise 11C Q4 Challenging

Given that $z = 3 + 4\mathrm{i}$ and $w = 5 - 12\mathrm{i}$, find the following.

(i) $|z|$ (ii) $|w|$ (iii) $|zw|$ (iv) $\left|\dfrac{z}{w}\right|$ (v) $\left|\dfrac{w}{z}\right|$

What do you notice?

💡 Hint

For parts (i) and (ii): $|a + b\mathrm{i}| = \sqrt{a^2 + b^2}$.

For part (iii): compute $zw$ first, then find $|zw|$. Alternatively, use the identity $|zw| = |z| \cdot |w|$.

For parts (iv) and (v): use $\left|\dfrac{z}{w}\right| = \dfrac{|z|}{|w|}$.

The “what do you notice?” question is asking you to spot the pattern in these identities.

👁 Reveal Solution

Parts (i) and (ii): Individual moduli

(i) $|z| = |3 + 4\mathrm{i}| = \sqrt{9 + 16} = \sqrt{25} = \boxed{5}$.

(ii) $|w| = |5 - 12\mathrm{i}| = \sqrt{25 + 144} = \sqrt{169} = \boxed{13}$.

Part (iii): Modulus of a product

$zw = (3 + 4\mathrm{i})(5 - 12\mathrm{i}) = 15 - 36\mathrm{i} + 20\mathrm{i} - 48\mathrm{i}^2 = 15 - 16\mathrm{i} + 48 = 63 - 16\mathrm{i}$.

$|zw| = \sqrt{63^2 + (-16)^2} = \sqrt{3969 + 256} = \sqrt{4225} = \boxed{65}$.

Check: $|z| \cdot |w| = 5 \times 13 = 65$. $\checkmark$

Parts (iv) and (v): Modulus of a quotient

(iv) $\left|\dfrac{z}{w}\right| = \dfrac{|z|}{|w|} = \dfrac{5}{13} = \boxed{\dfrac{5}{13}}$.

(v) $\left|\dfrac{w}{z}\right| = \dfrac{|w|}{|z|} = \dfrac{13}{5} = \boxed{\dfrac{13}{5}}$.

What we notice

$$|zw| = |z| \cdot |w|, \qquad \left|\dfrac{z}{w}\right| = \dfrac{|z|}{|w|}.$$

The modulus of a product is the product of the moduli, and the modulus of a quotient is the quotient of the moduli. This mirrors the rules for exponents: $\log(zw) = \log z + \log w$.

Exercise 11H Q12 Advanced

Given that $z = -2 + \mathrm{i}$ is a root of the equation $z^4 + az^3 + bz^2 + 10z + 25 = 0$, find the values of $a$ and $b$, and solve the equation.

💡 Hint 1

The coefficients are real ($a, b$ are real numbers), so the conjugate root theorem applies.

If $z = -2 + \mathrm{i}$ is a root, then $z^* = -2 - \mathrm{i}$ is also a root.

Form the quadratic factor: $(z - (-2 + \mathrm{i}))(z - (-2 - \mathrm{i})) = (z + 2)^2 + 1 = z^2 + 4z + 5$.

💡 Hint 2

Since $z^4 + az^3 + bz^2 + 10z + 25 = (z^2 + 4z + 5)(z^2 + cz + d)$ for some real $c, d$, expand the right side and match coefficients.

The constant term gives $5d = 25$, so $d = 5$.

Then match the $z$ coefficient: $20 + 5c = 10$ to find $c$.

👁 Reveal Solution

Step 1: Identify the conjugate root

Since the coefficients are real and $z = -2 + \mathrm{i}$ is a root, $z^* = -2 - \mathrm{i}$ is also a root.

The quadratic factor is: $(z + 2 - \mathrm{i})(z + 2 + \mathrm{i}) = (z + 2)^2 - \mathrm{i}^2 = (z + 2)^2 + 1 = z^2 + 4z + 5$.

Step 2: Set up the factorisation

$z^4 + az^3 + bz^2 + 10z + 25 = (z^2 + 4z + 5)(z^2 + cz + d)$.

Expanding the right side: $z^4 + (c + 4)z^3 + (d + 4c + 5)z^2 + (4d + 5c)z + 5d$.

Step 3: Match coefficients

Constant: $5d = 25 \Rightarrow d = 5$.

$z$ term: $4d + 5c = 10 \Rightarrow 20 + 5c = 10 \Rightarrow c = -2$.

$z^3$ term: $a = c + 4 = -2 + 4 = \boxed{2}$.

$z^2$ term: $b = d + 4c + 5 = 5 - 8 + 5 = \boxed{2}$.

Step 4: Solve the remaining quadratic

The other factor is $z^2 - 2z + 5 = 0$.

$z = \dfrac{2 \pm \sqrt{4 - 20}}{2} = \dfrac{2 \pm \sqrt{-16}}{2} = \dfrac{2 \pm 4\mathrm{i}}{2} = 1 \pm 2\mathrm{i}$.

Final Answer

$$a = \boxed{2}, \quad b = \boxed{2}$$

$$\text{All four roots: } z = -2 + \mathrm{i}, \quad z = -2 - \mathrm{i}, \quad z = 1 + 2\mathrm{i}, \quad z = 1 - 2\mathrm{i}$$

Verification: $(-2+\mathrm{i})^2 = 3 - 4\mathrm{i}$, $(-2+\mathrm{i})^3 = -2 + 11\mathrm{i}$, $(-2+\mathrm{i})^4 = -7 - 24\mathrm{i}$.

$z^4 + 2z^3 + 2z^2 + 10z + 25 = (-7-24\mathrm{i}) + 2(-2+11\mathrm{i}) + 2(3-4\mathrm{i}) + 10(-2+\mathrm{i}) + 25 = (-7-4+6-20+25) + (-24+22-8+10)\mathrm{i} = 0 + 0\mathrm{i}$. $\checkmark$

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Exit Ticket

Before you leave, tick the objectives you feel confident you can achieve. Be honest — this guides next week's review.

I can perform all four operations on complex numbers in Cartesian form (add, subtract, multiply, divide using the conjugate) and simplify the result to $x + \mathrm{i}y$. I can use the conjugate identity $z \cdot z^* = |z|^2$ to clear complex denominators and compute $|zw| = |z||w|$ and $|z/w| = |z|/|w|$. I can interpret complex operations geometrically on an Argand diagram: conjugation as reflection, addition as translation, multiplication as rotation and scaling. I can apply the conjugate root theorem to solve polynomial equations with real coefficients, form quadratic factors, and find all roots including complex ones.