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Module 12 · L04 of 16 ~40 min ⚡ +90 XP available

Modulus and Argument

Every complex number lives in a two-dimensional plane — so it has both a length and a direction. The modulus $|z| = \sqrt{a^2 + b^2}$ measures distance from the origin; the argument $\arg(z)$ measures the angle from the positive real axis. Together they unlock polar form, multiplication by rotation, and de Moivre's theorem. The catch: $\arctan(b/a)$ alone gives the wrong answer in half the plane — you must reason about quadrants.

Today's hook — Find the principal argument of $z = -1 + i$. Your calculator's $\arctan(-1)$ gives $-\pi/4$. Why is that wrong, and what's the correct answer? Sketch $z$ on the Argand plane before reading on. Compare your answer after card 06.

0/5QUESTS

Recall — your gut answer first

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For $z = 1 + \sqrt{3}\,i$, predict (without a calculator) the values of $|z|$ and $\arg(z)$. Sketch it on the Argand plane first — which quadrant? What angle does it make with the positive real axis?

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The two moves for $|z|$ and $\arg(z)$

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Every problem in this lesson rewards two habits: apply Pythagoras to get the modulus, then place $z$ in its quadrant before using arctan to find the argument. The arctan function alone returns angles in $(-\pi/2, \pi/2)$ — half the plane. Quadrant reasoning rescues the rest.

The sketch–modulus–quadrant reading: (1) plot $z$ on the Argand plane and identify its quadrant, (2) compute $|z| = \sqrt{a^2 + b^2}$, (3) compute the reference angle $\alpha = \arctan(|b|/|a|)$ then adjust by quadrant to land in $(-\pi, \pi]$.

$|z| = \sqrt{a^2 + b^2}$ · $\arg(z) \in (-\pi, \pi]$ (principal value) · quadrant determines the sign + offset

$z = |z|(\cos\theta + i\sin\theta), \;\; \theta = \arg(z)$

$|z|$ uses Pythagoras

$|z| = \sqrt{a^2 + b^2}$ — the distance from $z$ to the origin on the Argand plane. Always non-negative; zero only when $z = 0$.

arg is defined modulo $2\pi$

If $\theta$ is an argument of $z$, then so is $\theta + 2k\pi$ for any integer $k$. The principal argument is the unique value in $(-\pi, \pi]$.

arctan alone is wrong in Q2 and Q3

$\arctan(b/a)$ returns a value in $(-\pi/2, \pi/2)$. For points in quadrants 2 or 3 (Re < 0), add or subtract $\pi$ to reach the principal argument.

What you'll master

Know

Key facts

$|z| = \sqrt{a^2 + b^2}$ for $z = a + bi$
$|z|^2 = z\bar{z}$
$\arg(z)$ is defined modulo $2\pi$; principal value lies in $(-\pi, \pi]$
$\tan(\arg z) = b/a$ — but the result depends on the quadrant

Understand

Concepts

Why $|z|$ is geometrically the distance from $z$ to the origin
Why argument is multivalued and why we choose a principal value
Why $\arctan(b/a)$ alone fails in quadrants 2 and 3

Can do

Skills

Compute $|z|$ for any complex $z$
Find the principal argument using sketches and quadrant adjustment
Express any non-zero $z$ in mod-arg form $|z|(\cos\theta + i\sin\theta)$

Key terms

Modulus ($|z|$)For $z = a + bi$, $|z| = \sqrt{a^2 + b^2}$. Geometrically: the distance from $z$ to the origin on the Argand plane. Always $\geq 0$; zero iff $z = 0$.

Argand planeThe complex plane with horizontal axis $\text{Re}(z)$ and vertical axis $\text{Im}(z)$. The complex number $z = a + bi$ corresponds to the point $(a, b)$.

Argument ($\arg z$)The angle (measured anticlockwise from the positive real axis) to the vector from $0$ to $z$. Defined modulo $2\pi$ — i.e., $\arg z$ is a set $\{\theta + 2k\pi : k \in \mathbb{Z}\}$.

Principal argumentThe unique value $\theta$ with $\arg(z) = \theta$ and $\theta \in (-\pi, \pi]$. In some texts $[0, 2\pi)$ is used; HSC convention is $(-\pi, \pi]$.

Reference angle$\alpha = \arctan(|b|/|a|) \in [0, \pi/2]$. The principal argument is built from $\alpha$ by quadrant: Q1 = $\alpha$; Q2 = $\pi - \alpha$; Q3 = $-(\pi - \alpha)$; Q4 = $-\alpha$.

Modulus-argument (polar) form$z = r(\cos\theta + i\sin\theta)$ where $r = |z|$ and $\theta = \arg z$. Equivalent to the Cartesian form via $a = r\cos\theta$, $b = r\sin\theta$.

MEX-N1NESA outcome (Complex Numbers I): uses Cartesian and modulus-argument forms, including the geometric interpretation of $|z|$ and $\arg z$, and converts between forms.

The modulus $|z|$

core concept

For $z = a + bi$ with $a, b \in \mathbb{R}$, the modulus is

$$|z| = \sqrt{a^2 + b^2}$$

Geometrically, $|z|$ is the distance from the point $(a, b)$ to the origin on the Argand plane — a direct application of Pythagoras. Three consequences:

$|z| \geq 0$ with equality iff $z = 0$.
$|z|^2 = a^2 + b^2 = z\bar{z}$ (linking modulus back to the conjugate identity from Lesson 03).
$|z\,w| = |z|\,|w|$ — the modulus of a product is the product of the moduli (proof in later lessons).

Distance interpretation. For two complex numbers $z_1, z_2$, the quantity $|z_1 - z_2|$ is the Euclidean distance between them on the Argand plane. This is why $|z - z_0| = r$ describes a circle of radius $r$ centred at $z_0$.

Quick reality check. For $z = 3 + 4i$: $|z| = \sqrt{9 + 16} = 5$. For $z = -3 + 4i$: $|z| = \sqrt{9 + 16} = 5$ also — modulus is unchanged by reflection. The arguments differ; the moduli match.

$|z| = \sqrt{a^2 + b^2}$ (Pythagoras on the Argand plane) · $|z|^2 = z\bar{z}$ (from Lesson 03) · $|z| \geq 0$; zero iff $z = 0$ · $|z_1 - z_2|$ = distance between $z_1$ and $z_2$

Pause — copy $|z| = \sqrt{a^2+b^2}$, the identity $|z|^2 = z\bar{z}$, the non-negativity of modulus, and $|z_1-z_2|$ as a distance into your book.

Quick check: Which of the following has the largest modulus?

The argument and quadrant care

core concept

We just saw that $|z| = \sqrt{a^2+b^2}$ gives the distance from the origin, with $|z|^2 = z\bar{z}$ and $|z_1-z_2|$ being the distance between two points. That raises a question: how do we find the direction (argument) of $z$, and how does the quadrant affect the answer? This card answers it → use the reference angle $\alpha = \arctan(|b/a|)$, then adjust sign and magnitude by quadrant to get $\theta \in (-\pi, \pi]$.

The argument $\arg(z)$ is the angle $\theta$, measured anticlockwise from the positive real axis, to the line segment from $0$ to $z$. Because rotating by $2\pi$ returns to the same point, $\arg z$ is defined only modulo $2\pi$. The principal argument is the unique value in $(-\pi, \pi]$.

Why $\arctan(b/a)$ alone is not enough. The calculator's $\arctan$ function returns values in $(-\pi/2, \pi/2)$ — only half the plane. For $z = -1 + i$ in quadrant 2, $b/a = -1$ so $\arctan(-1) = -\pi/4$ — but the true angle to $(-1, 1)$ is in the upper-left, i.e., $3\pi/4$. The fix: identify the quadrant first, then build the argument from a reference angle.

$$\alpha = \arctan\!\left(\tfrac{|b|}{|a|}\right) \in [0, \tfrac{\pi}{2}], \qquad \arg(z) = \begin{cases} \alpha & \text{Q1} \\ \pi - \alpha & \text{Q2} \\ -(\pi - \alpha) & \text{Q3} \\ -\alpha & \text{Q4} \end{cases}$$

Boundary cases. Positive real: $\arg = 0$. Negative real: $\arg = \pi$. Positive imaginary: $\arg = \pi/2$. Negative imaginary: $\arg = -\pi/2$. The number $z = 0$ has no argument.

Sketch first, compute second. A two-second sketch on the Argand plane eliminates 90% of quadrant errors. Mark $(a, b)$, identify the quadrant, then choose the formula. Never trust $\arctan(b/a)$ blindly.

Principal argument $\theta \in (-\pi, \pi]$ · Reference angle $\alpha = \arctan(|b|/|a|) \in [0, \pi/2]$ · Q1: $\theta = \alpha$; Q2: $\theta = \pi - \alpha$; Q3: $\theta = -(\pi - \alpha)$; Q4: $\theta = -\alpha$ · Always sketch $z$ on the Argand plane before computing $\arg z$

Pause — copy the principal argument range $\theta \in (-\pi,\pi]$, the reference-angle formula $\alpha = \arctan(|b/a|)$, and the four-quadrant adjustment rules (Q1: $\theta=\alpha$; Q2: $\theta=\pi-\alpha$; Q3: $\theta=-(\pi-\alpha)$; Q4: $\theta=-\alpha$) into your book.

Did you get this? True or false: the principal argument of $z = -1 + i$ is $-\pi/4$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · MODULUS AND ARGUMENT (Q1)

Find the modulus and principal argument of $z = 1 + \sqrt{3}\,i$.

$a = 1$, $b = \sqrt{3}$. Both positive, so $z$ lies in quadrant 1. $|z| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$.

Always sketch — Q1 means both real and imaginary parts positive. Pythagoras gives the modulus directly.

PROBLEM 2 · QUADRANT 2 ADJUSTMENT

Find the principal argument of $z = -1 + i$.

$a = -1 < 0$, $b = 1 > 0$, so $z$ is in quadrant 2 (upper-left). Sketch: point $(-1, 1)$.

Identifying the quadrant FIRST is the rule. Without this, $\arctan(b/a) = \arctan(-1) = -\pi/4$ — a Q4 angle, completely wrong.

PROBLEM 3 · QUADRANT 3 ADJUSTMENT

Find $|z|$ and $\arg(z)$ for $z = -\sqrt{3} - i$, and write $z$ in mod-arg form.

$a = -\sqrt{3} < 0$, $b = -1 < 0$, so $z$ is in quadrant 3. $|z| = \sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$.

Q3: both parts negative (lower-left). Modulus uses squares, so signs do not matter for $|z|$.

Fill the gap: The principal argument of $z = a + bi$ lies in the interval $($ $,$ $]$, and the modulus equals $\sqrt{a^2 + b^2}$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Using $\arctan(b/a)$ without checking the quadrant

For $z = -1 - i$ (Q3), $b/a = 1$ so $\arctan(1) = \pi/4$ — but the true argument is $-3\pi/4$. The arctan function's range $(-\pi/2, \pi/2)$ covers only Q1 and Q4. Always sketch and adjust for Q2 and Q3.

Trap 02

Treating $|z|$ as if it could be negative

$|z| = \sqrt{a^2 + b^2}$ is always non-negative. A statement like "$|z| = -3$" is meaningless. If a derivation gives a negative value, you've made an error — likely forgetting to take absolute value or square root.

Trap 03

Confusing $(-\pi, \pi]$ with $[0, 2\pi)$

HSC convention for the principal argument is $(-\pi, \pi]$ — negative values allowed, $\pi$ included, $-\pi$ excluded. For $z = -1$, $\arg z = \pi$ (not $-\pi$). Some calculators or other texts use $[0, 2\pi)$; convert if needed.

Did you get this? True or false: for any non-zero complex $z$, the principal argument $\arg(z)$ can equal $-\pi$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Find $|z|$ for each: (a) $3 + 4i$, (b) $-5 + 12i$, (c) $-7$, (d) $\sqrt{2} - \sqrt{2}\,i$.

Find the principal argument of: (a) $1 + i$, (b) $-1 - i$, (c) $-2i$.

For $z = -\sqrt{3} + i$, find $|z|$ and $\arg(z)$, then write $z$ in mod-arg form.

Identify all $z$ satisfying $|z| = 3$. Describe the set geometrically.

Convert from mod-arg form to Cartesian form: $z = 4(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3})$.

Odd one out: Three of these complex numbers have the same modulus. Which one does NOT?

Revisit your thinking

At the start you tried $\arg(-1 + i)$ and (likely) the calculator returned $-\pi/4$ — which was the wrong half of the plane.

The point $(-1, 1)$ sits in quadrant 2. The correct procedure: compute the reference angle $\alpha = \arctan(|1|/|-1|) = \pi/4$, then apply the Q2 rule $\arg(z) = \pi - \alpha = 3\pi/4$. The principal argument is $3\pi/4$, not $-\pi/4$. Sketching $z$ on the Argand plane before computing is the single habit that eliminates quadrant errors. Practise it on every problem until it's automatic — and once $|z|$ and $\arg z$ are second nature, polar form, multiplication-as-rotation and de Moivre's theorem become straightforward extensions.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Find $|z|$ and the principal argument of $z = -2 + 2i$. (2 marks)

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ApplyBand 43 marks

Q2. Write $z = -\sqrt{3} - i$ in modulus-argument form. Justify the choice of quadrant. (3 marks)

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AnalyseBand 53 marks

Q3. On the Argand plane, sketch the set of complex numbers satisfying $|z - 2| = 3$. Describe the set in words and find any two specific points belonging to it. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. (a) $|3 + 4i| = \sqrt{9 + 16} = 5$. (b) $|-5 + 12i| = \sqrt{25 + 144} = 13$. (c) $|-7| = 7$. (d) $|\sqrt{2} - \sqrt{2}i| = \sqrt{2 + 2} = 2$.

2. (a) $1 + i$ in Q1: $\arg = \pi/4$. (b) $-1 - i$ in Q3: reference $\alpha = \pi/4$, so $\arg = -(\pi - \pi/4) = -3\pi/4$. (c) $-2i$ on negative imaginary axis: $\arg = -\pi/2$.

3. $|z| = \sqrt{3 + 1} = 2$. Q2: $\alpha = \arctan(1/\sqrt{3}) = \pi/6$, so $\arg z = \pi - \pi/6 = 5\pi/6$. Mod-arg form: $z = 2\!\left(\cos\tfrac{5\pi}{6} + i\sin\tfrac{5\pi}{6}\right)$.

4. The set $\{z : |z| = 3\}$ is the circle of radius 3 centred at the origin in the Argand plane.

5. $\cos(2\pi/3) = -1/2$, $\sin(2\pi/3) = \sqrt{3}/2$. So $z = 4(-1/2 + i\sqrt{3}/2) = -2 + 2\sqrt{3}\,i$.

Q1 (2 marks): $|z| = \sqrt{(-2)^2 + 2^2} = 2\sqrt{2}$ [1]. Q2: reference $\alpha = \arctan(2/2) = \pi/4$, so $\arg(z) = \pi - \pi/4 = 3\pi/4$ [1].

Q2 (3 marks): $a = -\sqrt{3} < 0$ and $b = -1 < 0$, so $z$ is in quadrant 3 [1]. $|z| = \sqrt{3 + 1} = 2$ and reference angle $\alpha = \arctan(1/\sqrt{3}) = \pi/6$ [1]. Q3 rule: $\arg(z) = -(\pi - \alpha) = -5\pi/6$, so $z = 2\!\left(\cos(-5\pi/6) + i\sin(-5\pi/6)\right)$ [1].

Q3 (3 marks): $|z - 2|$ is the distance from $z$ to the complex number $2$ (i.e., the point $(2, 0)$) [1]. So $|z - 2| = 3$ describes the circle of radius 3 centred at $(2, 0)$ [1]. Sample points: $z = 5$, $z = -1$, $z = 2 + 3i$, $z = 2 - 3i$ (any two valid) [1].

Boss battle · The Quadrant Navigator

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Five timed questions on modulus, principal argument and mod-arg form. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering quick modulus / argument questions. Lighter alternative to the boss.

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