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.
+5 XP warm-up Expand in Cartesian form.
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?
+5 XP to read To multiply two complex numbers given in polar form, two habits do all the work: multiply the moduli and add the...
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 $\arg z_2$.
$z_1 z_2 = r_1 r_2 [\cos(\alpha+\beta) + i\sin(\alpha+\beta)]$
Let $z_1 = r_1(\cos\alpha + i\sin\alpha)$ and $z_2 = r_2(\cos\beta + i\sin\beta)$.
Let $z_1 = r_1(\cos\alpha + i\sin\alpha)$ and $z_2 = r_2(\cos\beta + i\sin\beta)$. Then
Proof. Expand the product directly:
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).
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?
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.
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 $\overrightarrow{Oz_1}$:
Special cases worth knowing by heart:
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
Worked examples · 3 in a row, reveal as you go
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.
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.
Let $z = 3 + 4i$. Describe the transformation that takes $z$ to $iz$, and to $(1+i)z$. Compute both in Cartesian form.
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
Misconceptions to fix · the 3 traps that cost marks
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.
Activities · practice with the ideas
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?
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.
Multiple choice, then short answer under exam conditions.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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)
Q2. Express $z = -1 + i\sqrt 3$ in polar form. Hence find $z^2$ in polar form and in Cartesian form. (3 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)
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].
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