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Module 12 · L07 of 16 ~40 min ⚡ +90 XP available

Multiplication in Polar Form

Multiplying complex numbers in Cartesian form is messy — four cross-products and a sign flip. In polar form it becomes effortless: multiply the moduli, add the arguments. Behind that one-line rule is the most powerful geometric idea in Module 12: complex multiplication is a rotation combined with a scaling. This lesson proves the rule from the addition formulas and shows you how to read every product as a transformation of the Argand plane.

Today's hook — Let $z_1 = 2(\cos 30^\circ + i\sin 30^\circ)$ and $z_2 = 3(\cos 60^\circ + i\sin 60^\circ)$. Without expanding into Cartesian form, predict $|z_1 z_2|$ and $\arg(z_1 z_2)$. Then describe geometrically what multiplying by $z_2$ does to the vector $z_1$ in the Argand plane. Check after card 05.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

Expand $(1 + i)(1 + i\sqrt 3)$ in Cartesian form. Then convert each factor to polar form and compute $|(1+i)(1+i\sqrt 3)|$ and $\arg((1+i)(1+i\sqrt 3))$ using whatever rule you remember. Do moduli multiply or add? Do arguments multiply or add?

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The two moves for multiplying in polar form

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To multiply two complex numbers given in polar form, two habits do all the work: multiply the moduli and add the arguments. That is the entire rule. Forgetting which operation pairs with which is the single most common error — anchor it now with the geometric picture.

The scale-then-rotate reading: multiplying $z_1$ by $z_2$ (1) scales the length of $z_1$ by the factor $|z_2|$, and (2) rotates $z_1$ anticlockwise about $O$ by the angle $\arg z_2$.

$z_1 z_2 = r_1 r_2 [\cos(\alpha+\beta) + i\sin(\alpha+\beta)]$

$|z_1 z_2| = |z_1||z_2|, \;\; \arg(z_1 z_2) = \arg z_1 + \arg z_2$

Moduli multiply

$|z_1 z_2| = |z_1| \cdot |z_2|$. The length of the product equals the product of the lengths — never the sum.

Arguments add

$\arg(z_1 z_2) = \arg z_1 + \arg z_2$ (mod $2\pi$). The angle of the product equals the sum of the angles.

Reduce to principal value

If $\alpha + \beta$ falls outside $(-\pi, \pi]$, add or subtract $2\pi$ to bring it back. The geometric direction is unchanged.

What you'll master

Know

Key facts

$z_1 z_2 = r_1 r_2[\cos(\alpha+\beta) + i\sin(\alpha+\beta)]$
$|z_1 z_2| = |z_1||z_2|$ (moduli multiply)
$\arg(z_1 z_2) = \arg z_1 + \arg z_2$ (arguments add, mod $2\pi$)
Multiplying by $z = r\,\text{cis}\,\theta$ is a rotation by $\theta$ and scaling by $r$

Understand

Concepts

Why the addition formulas for $\cos$ and $\sin$ produce the polar product rule
Why multiplication by $i$ is rotation by $\pi/2$ (and what $r=1$ guarantees)
Why argument may need adjustment by $\pm 2\pi$ to stay principal

Can do

Skills

Multiply two complex numbers given in polar form quickly and reliably
Convert between Cartesian and polar form to choose the easier multiplication
Describe a multiplication as a rotation-plus-scaling and predict the result geometrically

Key terms

Polar form$z = r(\cos\theta + i\sin\theta) = r\,\text{cis}\,\theta$, where $r = |z| \geq 0$ is the modulus and $\theta = \arg z$ is the argument.

Modulus ($|z|$)The distance from the origin to $z$ in the Argand plane. For $z = a + ib$, $|z| = \sqrt{a^2 + b^2}$. Moduli are non-negative reals.

Argument ($\arg z$)The anticlockwise angle from the positive real axis to the vector $\overrightarrow{Oz}$. The principal value lies in $(-\pi, \pi]$.

cis notation$\text{cis}\,\theta = \cos\theta + i\sin\theta$. Compact and useful for products: $\text{cis}\,\alpha \cdot \text{cis}\,\beta = \text{cis}(\alpha+\beta)$.

RotationMultiplying $z$ by $\text{cis}\,\theta$ (modulus 1) rotates $z$ anticlockwise about $O$ by the angle $\theta$ without changing $|z|$.

Scaling (dilation)Multiplying $z$ by a positive real $r$ scales the vector $\overrightarrow{Oz}$ by factor $r$ without changing the direction.

MEX-N1NESA outcome (Complex Numbers I): uses the polar form of complex numbers, including multiplication and division, and connects products to rotation and scaling in the Argand plane.

The polar product rule and its proof

core concept

Let $z_1 = r_1(\cos\alpha + i\sin\alpha)$ and $z_2 = r_2(\cos\beta + i\sin\beta)$. Then

$$z_1 z_2 = r_1 r_2 \big[\cos(\alpha+\beta) + i\sin(\alpha+\beta)\big].$$

Proof. Expand the product directly:

$z_1 z_2 = r_1 r_2 (\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta)$
$= r_1 r_2 \big[\cos\alpha\cos\beta + i\cos\alpha\sin\beta + i\sin\alpha\cos\beta + i^2 \sin\alpha\sin\beta\big]$
$= r_1 r_2 \big[(\cos\alpha\cos\beta - \sin\alpha\sin\beta) + i(\sin\alpha\cos\beta + \cos\alpha\sin\beta)\big]$
Now use the addition formulas: $\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$ and $\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$.
$\therefore z_1 z_2 = r_1 r_2 \big[\cos(\alpha+\beta) + i\sin(\alpha+\beta)\big]$. $\blacksquare$

Reading off: $|z_1 z_2| = r_1 r_2 = |z_1||z_2|$ and $\arg(z_1 z_2) = \alpha + \beta = \arg z_1 + \arg z_2$ (reduce mod $2\pi$ to the principal value if needed).

Connecting to the hook. $z_1 = 2\,\text{cis}\,30^\circ$ and $z_2 = 3\,\text{cis}\,60^\circ$. By the rule: $|z_1 z_2| = 2 \cdot 3 = 6$ and $\arg(z_1 z_2) = 30^\circ + 60^\circ = 90^\circ$. So $z_1 z_2 = 6\,\text{cis}\,90^\circ = 6i$. Geometrically, $z_2$ scales $z_1$ by a factor of 3 and rotates it $60^\circ$ anticlockwise — landing on the positive imaginary axis.

Polar product rule: $z_1 z_2 = r_1 r_2 \,\text{cis}(\alpha+\beta)$ · Moduli multiply: $|z_1 z_2| = |z_1||z_2|$ · Arguments add: $\arg(z_1 z_2) = \arg z_1 + \arg z_2$ (mod $2\pi$) · Proof uses $\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$ and $\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$

Pause — copy the polar product rule $z_1 z_2 = r_1 r_2\,\text{cis}(\alpha+\beta)$, the laws $|z_1 z_2| = |z_1||z_2|$ and $\arg(z_1 z_2) = \arg z_1 + \arg z_2$, and the proof strategy into your book.

Quick check: If $z_1 = 4\,\text{cis}\tfrac{\pi}{6}$ and $z_2 = 5\,\text{cis}\tfrac{\pi}{4}$, what is $z_1 z_2$ in polar form?

Geometric interpretation: rotation + scaling

core concept

We just saw the polar product rule $z_1 z_2 = r_1 r_2\,\text{cis}(\alpha+\beta)$ — moduli multiply, arguments add — proved by the $\cos/\sin$ addition formulas. That raises a question: what does multiplication by a fixed complex number actually do to every point in the plane? This card answers it → multiplying by $z_2 = r_2\,\text{cis}\beta$ rotates by $\beta$ and scales by $r_2$; multiplying by $i$ rotates $90°$ anticlockwise.

The product rule has a clean geometric reading. Multiplying $z_1$ by $z_2 = r_2 \,\text{cis}\,\beta$ does two things to the vector $\overrightarrow{Oz_1}$:

Rotation by the angle $\beta = \arg z_2$ anticlockwise about the origin.
Scaling (dilation from $O$) by the factor $r_2 = |z_2|$.

Special cases worth knowing by heart:

Multiplying by $i = \text{cis}\tfrac{\pi}{2}$: rotation by $90^\circ$ anticlockwise, no scaling.
Multiplying by $-1 = \text{cis}\,\pi$: rotation by $180^\circ$, no scaling (the point is sent to its diametrical opposite).
Multiplying by a positive real $r$: pure scaling, no rotation.
Multiplying by $\text{cis}\,\theta$ (any $\theta$, modulus 1): pure rotation by $\theta$, no scaling.

$$z \mapsto z \cdot \text{cis}\,\theta \;\;\text{is rotation by } \theta \text{ about } O.$$

Common mistake. Students often write $|z_1 z_2| = |z_1| + |z_2|$ or $\arg(z_1 z_2) = \arg z_1 \cdot \arg z_2$ when tired. The correct pairing is the opposite: moduli multiply, arguments add. Memorise this with the slogan: "operation flips when you move to polar form."

Multiplication by $z_2$ = rotation by $\arg z_2$ + scaling by $|z_2|$ · $\times i$ = rotate $90^\circ$ anticlockwise · $\times (-1)$ = rotate $180^\circ$ · $\times \text{cis}\,\theta$ (modulus 1) = pure rotation by $\theta$ · Slogan: moduli multiply, arguments add

Pause — copy the geometric interpretation (multiply by $z_2$ = rotate by $\arg z_2$ + scale by $|z_2|$), the special cases $\times i = 90°$ anticlockwise and $\times\text{cis}\theta$ = pure rotation, and the slogan "moduli multiply, arguments add" into your book.

Did you get this? True or false: multiplying any complex number $z$ by $i$ rotates the point representing $z$ by $90^\circ$ anticlockwise about the origin and leaves its modulus unchanged.

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

PROBLEM 1 · STRAIGHT APPLICATION

Given $z_1 = 2\,\text{cis}\tfrac{\pi}{3}$ and $z_2 = 5\,\text{cis}\tfrac{\pi}{6}$, find $z_1 z_2$ in polar form and in Cartesian form.

Apply the rule: $|z_1 z_2| = 2 \cdot 5 = 10$ and $\arg(z_1 z_2) = \tfrac{\pi}{3} + \tfrac{\pi}{6} = \tfrac{2\pi + \pi}{6} = \tfrac{\pi}{2}$.

Moduli multiply, arguments add — always pair the two operations like this when working in polar form.

PROBLEM 2 · CARTESIAN TO POLAR THEN MULTIPLY

Compute $(1 + i)(1 + i\sqrt 3)$ by first converting each factor to polar form. Give the answer in polar form, then in Cartesian form.

Convert $1+i$: $|1+i| = \sqrt{1^2+1^2} = \sqrt 2$; $\arg(1+i) = \tfrac{\pi}{4}$. So $1 + i = \sqrt 2 \,\text{cis}\tfrac{\pi}{4}$. Similarly $|1 + i\sqrt 3| = \sqrt{1 + 3} = 2$ and $\arg = \tan^{-1}\sqrt 3 = \tfrac{\pi}{3}$, so $1 + i\sqrt 3 = 2\,\text{cis}\tfrac{\pi}{3}$.

For points in the first quadrant, $\arg z = \tan^{-1}(b/a)$ directly. For other quadrants you must adjust by $\pm\pi$.

PROBLEM 3 · GEOMETRIC INTERPRETATION

Let $z = 3 + 4i$. Describe the transformation that takes $z$ to $iz$, and to $(1+i)z$. Compute both in Cartesian form.

$iz = i(3 + 4i) = 3i + 4i^2 = -4 + 3i$. Since $|i| = 1$, $\arg i = \tfrac{\pi}{2}$: multiplication by $i$ is a pure rotation $90^\circ$ anticlockwise about $O$, no scaling. Check: $|z| = 5 = |iz|$. $\checkmark$

The point $(3,4)$ rotates to $(-4,3)$ — confirming the $90^\circ$ rotation rule $(a,b) \mapsto (-b,a)$.

Fill the gap: If $z_1 = r_1\,\text{cis}\,\alpha$ and $z_2 = r_2\,\text{cis}\,\beta$, then $z_1 z_2 = $ $\,\text{cis}\,($ $)$. Moduli multiply, arguments add.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Swapping the operations

A common slip is writing $|z_1 z_2| = |z_1| + |z_2|$ or $\arg(z_1 z_2) = \arg z_1 \cdot \arg z_2$. The correct pairing is the opposite: moduli multiply, arguments add. Anchor it geometrically — lengths multiply, angles add.

Trap 02

Argument out of principal range

If $\alpha + \beta$ falls outside $(-\pi, \pi]$, the answer is geometrically correct but not in principal form. Adjust by $\pm 2\pi$. Example: $\arg z_1 = \tfrac{2\pi}{3}$, $\arg z_2 = \tfrac{2\pi}{3}$ gives $\tfrac{4\pi}{3}$ which should be written as $-\tfrac{2\pi}{3}$.

Trap 03

Reading the argument from the wrong quadrant

For a point not in the first quadrant, $\arg z = \tan^{-1}(b/a)$ gives the wrong answer. You must check the signs of $a$ and $b$ to place $z$ in the correct quadrant, then adjust the angle by $\pm \pi$. This pollutes every subsequent product calculation.

Did you get this? True or false: if $z_1 = 3\,\text{cis}\tfrac{2\pi}{3}$ and $z_2 = 4\,\text{cis}\tfrac{3\pi}{4}$, then $z_1 z_2 = 12\,\text{cis}\tfrac{17\pi}{12}$, and this argument is in principal form.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Compute $3\,\text{cis}\tfrac{\pi}{5} \cdot 4\,\text{cis}\tfrac{2\pi}{5}$ in polar form. Then express in Cartesian form.

Convert $\sqrt 3 + i$ to polar form. Then compute $(\sqrt 3 + i)^2$ using polar form and convert back to Cartesian. Verify by direct expansion.

Let $z = 2 - 2i$. Describe the geometric effect of multiplying $z$ by $-i$ and compute $-iz$.

If $z_1 = 5\,\text{cis}\tfrac{5\pi}{6}$ and $z_2 = 2\,\text{cis}\tfrac{3\pi}{4}$, find $z_1 z_2$ in polar form with argument in $(-\pi, \pi]$.

Prove using the polar product rule: $|z_1 z_2 z_3| = |z_1||z_2||z_3|$ and $\arg(z_1 z_2 z_3) = \arg z_1 + \arg z_2 + \arg z_3$ (mod $2\pi$).

Odd one out: Three of the following statements about $z_1 z_2$ (with $z_1 = r_1\,\text{cis}\,\alpha$, $z_2 = r_2\,\text{cis}\,\beta$) are correct. Which one is NOT?

Revisit your thinking

Earlier you predicted $|z_1 z_2|$ and $\arg(z_1 z_2)$ for $z_1 = 2\,\text{cis}\,30^\circ$ and $z_2 = 3\,\text{cis}\,60^\circ$.

By the polar product rule: $|z_1 z_2| = 2 \cdot 3 = 6$ and $\arg(z_1 z_2) = 30^\circ + 60^\circ = 90^\circ$, giving $z_1 z_2 = 6\,\text{cis}\,90^\circ = 6i$. Geometrically, multiplication by $z_2$ scales the vector $\overrightarrow{Oz_1}$ by factor 3 and rotates it $60^\circ$ anticlockwise — sending it from a length-2 vector at $30^\circ$ to a length-6 vector at $90^\circ$, exactly on the positive imaginary axis. Reading every product as "scale then rotate" is the single most useful skill in Module 12.

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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 the product $4\,\text{cis}\tfrac{\pi}{12} \cdot 3\,\text{cis}\tfrac{\pi}{4}$ in polar form, with argument in $(-\pi, \pi]$. (2 marks)

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ApplyBand 43 marks

Q2. Express $z = -1 + i\sqrt 3$ in polar form. Hence find $z^2$ in polar form and in Cartesian form. (3 marks)

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AnalyseBand 53 marks

Q3. Let $w = \text{cis}\tfrac{\pi}{6}$. Describe the geometric effect of multiplying any non-zero complex number $z$ by $w$. Hence find the image of the point $z = 2 + 2i$ under multiplication by $w$, giving your answer in Cartesian form (exact values). (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. Modulus $3 \cdot 4 = 12$; argument $\tfrac{\pi}{5} + \tfrac{2\pi}{5} = \tfrac{3\pi}{5}$. Polar: $12\,\text{cis}\tfrac{3\pi}{5}$. Cartesian: $12\cos\tfrac{3\pi}{5} + 12i\sin\tfrac{3\pi}{5} \approx -3.71 + 11.41 i$.

2. $\sqrt 3 + i = 2\,\text{cis}\tfrac{\pi}{6}$. $(\sqrt 3 + i)^2 = 4\,\text{cis}\tfrac{\pi}{3} = 4(\tfrac{1}{2} + i\tfrac{\sqrt 3}{2}) = 2 + 2i\sqrt 3$. Direct expansion: $(\sqrt 3)^2 + 2\sqrt 3 i + i^2 = 3 - 1 + 2i\sqrt 3 = 2 + 2i\sqrt 3$. $\checkmark$

3. $-i = \text{cis}(-\tfrac{\pi}{2})$: rotation $90^\circ$ clockwise, no scaling. $-iz = -i(2 - 2i) = -2i + 2i^2 = -2 - 2i$. Modulus check: $|z| = 2\sqrt 2 = |-iz|$. $\checkmark$

4. $|z_1 z_2| = 10$. Argument: $\tfrac{5\pi}{6} + \tfrac{3\pi}{4} = \tfrac{10\pi}{12} + \tfrac{9\pi}{12} = \tfrac{19\pi}{12}$, outside $(-\pi, \pi]$. Subtract $2\pi = \tfrac{24\pi}{12}$: $\tfrac{19\pi}{12} - \tfrac{24\pi}{12} = -\tfrac{5\pi}{12}$. Answer: $10\,\text{cis}(-\tfrac{5\pi}{12})$.

5. By the polar product rule applied twice (using associativity): $z_1 z_2 z_3 = (z_1 z_2) z_3$ has modulus $|z_1 z_2| \cdot |z_3| = |z_1||z_2||z_3|$, and argument $\arg(z_1 z_2) + \arg z_3 = \arg z_1 + \arg z_2 + \arg z_3$ (mod $2\pi$). $\blacksquare$

Q1 (2 marks): Moduli multiply: $4 \cdot 3 = 12$ [1]. Arguments add: $\tfrac{\pi}{12} + \tfrac{\pi}{4} = \tfrac{\pi}{12} + \tfrac{3\pi}{12} = \tfrac{4\pi}{12} = \tfrac{\pi}{3}$. Answer: $12\,\text{cis}\tfrac{\pi}{3}$ [1].

Q2 (3 marks): $|z| = \sqrt{(-1)^2 + (\sqrt 3)^2} = 2$; $z$ is in the second quadrant ($a < 0, b > 0$). Reference angle $\tan^{-1}\sqrt 3 = \tfrac{\pi}{3}$, so $\arg z = \pi - \tfrac{\pi}{3} = \tfrac{2\pi}{3}$, giving $z = 2\,\text{cis}\tfrac{2\pi}{3}$ [1]. Then $z^2 = 4\,\text{cis}\tfrac{4\pi}{3} = 4\,\text{cis}(-\tfrac{2\pi}{3})$ in principal form [1]. Cartesian: $4(\cos(-\tfrac{2\pi}{3}) + i\sin(-\tfrac{2\pi}{3})) = 4(-\tfrac{1}{2} - i\tfrac{\sqrt 3}{2}) = -2 - 2i\sqrt 3$ [1].

Q3 (3 marks): $|w| = 1$ and $\arg w = \tfrac{\pi}{6}$, so multiplication by $w$ is a pure rotation of $\tfrac{\pi}{6}$ anticlockwise about $O$ (no scaling) [1]. $z = 2 + 2i = 2\sqrt 2\,\text{cis}\tfrac{\pi}{4}$, so $wz = 2\sqrt 2\,\text{cis}(\tfrac{\pi}{4} + \tfrac{\pi}{6}) = 2\sqrt 2\,\text{cis}\tfrac{5\pi}{12}$ [1]. Cartesian: $wz = 2\sqrt 2 \cos\tfrac{5\pi}{12} + 2\sqrt 2 i\sin\tfrac{5\pi}{12}$. Using $\cos\tfrac{5\pi}{12} = \tfrac{\sqrt 6 - \sqrt 2}{4}$ and $\sin\tfrac{5\pi}{12} = \tfrac{\sqrt 6 + \sqrt 2}{4}$, $wz = \tfrac{\sqrt{12} - 2}{2} + i \tfrac{\sqrt{12} + 2}{2} = (\sqrt 3 - 1) + i(\sqrt 3 + 1)$ [1].

Boss battle · The Polar Multiplier

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Five timed questions on polar multiplication, the scale-rotate interpretation, and principal arguments. 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 polar-form questions. Lighter alternative to the boss.

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Module overview · Extension 2 · Checkpoint 1