9709 P3 §3.9 — Complex Numbers in Cartesian Form

💡 Lead In: When $x^2 + 1 = 0$ Has No Real Solution

The Scenario

You have solved quadratic equations for years. The equation $x^2 = 4$ has two solutions: $x = 2$ and $x = -2$. The equation $x^2 = 0$ has one solution: $x = 0$. These are easy because real squares are never negative: for every real number $x$, we have $x^2 \geq 0$.

But now consider $x^2 + 1 = 0$, that is, $x^2 = -1$. The quadratic formula gives $x = \dfrac{\pm\sqrt{-4}}{2}$, and the discriminant is negative. There is no real number whose square is $-1$. The equation is “broken” on the real number line.

The solution is bold: invent a new number. Call it $\mathrm{i}$, and declare that $\mathrm{i}^2 = -1$. Then $x^2 + 1 = (x + \mathrm{i})(x - \mathrm{i}) = 0$ gives $x = \mathrm{i}$ or $x = -\mathrm{i}$. The equation is no longer broken — we have just extended the number system from the one-dimensional real line to the two-dimensional complex plane.

The Question

We have invented a new number $\mathrm{i}$ with $\mathrm{i}^2 = -1$, and placed it on a new “imaginary” axis perpendicular to the real line. Every point in this plane is a complex number $z = x + \mathrm{i}y$, where $x$ and $y$ are real.

How do we describe these complex numbers precisely? What are their “parts”, their distance from the origin, their angle, their mirror image, and when are two of them equal?

💡 The Bridge (click to reveal)

A complex number has the form $z = x + \mathrm{i}y$ where $x, y \in \mathbb{R}$.

The real part is $\text{Re}(z) = x$ and the imaginary part is $\text{Im}(z) = y$ (a real number, not $y\mathrm{i}$).

The modulus $|z| = \sqrt{x^2 + y^2}$ is the distance from the origin (Pythagoras). The argument $\arg z$ is the angle from the positive real axis.

The complex conjugate is $z^* = x - \mathrm{i}y$ (reflect across the real axis).

Two complex numbers are equal if and only if both their real parts and their imaginary parts match: $z_1 = z_2 \iff \text{Re}(z_1) = \text{Re}(z_2)$ and $\text{Im}(z_1) = \text{Im}(z_2)$.

📖 Key Concepts: Describing Complex Numbers

A complex number is $z = x + \mathrm{i}y$, where $x, y \in \mathbb{R}$ and $\mathrm{i}^2 = -1$. We describe it using five key ideas: real part, imaginary part, modulus, argument, and conjugate.

Key relation 1 — Cartesian form:

Every complex number can be written as:

$$\boxed{\;z = x + \mathrm{i}y\;}$$

Here $x = \text{Re}(z)$ is the real part and $y = \text{Im}(z)$ is the imaginary part (note: $\text{Im}(z) = y$, a real number, not $y\mathrm{i}$).

Key relation 2 — Modulus and argument:

The modulus $|z|$ is the distance from the origin to the point $z$ on the Argand diagram (Pythagoras):

$$\boxed{\;|z| = \sqrt{x^2 + y^2}\;}$$

The argument $\arg z$ is the angle (in radians) from the positive real axis to the line $Oz$, measured counter-clockwise:

$$\boxed{\;\arg z = \tan^{-1}\!\left(\dfrac{y}{x}\right) \quad (\text{adjusted for quadrant})\;}$$

Key relation 3 — Complex conjugate:

The complex conjugate of $z = x + \mathrm{i}y$ is obtained by changing the sign of the imaginary part:

$$\boxed{\;z^* = x - \mathrm{i}y\;}$$

Geometrically, $z^*$ is the reflection of $z$ across the real axis. They have the same real part but opposite imaginary parts.

Key relation 4 — Equality of complex numbers:

Two complex numbers are equal if and only if both their real parts and their imaginary parts are equal:

$$\boxed{\;z_1 = z_2 \iff \text{Re}(z_1) = \text{Re}(z_2) \;\text{and}\; \text{Im}(z_1) = \text{Im}(z_2)\;}$$

This means $x_1 + \mathrm{i}y_1 = x_2 + \mathrm{i}y_2$ if and only if $x_1 = x_2$ and $y_1 = y_2$. One complex equation gives two real equations.

Group · Ch12 Activity 1: Explanation — Building the Complex Number System

The teacher gives a structured explanation of complex numbers, linking each new idea to something students already know. Students follow along and take notes.

Roadmap (30 seconds): “Today we will: (1) define a complex number $z = x + \mathrm{i}y$, (2) identify its real and imaginary parts, (3) compute its modulus and argument, (4) find its conjugate, and (5) decide when two complex numbers are equal. By the end you will be able to describe any complex number completely.”

Step 1 (Define $z = x + \mathrm{i}y$): Write $z = x + \mathrm{i}y$ on the board. “This is just like a coordinate $(x, y)$ on a plane — the real axis is horizontal, the imaginary axis is vertical. The number $\mathrm{i}$ lives at $(0, 1)$, and $\mathrm{i}^2 = -1$.” Example: $z = 3 + 2\mathrm{i}$ is the point 3 along, 2 up.

Step 2 (Real and imaginary parts): For $z = 3 + 2\mathrm{i}$, we have $\text{Re}(z) = 3$ and $\text{Im}(z) = 2$. “The imaginary part is the number $2$, not $2\mathrm{i}$ — the $\mathrm{i}$ is just the label for the axis.” Link: this is like reading off the $x$-coordinate and $y$-coordinate of a point.

Step 3 (Modulus $|z|$): “How far is $z = 3 + 2\mathrm{i}$ from the origin?” Use Pythagoras: $|z| = \sqrt{3^2 + 2^2} = \sqrt{13}$. Link: this is exactly the distance formula $\sqrt{x^2 + y^2}$ from coordinate geometry.

Step 4 (Argument $\arg z$): “What angle does the line $Oz$ make with the positive real axis?” Use $\tan^{-1}(y/x)$: $\arg(3+2\mathrm{i}) = \tan^{-1}(2/3) \approx 0.588$ radians. Link: this is the same as finding the direction angle of a vector. Caution: adjust for quadrant when $x < 0$.

Step 5 (Conjugate $z^*$): For $z = 3 + 2\mathrm{i}$, the conjugate is $z^* = 3 - 2\mathrm{i}$. “Flip the sign of the imaginary part — reflect across the real axis.” Analogy: like a mirror image in the real axis.

Step 6 (Equality): “When are $z_1 = a + \mathrm{i}b$ and $z_2 = c + \mathrm{i}d$ equal?” Both parts must match: $a = c$ AND $b = d$. “One complex equation gives two real equations — this is the key to solving problems with unknowns.” Example: if $z_1 = z_2$ and $z_1 = (p+1) + 3\mathrm{i}$, $z_2 = 4 + q\mathrm{i}$, then $p+1 = 4$ and $q = 3$, so $p = 3, q = 3$.

Immediate check (Think-Pair-Share, 90 seconds): “For $z = 5 - 12\mathrm{i}$, write down $\text{Re}(z)$, $\text{Im}(z)$, $|z|$, $z^*$.” Give 60 seconds of silence, then pairs compare for 30 seconds. Cold-call one pair. (Answers: $\text{Re} = 5$, $\text{Im} = -12$, $|z| = 13$, $z^* = 5 + 12\mathrm{i}$.)

Individual · Ch31 Activity 2: Learning from Experience — From Real to Complex

Reflect on your experience with real quadratic equations, then generalise to complex numbers. You will move through the reflection cycle: concrete experience → reflection → conceptualisation → active experimentation.

Concrete experience (2 minutes): On your paper, solve these three quadratic equations using the quadratic formula $z = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$z^2 - 5z + 6 = 0$ (discriminant $= 1$, two real roots)
$z^2 - 4z + 4 = 0$ (discriminant $= 0$, one repeated root)
$z^2 + 1 = 0$ (discriminant $= -4$, ???)

Write down what happens in the third case. Where does the formula “break”?

Reflective observation (2 minutes): Answer these prompts in your own words:

Why can you not take $\sqrt{-4}$ in the real number system?
What would you need to “invent” to make the formula work?
If $\mathrm{i}^2 = -1$, what is $\sqrt{-4}$? (Hint: $-4 = 4 \times (-1)$.)

Abstract conceptualisation (2 minutes): Now extend your knowledge. A complex number $z = x + \mathrm{i}y$ has a real part $x$ and imaginary part $y$. Complete this table on your paper:

$z$	Re($z$)	Im($z$)	$\|z\|$	$z^*$
$3 + 4\mathrm{i}$	?	?	?	?
$-2 + 5\mathrm{i}$	?	?	?	?
$7$	?	?	?	?
$-6\mathrm{i}$	?	?	?	?

Active experimentation (2 minutes): Now go back to the “broken” equation $z^2 + 1 = 0$. Using $\mathrm{i} = \sqrt{-1}$:

Write $\sqrt{-4} = \sqrt{4 \times (-1)} = 2\sqrt{-1} = 2\mathrm{i}$.
Apply the quadratic formula: $z = \dfrac{0 \pm \sqrt{-4}}{2} = \dfrac{\pm 2\mathrm{i}}{2} = \pm\mathrm{i}$.
Verify: does $\mathrm{i}^2 + 1 = -1 + 1 = 0$? Does $(-\mathrm{i})^2 + 1 = -1 + 1 = 0$?
Try one yourself: solve $z^2 + 4 = 0$. (Answer: $z = \pm 2\mathrm{i}$.)

👁 Sample table answers

$z$	Re($z$)	Im($z$)	$\|z\|$	$z^*$
$3 + 4\mathrm{i}$	3	4	5	$3 - 4\mathrm{i}$
$-2 + 5\mathrm{i}$	-2	5	$\sqrt{29}$	$-2 - 5\mathrm{i}$
$7$	7	0	7	7
$-6\mathrm{i}$	0	-6	6	$6\mathrm{i}$

Note: real numbers are complex numbers with $\text{Im}(z) = 0$. Purely imaginary numbers have $\text{Re}(z) = 0$.

Group · Guided Worked Example 1: Arithmetic in Cartesian Form Exercise 11A Q1

Express the following in the form $x + \mathrm{i}y$.

(i) $(8 + 6\mathrm{i}) + (6 + 4\mathrm{i})$ (ii) $(9 - 3\mathrm{i}) + (-4 + 5\mathrm{i})$ (iii) $(2 + 7\mathrm{i}) - (5 + 3\mathrm{i})$ (iv) $(5 - \mathrm{i}) - (6 - 2\mathrm{i})$

(v) $3(4 + 6\mathrm{i}) + 9(1 - 2\mathrm{i})$ (vi) $3\mathrm{i}(7 - 4\mathrm{i})$ (vii) $(9 + 2\mathrm{i})(1 + 3\mathrm{i})$ (viii) $(4 - \mathrm{i})(3 + 2\mathrm{i})$

(ix) $(7 + 3\mathrm{i})^2$ (x) $(8 + 6\mathrm{i})(8 - 6\mathrm{i})$ (xi) $(1 + 2\mathrm{i})(3 - 4\mathrm{i})(5 + 6\mathrm{i})$ (xii) $(3 + 2\mathrm{i})^3$

👁 Reveal Solution

Parts (i)–(iv): Addition and subtraction

(i) $(8+6\mathrm{i}) + (6+4\mathrm{i}) = (8+6) + (6+4)\mathrm{i} = \boxed{14 + 10\mathrm{i}}$

(ii) $(9-3\mathrm{i}) + (-4+5\mathrm{i}) = (9-4) + (-3+5)\mathrm{i} = \boxed{5 + 2\mathrm{i}}$

(iii) $(2+7\mathrm{i}) - (5+3\mathrm{i}) = (2-5) + (7-3)\mathrm{i} = \boxed{-3 + 4\mathrm{i}}$

(iv) $(5-\mathrm{i}) - (6-2\mathrm{i}) = (5-6) + (-1+2)\mathrm{i} = \boxed{-1 + \mathrm{i}}$

Parts (v)–(viii): Scalar multiply and multiply

(v) $3(4+6\mathrm{i}) + 9(1-2\mathrm{i}) = (12+18\mathrm{i}) + (9-18\mathrm{i}) = \boxed{21}$ (the imaginary parts cancel!)

(vi) $3\mathrm{i}(7-4\mathrm{i}) = 21\mathrm{i} - 12\mathrm{i}^2 = 21\mathrm{i} + 12 = \boxed{12 + 21\mathrm{i}}$

(vii) $(9+2\mathrm{i})(1+3\mathrm{i}) = 9 + 27\mathrm{i} + 2\mathrm{i} + 6\mathrm{i}^2 = 9 + 29\mathrm{i} - 6 = \boxed{3 + 29\mathrm{i}}$

(viii) $(4-\mathrm{i})(3+2\mathrm{i}) = 12 + 8\mathrm{i} - 3\mathrm{i} - 2\mathrm{i}^2 = 12 + 5\mathrm{i} + 2 = \boxed{14 + 5\mathrm{i}}$

Parts (ix)–(xii): Powers and products

(ix) $(7+3\mathrm{i})^2 = 49 + 42\mathrm{i} + 9\mathrm{i}^2 = 49 + 42\mathrm{i} - 9 = \boxed{40 + 42\mathrm{i}}$

(x) $(8+6\mathrm{i})(8-6\mathrm{i}) = 64 - 36\mathrm{i}^2 = 64 + 36 = \boxed{100}$ (a real number — this is $z \cdot z^* = |z|^2$!)

(xi) First: $(1+2\mathrm{i})(3-4\mathrm{i}) = 3 - 4\mathrm{i} + 6\mathrm{i} - 8\mathrm{i}^2 = 3 + 2\mathrm{i} + 8 = 11 + 2\mathrm{i}$. Then: $(11+2\mathrm{i})(5+6\mathrm{i}) = 55 + 66\mathrm{i} + 10\mathrm{i} + 12\mathrm{i}^2 = 55 + 76\mathrm{i} - 12 = \boxed{43 + 76\mathrm{i}}$

(xii) First: $(3+2\mathrm{i})^2 = 9 + 12\mathrm{i} + 4\mathrm{i}^2 = 5 + 12\mathrm{i}$. Then: $(5+12\mathrm{i})(3+2\mathrm{i}) = 15 + 10\mathrm{i} + 36\mathrm{i} + 24\mathrm{i}^2 = 15 + 46\mathrm{i} - 24 = \boxed{-9 + 46\mathrm{i}}$

Individual · Guided Worked Example 2: Quadratic Equations with Complex Roots Exercise 11A Q2

Solve each of the following equations, and check the roots in each case.

(i) $z^2 + 2z + 2 = 0$ (ii) $z^2 - 2z + 5 = 0$ (iii) $z^2 - 4z + 13 = 0$ (iv) $z^2 + 6z + 34 = 0$

(v) $4z^2 - 4z + 17 = 0$ (vi) $z^2 + 4z + 6 = 0$

👁 Reveal Solution

Part (i): $z^2 + 2z + 2 = 0$

$z = \dfrac{-2 \pm \sqrt{4 - 8}}{2} = \dfrac{-2 \pm \sqrt{-4}}{2} = \dfrac{-2 \pm 2\mathrm{i}}{2}$

$\boxed{z = -1 \pm \mathrm{i}}$

Check $z = -1 + \mathrm{i}$: $(-1+\mathrm{i})^2 + 2(-1+\mathrm{i}) + 2 = 1 - 2\mathrm{i} - 1 - 2 + 2\mathrm{i} + 2 = 0$ ✓

Part (iii): $z^2 - 4z + 13 = 0$

$z = \dfrac{4 \pm \sqrt{16 - 52}}{2} = \dfrac{4 \pm \sqrt{-36}}{2} = \dfrac{4 \pm 6\mathrm{i}}{2}$

$\boxed{z = 2 \pm 3\mathrm{i}}$

Check $z = 2 + 3\mathrm{i}$: $(2+3\mathrm{i})^2 - 4(2+3\mathrm{i}) + 13 = 4 + 12\mathrm{i} - 9 - 8 - 12\mathrm{i} + 13 = 0$ ✓

Part (v): $4z^2 - 4z + 17 = 0$

$z = \dfrac{4 \pm \sqrt{16 - 272}}{8} = \dfrac{4 \pm \sqrt{-256}}{8} = \dfrac{4 \pm 16\mathrm{i}}{8}$

$\boxed{z = \dfrac{1}{2} \pm 2\mathrm{i}}$

Check $z = \dfrac{1}{2} + 2\mathrm{i}$: $4\left(\dfrac{1}{4} + 2\mathrm{i} - 4\right) - 4\left(\dfrac{1}{2} + 2\mathrm{i}\right) + 17 = 4\left(\dfrac{1}{4} + 2\mathrm{i} - 4\right) - 2 - 8\mathrm{i} + 17$. Let's recompute: $\left(\dfrac{1}{2}+2\mathrm{i}\right)^2 = \dfrac{1}{4} + 2\mathrm{i} + 4\mathrm{i}^2 = \dfrac{1}{4} + 2\mathrm{i} - 4 = -\dfrac{15}{4} + 2\mathrm{i}$. Then $4\left(-\dfrac{15}{4} + 2\mathrm{i}\right) - 4\left(\dfrac{1}{2}+2\mathrm{i}\right) + 17 = -15 + 8\mathrm{i} - 2 - 8\mathrm{i} + 17 = 0$ ✓

Part (vi): $z^2 + 4z + 6 = 0$

$z = \dfrac{-4 \pm \sqrt{16 - 24}}{2} = \dfrac{-4 \pm \sqrt{-8}}{2} = \dfrac{-4 \pm 2\sqrt{2}\,\mathrm{i}}{2}$

$\boxed{z = -2 \pm \sqrt{2}\,\mathrm{i}}$

Remaining parts (ii) and (iv)

(ii) $z^2 - 2z + 5 = 0$: $z = \dfrac{2 \pm \sqrt{4-20}}{2} = \dfrac{2 \pm 4\mathrm{i}}{2} = \boxed{1 \pm 2\mathrm{i}}$

(iv) $z^2 + 6z + 34 = 0$: $z = \dfrac{-6 \pm \sqrt{36-136}}{2} = \dfrac{-6 \pm 10\mathrm{i}}{2} = \boxed{-3 \pm 5\mathrm{i}}$

Pattern: when the discriminant is negative, the roots are always complex conjugates of each other (same real part, opposite imaginary parts). This is because the $\pm$ in the formula only affects the imaginary part.

✏️ Concept Quiz

Score: 0 / 4

1. If $z = 5 - 3\mathrm{i}$, what are $\text{Re}(z)$ and $\text{Im}(z)$?

$\text{Re}(z) = -3$, $\text{Im}(z) = 5$

$\text{Re}(z) = 5$, $\text{Im}(z) = -3$

$\text{Re}(z) = 5$, $\text{Im}(z) = 3$

$\text{Re}(z) = 5\mathrm{i}$, $\text{Im}(z) = -3$

For $z = 5 - 3\mathrm{i}$, the real part is the number before $\mathrm{i}$, which is $5$, and the imaginary part is the coefficient of $\mathrm{i}$, which is $-3$ (a real number, not $-3\mathrm{i}$). The trap "Re = -3, Im = 5" swaps the two. The trap "Im = 3" forgets the minus sign. The trap "Re = 5i" incorrectly includes $\mathrm{i}$ in the real part.

2. If $z = 3 + 4\mathrm{i}$, what is $|z|$?

$7$

$25$

$5$

$\sqrt{7}$

$|z| = \sqrt{x^2 + y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. This is the distance from the origin to the point $(3, 4)$, a 3-4-5 Pythagorean triple. The trap $7$ adds $3+4$ instead of using Pythagoras. The trap $25$ forgets to take the square root. The trap $\sqrt{7}$ computes $\sqrt{3+4}$ instead of $\sqrt{3^2+4^2}$.

3. If $z = -2 + 5\mathrm{i}$, what is $z^*$?

$2 + 5\mathrm{i}$

$-2 + 5\mathrm{i}$

$2 - 5\mathrm{i}$

$-2 - 5\mathrm{i}$

The conjugate $z^* = x - \mathrm{i}y$ changes the sign of the imaginary part only. For $z = -2 + 5\mathrm{i}$, we have $x = -2$ and $y = 5$, so $z^* = -2 - 5\mathrm{i}$. The real part $-2$ stays the same; only the $+5\mathrm{i}$ becomes $-5\mathrm{i}$. The trap $2+5\mathrm{i}$ also flips the sign of the real part. The trap $2-5\mathrm{i}$ flips both signs. The trap $-2+5\mathrm{i}$ is just $z$ itself (no change).

4. If $z_1 = (a+2) + (b-3)\mathrm{i}$ and $z_2 = 5 + 7\mathrm{i}$ are equal, what are $a$ and $b$?

$a = 7$, $b = 4$

$a = 3$, $b = 10$

$a = 3$, $b = -4$

$a = 7$, $b = 10$

Equality of complex numbers requires both real and imaginary parts to match. Real parts: $a + 2 = 5$, so $a = 3$. Imaginary parts: $b - 3 = 7$, so $b = 10$. The trap $a=7, b=4$ solves $a+2=5$ as $a=7$ and $b-3=7$ as $b=4$ (arithmetic errors). The trap $a=3, b=-4$ gets $a$ right but solves $b-3=7$ as $b=-4$ (sign error). The trap $a=7, b=10$ gets $b$ right but $a$ wrong.

🎉

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📝 More Practice

Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.

Exercise 11A Q3 Standard

Solve each of the following equations.

(i) $z^2 - 4\mathrm{i}z - 4 = 0$ (ii) $z^2 - 2\mathrm{i}z + 15 = 0$ (iii) $z^2 - 2\mathrm{i}z - 2 = 0$

(iv) $z^2 + 6\mathrm{i}z - 13 = 0$ (v) $z^2 + 8\mathrm{i}z - 17 = 0$ (vi) $z^2 + \mathrm{i}z + 6 = 0$

💡 Hint

Use the quadratic formula $z = \dfrac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ with $A = 1$, $B = -4\mathrm{i}$ (for part (i)), $C = -4$.

The discriminant will involve $\mathrm{i}^2 = -1$, so simplify carefully.

For example, in (i): $B^2 = (-4\mathrm{i})^2 = 16\mathrm{i}^2 = -16$.

👁 Reveal Solution

Part (i): $z^2 - 4\mathrm{i}z - 4 = 0$

$z = \dfrac{4\mathrm{i} \pm \sqrt{(-4\mathrm{i})^2 - 4(1)(-4)}}{2} = \dfrac{4\mathrm{i} \pm \sqrt{-16 + 16}}{2} = \dfrac{4\mathrm{i} \pm 0}{2}$

$\boxed{z = 2\mathrm{i}}$ (repeated root)

Check: $(2\mathrm{i})^2 - 4\mathrm{i}(2\mathrm{i}) - 4 = -4 + 8 - 4 = 0$ ✓

Part (iii): $z^2 - 2\mathrm{i}z - 2 = 0$

$z = \dfrac{2\mathrm{i} \pm \sqrt{(-2\mathrm{i})^2 - 4(1)(-2)}}{2} = \dfrac{2\mathrm{i} \pm \sqrt{-4 + 8}}{2} = \dfrac{2\mathrm{i} \pm 2}{2}$

$\boxed{z = 1 + \mathrm{i} \quad \text{or} \quad z = -1 + \mathrm{i}}$

Part (vi): $z^2 + \mathrm{i}z + 6 = 0$

$z = \dfrac{-\mathrm{i} \pm \sqrt{\mathrm{i}^2 - 24}}{2} = \dfrac{-\mathrm{i} \pm \sqrt{-1 - 24}}{2} = \dfrac{-\mathrm{i} \pm \sqrt{-25}}{2} = \dfrac{-\mathrm{i} \pm 5\mathrm{i}}{2}$

$\boxed{z = 2\mathrm{i} \quad \text{or} \quad z = -3\mathrm{i}}$

Check $z = 2\mathrm{i}$: $(2\mathrm{i})^2 + \mathrm{i}(2\mathrm{i}) + 6 = -4 - 2 + 6 = 0$ ✓

Remaining parts (ii), (iv), (v)

(ii) $z^2 - 2\mathrm{i}z + 15 = 0$: $z = \dfrac{2\mathrm{i} \pm \sqrt{-4 - 60}}{2} = \dfrac{2\mathrm{i} \pm 8\mathrm{i}}{2} = \boxed{5\mathrm{i} \text{ or } -3\mathrm{i}}$

(iv) $z^2 + 6\mathrm{i}z - 13 = 0$: $z = \dfrac{-6\mathrm{i} \pm \sqrt{-36 + 52}}{2} = \dfrac{-6\mathrm{i} \pm 4}{2} = \boxed{2 - 3\mathrm{i} \text{ or } -2 - 3\mathrm{i}}$

(v) $z^2 + 8\mathrm{i}z - 17 = 0$: $z = \dfrac{-8\mathrm{i} \pm \sqrt{-64 + 68}}{2} = \dfrac{-8\mathrm{i} \pm 2}{2} = \boxed{1 - 4\mathrm{i} \text{ or } -1 - 4\mathrm{i}}$

Exercise 11A Q4 Medium

Given that $z = 2 + 3\mathrm{i}$ and $w = 6 - 4\mathrm{i}$, find the following.

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

(vii) $\text{Im}(z + z^*)$ (viii) $\text{Re}(w - w^*)$ (ix) $zz^* - ww^*$ (x) $(z^3)^*$ (xi) $(z^*)^3$ (xii) $zw^* - z^*w$

💡 Hint

Recall: $z^* = x - \mathrm{i}y$ (change the sign of the imaginary part).

$\text{Re}(z) = x$ and $\text{Im}(z) = y$ (real numbers, without $\mathrm{i}$).

For powers: first compute $z^2 = z \cdot z$, then $z^3 = z^2 \cdot z$, using $\mathrm{i}^2 = -1$.

For (xii): compute $zw^*$ and $z^*w$ separately, then subtract.

👁 Reveal Solution

Parts (i)–(vi): Basic operations

(i) $\text{Re}(z) = \boxed{2}$ (ii) $\text{Im}(w) = \boxed{-4}$

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

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

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

Parts (vii)–(ix): Properties of conjugates

(vii) $z + z^* = (2+3\mathrm{i}) + (2-3\mathrm{i}) = 4$, so $\text{Im}(z + z^*) = \boxed{0}$ (a complex number plus its conjugate is always real)

(viii) $w - w^* = (6-4\mathrm{i}) - (6+4\mathrm{i}) = -8\mathrm{i}$, so $\text{Re}(w - w^*) = \boxed{0}$ (a complex number minus its conjugate is always purely imaginary)

(ix) $zz^* = |z|^2 = 2^2 + 3^2 = 13$ and $ww^* = |w|^2 = 6^2 + 4^2 = 52$, so $zz^* - ww^* = 13 - 52 = \boxed{-39}$

Parts (x)–(xi): Powers and conjugates

First compute $z^3$. We have $z^2 = (2+3\mathrm{i})^2 = 4 + 12\mathrm{i} + 9\mathrm{i}^2 = -5 + 12\mathrm{i}$.

Then $z^3 = z^2 \cdot z = (-5+12\mathrm{i})(2+3\mathrm{i}) = -10 - 15\mathrm{i} + 24\mathrm{i} + 36\mathrm{i}^2 = -10 + 9\mathrm{i} - 36 = -46 + 9\mathrm{i}$.

(x) $(z^3)^* = \boxed{-46 - 9\mathrm{i}}$

(xi) $(z^*)^3$: $z^* = 2 - 3\mathrm{i}$, $(z^*)^2 = 4 - 12\mathrm{i} + 9\mathrm{i}^2 = -5 - 12\mathrm{i}$, $(z^*)^3 = (-5-12\mathrm{i})(2-3\mathrm{i}) = -10 + 15\mathrm{i} - 24\mathrm{i} + 36\mathrm{i}^2 = -10 - 9\mathrm{i} - 36 = \boxed{-46 - 9\mathrm{i}}$

Observation: $(z^3)^* = (z^*)^3$ — conjugating a power is the same as powering the conjugate.

Part (xii): $zw^* - z^*w$

$zw^* = (2+3\mathrm{i})(6+4\mathrm{i}) = 12 + 8\mathrm{i} + 18\mathrm{i} + 12\mathrm{i}^2 = 12 + 26\mathrm{i} - 12 = 26\mathrm{i}$

$z^*w = (2-3\mathrm{i})(6-4\mathrm{i}) = 12 - 8\mathrm{i} - 18\mathrm{i} + 12\mathrm{i}^2 = 12 - 26\mathrm{i} - 12 = -26\mathrm{i}$

$zw^* - z^*w = 26\mathrm{i} - (-26\mathrm{i}) = \boxed{52\mathrm{i}}$

Observation: $zw^* - z^*w$ is always purely imaginary.

Exercise 11A Q5 Medium

Let $z = x + \mathrm{i}y$. Show that $(z^*)^* = z$.

💡 Hint

Start with $z = x + \mathrm{i}y$. What is $z^*$?

Then take the conjugate of $z^*$. The conjugate changes the sign of the imaginary part.

Apply the definition twice and see what you get.

👁 Reveal Solution

Step 1: Write down $z$ and $z^*$

Let $z = x + \mathrm{i}y$ where $x, y \in \mathbb{R}$.

By definition, the conjugate changes the sign of the imaginary part:

$z^* = x - \mathrm{i}y$

Step 2: Take the conjugate of $z^*$

Now apply the conjugate to $z^* = x - \mathrm{i}y$. The real part is $x$ and the imaginary part is $-y$.

Changing the sign of the imaginary part: $-(-y) = y$.

$(z^*)^* = x + \mathrm{i}y = z$

Conclusion

$\boxed{(z^*)^* = z}$ $\blacksquare$

Geometric interpretation: reflecting a point across the real axis twice returns it to its original position.

Exercise 11A Q7 Challenging

Given that the complex numbers $z_1 = a^2 + (3 + 2b)\mathrm{i}$ and $z_2 = (5a - 4) + b^2\mathrm{i}$ are equal, find the possible values of $a$ and $b$. Hence list the possible pairs of complex numbers $z_1$ and $z_2$.

💡 Hint

Use the equality condition: $z_1 = z_2$ requires both real parts and imaginary parts to match.

Real parts: $a^2 = 5a - 4$.

Imaginary parts: $3 + 2b = b^2$.

Solve each equation separately (they are independent). Each is a quadratic in one variable.

👁 Reveal Solution

Step 1: Apply the equality condition

$z_1 = z_2$ requires $\text{Re}(z_1) = \text{Re}(z_2)$ and $\text{Im}(z_1) = \text{Im}(z_2)$:

$a^2 = 5a - 4$ (real parts) (1)

$3 + 2b = b^2$ (imaginary parts) (2)

Step 2: Solve equation (1) for $a$

$a^2 - 5a + 4 = 0 \Rightarrow (a - 1)(a - 4) = 0$

$a = 1$ or $a = 4$

Step 3: Solve equation (2) for $b$

$b^2 - 2b - 3 = 0 \Rightarrow (b - 3)(b + 1) = 0$

$b = 3$ or $b = -1$

Step 4: List all possible pairs $(a, b)$

Since $a$ and $b$ are independent, there are $2 \times 2 = 4$ possible pairs:

$(a, b) = (1, 3), \quad (1, -1), \quad (4, 3), \quad (4, -1)$

Step 5: Find the corresponding complex numbers

For each pair, compute $z_1 = z_2$:

$(a, b) = (1, 3)$: $z_1 = z_2 = 1 + 9\mathrm{i}$

$(a, b) = (1, -1)$: $z_1 = z_2 = 1 + 1\mathrm{i}$

$(a, b) = (4, 3)$: $z_1 = z_2 = 16 + 9\mathrm{i}$

$(a, b) = (4, -1)$: $z_1 = z_2 = 16 + 1\mathrm{i}$

$\boxed{\text{Pairs: } (1+9\mathrm{i}),\; (1+\mathrm{i}),\; (16+9\mathrm{i}),\; (16+\mathrm{i})}$

🎫

Exit Ticket

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

I can express a complex number in the form $z = x + \mathrm{i}y$ and identify its real part $\text{Re}(z) = x$ and imaginary part $\text{Im}(z) = y$ (a real number, not $y\mathrm{i}$). I can calculate the modulus $|z| = \sqrt{x^2 + y^2}$ (distance from origin) and the argument $\arg z$ (angle from the positive real axis) of a complex number, and interpret them geometrically on an Argand diagram. I can find the complex conjugate $z^* = x - \mathrm{i}y$ and understand it as a reflection of $z$ across the real axis. I can use the equality condition ($z_1 = z_2 \iff \text{Re}(z_1) = \text{Re}(z_2)$ and $\text{Im}(z_1) = \text{Im}(z_2)$) to solve problems with unknown real coefficients, obtaining two real equations from one complex equation.