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hscscience Ext 2 · Y12
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Module 12 · L06 of 16 ~40 min ⚡ +90 XP available

Polar (Mod–Arg) Form

Rectangular form $z = a + bi$ is great for addition. Polar form $z = r(\cos\theta + i\sin\theta)$ is great for everything else, multiplication, division, powers, roots. Switching between the two is a one-page skill that unlocks de Moivre, $n$th roots of unity, and the geometric meaning of multiplication as "scale by $r$, rotate by $\theta$".

Today's hook, Convert $z = 1 + \sqrt{3}\,i$ to polar form. Before reading on, find its modulus and its argument (in radians, in the interval $(-\pi, \pi]$). Sketch the position vector, what angle does it make with the positive real axis? Compare your answer after card 05.
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Recall, your gut answer first

+5 XP warm-up For the point $(1, 1)$ in the Cartesian plane, its polar coordinates are $r = \sqrt{2}$, $\theta = \pi/4$.

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Recall, your gut answer first
+5 XP warm-up

For the point $(1, 1)$ in the Cartesian plane, its polar coordinates are $r = \sqrt{2}$, $\theta = \pi/4$. Before checking what does this say about the complex number $z = 1 + i$? Predict the polar form $z = r(\cos\theta + i\sin\theta)$ and verify by expanding.

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The two moves for polar conversion

+5 XP to read Every conversion between rectangular and polar form rewards two habits: compute first (the length), then work...

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The two moves for polar conversion
+5 XP to read

Every conversion between rectangular and polar form rewards two habits: compute $r = |z|$ first (the length), then work out $\theta = \arg z$ from the quadrant of the point (the angle). Skip the quadrant check and you will land in the wrong half-plane.

The length-then-angle reading: (1) compute $r = \sqrt{a^2 + b^2}$, (2) find a reference angle from $\tan\alpha = |b/a|$, (3) place $\theta$ in the correct quadrant of the Argand diagram and choose the value in the principal interval $(-\pi, \pi]$.

Rectangular: $z = a + bi$  ·  Polar: $z = r(\cos\theta + i\sin\theta)$

Polar conversion flow migrated to sequence layout
$r = \sqrt{a^2+b^2}, \;\; \tan\theta = b/a$ (with quadrant)
$r \geq 0$ always
The modulus is a length, it is never negative. If you ever get $r < 0$, you have made a sign error.
$\arg z \in (-\pi, \pi]$
The principal argument is the unique angle in $(-\pi, \pi]$. Other angles differ by integer multiples of $2\pi$ but are not principal.
Sketch first, then compute
Plot $(a, b)$ on the Argand diagram before reaching for $\tan^{-1}$. The picture tells you which quadrant $\theta$ lives in, the calculator does not.
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What you'll master
Know

Key facts

  • Polar (mod–arg) form: $z = r(\cos\theta + i\sin\theta)$ where $r = |z|$ and $\theta = \arg z$
  • Modulus: $r = \sqrt{a^2 + b^2}$
  • Argument: $\tan\theta = b/a$, with quadrant determined by the signs of $a$ and $b$
  • Principal argument: unique $\theta$ in $(-\pi, \pi]$
Understand

Concepts

  • Why polar form encodes geometry: $r$ is length, $\theta$ is direction
  • Why a single $\tan^{-1}$ value is not always the argument, quadrants matter
  • Why the argument is defined only up to multiples of $2\pi$ (and how the principal value fixes this)
Can do

Skills

  • Convert any complex number from rectangular to polar form (with correct quadrant)
  • Convert any polar form back to rectangular form by expanding $r\cos\theta + ir\sin\theta$
  • Read $r$ and $\theta$ directly from an Argand-diagram sketch
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Key terms
Polar (mod–arg) formThe representation $z = r(\cos\theta + i\sin\theta)$, in which $r$ encodes length and $\theta$ encodes direction.
Modulus $r = |z|$The distance from the origin to $z$ on the Argand diagram: $r = \sqrt{a^2 + b^2}$. Always non-negative.
Argument $\arg z$The angle (measured anticlockwise from the positive real axis) of the position vector of $z$. Defined only up to $2\pi$.
Principal argumentThe unique value of $\arg z$ in the interval $(-\pi, \pi]$. NESA uses this convention throughout Extension 2.
Reference angle $\alpha$The acute angle between the position vector and the real axis: $\tan\alpha = |b/a|$. Used as a stepping stone to the argument.
Rectangular formThe representation $z = a + bi$ in real and imaginary parts. Optimal for addition and subtraction.
MEX-N1NESA outcome (Complex Numbers I): uses the modulus–argument form $r(\cos\theta + i\sin\theta)$ to represent complex numbers and converts between this and rectangular form.
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Polar form: $z = r(\cos\theta + i\sin\theta)$

Any non-zero complex number $z = a + bi$ can be described by its position vector $\vec{OZ}$: a length $r = |z|$ and an angle $\theta = \arg z$ measured anticlockwise from the positive real axis.

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Polar form: $z = r(\cos\theta + i\sin\theta)$
core concept

Any non-zero complex number $z = a + bi$ can be described by its position vector $\vec{OZ}$: a length $r = |z|$ and an angle $\theta = \arg z$ measured anticlockwise from the positive real axis. Reading the components of $\vec{OZ}$ along the two axes gives:

$a = r\cos\theta$  and  $b = r\sin\theta$, so $z = a + bi = r\cos\theta + i\,r\sin\theta = r(\cos\theta + i\sin\theta)$.

  • $r = |z| = \sqrt{a^2 + b^2}$ the modulus is the distance from the origin to $z$.
  • $\theta = \arg z$ the angle of the position vector from the positive real axis. NESA convention: take $\theta \in (-\pi, \pi]$ (the principal argument).
  • $\tan\theta = b/a$ but only after checking the quadrant; otherwise the calculator will return an angle that is $\pi$ off.

Worked through the hook: $z = 1 + \sqrt{3}\,i$. Then $r = \sqrt{1 + 3} = 2$. The point $(1, \sqrt{3})$ is in quadrant 1 (both coordinates positive), so $\theta$ is acute with $\tan\theta = \sqrt{3}/1 = \sqrt{3}$, giving $\theta = \pi/3$. So $z = 2(\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3})$.

Connecting to later work. Polar form makes multiplication geometric: $r_1(\cos\theta_1 + i\sin\theta_1) \cdot r_2(\cos\theta_2 + i\sin\theta_2) = r_1 r_2 (\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2))$, moduli multiply, arguments add. This is the key to de Moivre's theorem and to roots of unity.

$z = r(\cos\theta + i\sin\theta)$, $r = |z|$, $\theta = \arg z$ · $a = r\cos\theta$, $b = r\sin\theta$ · $r = \sqrt{a^2 + b^2}$; $\tan\theta = b/a$ with quadrant check · Principal argument: $\theta \in (-\pi, \pi]$

Pause, copy polar form $z = r(\cos\theta+i\sin\theta)$, the component formulas $a = r\cos\theta$, $b = r\sin\theta$, and the principal argument range $(-\pi,\pi]$ into your book.

Quick check: The complex number $z = 1 + \sqrt{3}\,i$ in polar form is:

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Converting between rectangular and polar

We just saw that polar form $z = r(\cos\theta + i\sin\theta)$ uses $r = |z|$ and $\theta = \arg z$, with $a = r\cos\theta$ and $b = r\sin\theta$.

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Converting between rectangular and polar
core concept

We just saw that polar form $z = r(\cos\theta + i\sin\theta)$ uses $r = |z|$ and $\theta = \arg z$, with $a = r\cos\theta$ and $b = r\sin\theta$. That raises a question: how do we reliably convert in both directions, especially avoiding the $\arctan$ quadrant trap? This card answers it → sketch first, find the reference angle $\alpha = \arctan(|b/a|)$, then choose $\theta \in (-\pi,\pi]$ by quadrant.

Rectangular to polar ($a + bi \to r(\cos\theta + i\sin\theta)$):

  1. Sketch $(a, b)$ on the Argand diagram to see the quadrant.
  2. Compute $r = \sqrt{a^2 + b^2}$.
  3. Find the reference angle $\alpha$ from $\tan\alpha = |b/a|$.
  4. Place $\theta$ in the correct quadrant in $(-\pi, \pi]$: Q1 gives $\theta = \alpha$; Q2 gives $\theta = \pi - \alpha$; Q3 gives $\theta = -(\pi - \alpha)$; Q4 gives $\theta = -\alpha$.

Polar to rectangular ($r(\cos\theta + i\sin\theta) \to a + bi$): expand directly: $a = r\cos\theta$, $b = r\sin\theta$. No quadrant fuss needed.

$$r = \sqrt{a^2 + b^2}, \qquad \tan\theta = \tfrac{b}{a} \;\text{(with quadrant)}, \qquad a = r\cos\theta, \quad b = r\sin\theta$$
Common mistake. Using $\tan^{-1}(b/a)$ blindly. The calculator's $\tan^{-1}$ returns a value in $(-\pi/2, \pi/2)$, so for $z$ in quadrants 2 or 3 you must add or subtract $\pi$. A sketch is the safest fix.

Rect $\to$ polar: sketch, find $r$, find reference angle, choose quadrant, write $\theta \in (-\pi, \pi]$ · Polar $\to$ rect: $a = r\cos\theta$, $b = r\sin\theta$, then $z = a + bi$ · $\tan^{-1}$ alone gives Q1/Q4, manually fix for Q2/Q3 by adding/subtracting $\pi$ · Reference angle $\alpha$ is always acute and computed from $|b/a|$

Pause, copy the two-direction conversion procedure: rect→polar (sketch, $r$, reference angle, quadrant choice) and polar→rect ($a = r\cos\theta$, $b = r\sin\theta$, write $a+bi$), and the warning that $\arctan$ alone gives Q1/Q4 only into your book.

Did you get this? True or false: for $z = -1 - i$, the principal argument is $\arg z = -3\pi/4$.

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

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

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

PROBLEM 1 · QUADRANT 1 CONVERSION

Express $z = \sqrt{3} + i$ in polar form $r(\cos\theta + i\sin\theta)$ with $\theta \in (-\pi, \pi]$.

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Modulus: $r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2$.
Always compute $r$ first. The expression simplifies cleanly when $a$ and $b$ involve $\sqrt{3}$ or $1$, a signal that $\theta$ will be a standard angle.
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PROBLEM 2 · QUADRANT 3 CONVERSION (NEGATIVE ARGUMENT)

Express $z = -1 - \sqrt{3}\,i$ in polar form with principal argument $\theta \in (-\pi, \pi]$.

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Modulus: $r = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2$.
Signs disappear under squaring, the modulus calculation is the same in every quadrant.
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PROBLEM 3 · POLAR TO RECTANGULAR

Express $z = 4\bigl(\cos\tfrac{5\pi}{6} + i\sin\tfrac{5\pi}{6}\bigr)$ in rectangular form $a + bi$.

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Identify: $r = 4$, $\theta = 5\pi/6$. Sketch: $5\pi/6$ is in quadrant 2 (between $\pi/2$ and $\pi$), so the point is to the left and above.
A quick sketch tells you to expect a negative real part and a positive imaginary part, a sanity check for the final answer.
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Fill the gap: For $z = a + bi$, the modulus is $r = $ and the principal argument satisfies $\tan\theta = $ with the quadrant determined by the signs of $a$ and $b$.

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Misconceptions to fix · the 3 traps that cost marks

Misconceptions to fix · the 3 traps that cost marks

Misconceptions to fix · the 3 traps that cost marks

Trap 01
Blindly using $\tan^{-1}(b/a)$
The calculator's $\tan^{-1}$ returns angles in $(-\pi/2, \pi/2)$ only, i.e., quadrants 1 and 4. For points in Q2 or Q3 you must add or subtract $\pi$. Sketch first.
Trap 02
Negative modulus
$r = |z| \geq 0$ always. A "polar form" with $r < 0$ is not in NESA standard form. If your algebra produces a negative $r$, factor out the sign and rewrite.
Trap 03
Argument outside $(-\pi, \pi]$
NESA uses the principal argument: $\theta \in (-\pi, \pi]$. An answer like $\theta = 7\pi/4$ is not principal, subtract $2\pi$ to bring it into range ($7\pi/4 - 2\pi = -\pi/4$).

Did you get this? True or false: the polar form $z = -3(\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4})$ is in standard NESA polar form because it has $r$ and $\theta$ specified.

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Revisit your thinking

Activities · practice with the ideas

Activities · practice with the ideas

1

Convert to polar form with principal argument: (a) $z = 1 + i$, (b) $z = -1 + i$, (c) $z = -1 - i$, (d) $z = 1 - i$. Note the pattern of arguments.

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Find the modulus and principal argument of $z = -2\sqrt{3} + 2i$, then write in polar form.

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Convert these polar forms to rectangular form: (a) $3(\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2})$, (b) $5(\cos\pi + i\sin\pi)$, (c) $\sqrt{2}(\cos(-\tfrac{\pi}{4}) + i\sin(-\tfrac{\pi}{4}))$.

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The polar form of $z$ is $z = 6(\cos\tfrac{7\pi}{4} + i\sin\tfrac{7\pi}{4})$. Is this in NESA standard form? If not, rewrite with principal argument.

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Show that $z = r(\cos\theta + i\sin\theta)$ and $\bar z = r(\cos(-\theta) + i\sin(-\theta))$, i.e., conjugation negates the argument while preserving the modulus.

Odd one out: Three of these are valid polar forms of $z = -i$ (principal argument). Which one is NOT?

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Revisit your thinking

Earlier you converted $z = 1 + \sqrt{3}\,i$ to polar form, finding $r$ and the principal argument $\theta \in (-\pi, \pi]$.

The modulus is $r = \sqrt{1 + 3} = 2$. The point $(1, \sqrt{3})$ lies in quadrant 1, with reference angle $\tan\alpha = \sqrt{3}$, so $\alpha = \pi/3$. Since Q1, $\theta = \pi/3$. So $z = 2(\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3})$. The geometric picture is a vector of length 2 making an angle of $60°$ with the positive real axis, exactly what a sketch on the Argand diagram shows. Reading $r$ and $\theta$ off the diagram is faster than algebra once you trust the picture.

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Show what you have learned

Multiple choice, then short answer under exam conditions.

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Multiple choice
+5 XP per correct · +25 XP all-correct

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

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Short answer
ApplyBand 32 marks

Q1. Express $z = -2 + 2i$ in polar form $r(\cos\theta + i\sin\theta)$ with $\theta \in (-\pi, \pi]$. (2 marks)

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ApplyBand 43 marks

Q2. Convert $z = 4(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3})$ to rectangular form $a + bi$, showing exact values. (3 marks)

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AnalyseBand 53 marks

Q3. Let $z = -\sqrt{3} - i$. (a) Find $|z|$ and $\arg z$ (principal). (b) Hence write $z$ in polar form. (c) Verify by expanding the polar form back to rectangular. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. All have $r = \sqrt{2}$. (a) Q1: $\theta = \pi/4$. (b) Q2: $\theta = 3\pi/4$. (c) Q3: $\theta = -3\pi/4$. (d) Q4: $\theta = -\pi/4$. Arguments differ by $\pi/2$ as the point rotates around the origin.

2. $r = \sqrt{12 + 4} = 4$. Point $(-2\sqrt{3}, 2)$ in Q2; $\tan\alpha = 2/(2\sqrt{3}) = 1/\sqrt{3}$, so $\alpha = \pi/6$. Q2: $\theta = \pi - \pi/6 = 5\pi/6$. Polar: $z = 4(\cos\tfrac{5\pi}{6} + i\sin\tfrac{5\pi}{6})$.

3. (a) $3(0 + i \cdot 1) = 3i$. (b) $5(-1 + 0) = -5$. (c) $\sqrt{2}(\tfrac{\sqrt{2}}{2} - i\tfrac{\sqrt{2}}{2}) = 1 - i$.

4. $7\pi/4 \notin (-\pi, \pi]$. Subtract $2\pi$: $7\pi/4 - 2\pi = -\pi/4$. Standard form: $z = 6(\cos(-\tfrac{\pi}{4}) + i\sin(-\tfrac{\pi}{4}))$.

5. $\bar z = r\cos\theta - i\,r\sin\theta = r\cos(-\theta) + i\,r\sin(-\theta)$ using $\cos$ even, $\sin$ odd. So $|\bar z| = r$ and $\arg\bar z = -\theta$.

Q1 (2 marks): $r = \sqrt{4 + 4} = 2\sqrt{2}$ [1]. Point in Q2, reference angle $\pi/4$, so $\theta = \pi - \pi/4 = 3\pi/4$. Polar: $z = 2\sqrt{2}(\cos\tfrac{3\pi}{4} + i\sin\tfrac{3\pi}{4})$ [1].

Q2 (3 marks): $\cos(2\pi/3) = -1/2$ [1]; $\sin(2\pi/3) = \sqrt{3}/2$ [1]. Therefore $z = 4(-1/2 + i\sqrt{3}/2) = -2 + 2\sqrt{3}\,i$ [1].

Q3 (3 marks): (a) $|z| = \sqrt{3 + 1} = 2$; point in Q3 with reference angle $\pi/6$, so $\arg z = -(π - π/6) = -5\pi/6$ [1]. (b) $z = 2(\cos(-\tfrac{5\pi}{6}) + i\sin(-\tfrac{5\pi}{6}))$ [1]. (c) Expand: $2(-\sqrt{3}/2 - i \cdot 1/2) = -\sqrt{3} - i$ ✓ matches original [1].