Polar and Euler Form of Complex Numbers

Cartesian form $a + bi$ tells you how far to go right and up. Polar form describes the same point by its distance from the origin and its direction instead. That sounds like a small change, but it makes multiplication beautifully simple: multiply the distances and add the angles. This page covers subtopic AHL 1.13 for both AA HL and AI HL. All angles are in radians.

Key ideas

Modulus–argument (polar) form

If $z$ has modulus $r = |z|$ and argument $\theta = \arg z$, then the point $z$ is at $a = r\cos\theta$, $b = r\sin\theta$. So

$$z = r(\cos\theta + i\sin\theta) = r\,\text{cis}\,\theta$$

"$\text{cis}\,\theta$" is short for $\cos\theta + i\sin\theta$.

$z = -2 + 2i$ has $r = 2\sqrt{2}$ and $\theta = \dfrac{3\pi}{4}$, so $z = 2\sqrt{2}\,\text{cis}\,\dfrac{3\pi}{4}$.

Euler (exponential) form

Euler’s formula connects the exponential function with sine and cosine:

$$e^{i\theta} = \cos\theta + i\sin\theta$$

So any complex number can also be written in Euler form (the AI guide calls it exponential form):

$$z = re^{i\theta}$$

Putting $\theta = \pi$ gives $e^{i\pi} = -1$, or $e^{i\pi} + 1 = 0$: one equation linking $e$, $i$, $\pi$, $1$ and $0$.

Converting between forms

From	To	How
$a + bi$	$r\,\text{cis}\,\theta$ or $re^{i\theta}$	$r = \sqrt{a^2 + b^2}$; find $\theta$ from a sketch (correct quadrant)
$r\,\text{cis}\,\theta$ or $re^{i\theta}$	$a + bi$	$a = r\cos\theta$, $b = r\sin\theta$

• Give $\theta$ as the principal argument, $-\pi \lt \theta \le \pi$, unless a question says otherwise. Adding any multiple of $2\pi$ to $\theta$ gives the same point, so $2\,\text{cis}\,\dfrac{7\pi}{4} = 2\,\text{cis}\left(-\dfrac{\pi}{4}\right)$.
• The modulus $r$ must be positive. $-3\,\text{cis}\,\dfrac{\pi}{6}$ is not in polar form; rewrite it as $3\,\text{cis}\left(-\dfrac{5\pi}{6}\right)$.
• Your GDC can convert in both directions (look for its polar/rectangular or $re^{i\theta}$ / $a + bi$ setting). The AI course expects conversions both by hand and with technology; for AA, make sure you can do the special angles by hand.

Products, quotients and powers

For $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$, the exponent laws give

$$z_1 z_2 = r_1 r_2\, e^{i(\theta_1 + \theta_2)} \qquad\qquad \frac{z_1}{z_2} = \frac{r_1}{r_2}\, e^{i(\theta_1 - \theta_2)}$$

In words: multiply the moduli and add the arguments; divide the moduli and subtract the arguments. The same rules hold in cis notation. (You can prove the product rule by expanding $(\cos\theta_1 + i\sin\theta_1)(\cos\theta_2 + i\sin\theta_2)$ and using the compound angle formulas.)

Repeating the product rule gives integer powers:

$$z^n = r^n e^{in\theta} = r^n\,\text{cis}\,n\theta$$

This is De Moivre’s theorem, which AA HL studies further (including roots). (AI HL) In AI exams you calculate products, quotients and integer powers in polar or exponential form, but you won’t be asked to find roots of complex numbers.

Also useful: the conjugate of $re^{i\theta}$ is $re^{-i\theta}$ (same modulus, opposite argument).

After adding or subtracting arguments, the angle may fall outside $-\pi \lt \theta \le \pi$. Add or subtract $2\pi$ to bring it back.

Geometric interpretation

• Adding complex numbers is vector addition on the Argand diagram (tip to tail, or the diagonal of a parallelogram). Subtracting $z_2$ from $z_1$ gives the vector from $z_2$ to $z_1$, so $|z_1 - z_2|$ is the distance between the two points.
• Multiplying by $w = r e^{i\alpha}$ rotates a point anticlockwise about the origin by $\alpha$ and stretches (enlarges) its distance from the origin by the factor $r$.
• In particular, multiplying by $i = e^{i\pi/2}$ is a rotation of $\dfrac{\pi}{2}$ anticlockwise, and multiplying by $-1 = e^{i\pi}$ is a half-turn.
• Dividing by $w$ undoes this: rotate by $-\alpha$ and divide the distance by $r$.

Adding sinusoidal functions (AI HL)

Two waves with the same frequency but different amplitudes and phase shifts always add up to a single wave of that frequency. Complex numbers find it quickly. Since $\cos(\omega t + \alpha)$ is the real part of $e^{i(\omega t + \alpha)} = e^{i\alpha}e^{i\omega t}$,

$$A_1\cos(\omega t + \alpha_1) + A_2\cos(\omega t + \alpha_2) = \text{Re}\left[\left(A_1 e^{i\alpha_1} + A_2 e^{i\alpha_2}\right) e^{i\omega t}\right]$$

So:

1. Write each wave as a complex number $A e^{i\alpha}$ (its amplitude and phase; engineers call this a phasor).
2. Add the complex numbers, in Cartesian form, to get $R e^{i\beta}$.
3. The sum is $R\cos(\omega t + \beta)$.

The same method works for sines, using imaginary parts: $A_1\sin(\omega t + \alpha_1) + A_2\sin(\omega t + \alpha_2) = R\sin(\omega t + \beta)$ with the same $R e^{i\beta}$. This is how voltages and currents in AC circuits are combined. On a GDC, you can add the numbers in polar form directly.

Worked examples

Example 1: Converting between forms

• (a) Write $z = -2 + 2i$ in polar form and in Euler form.
• (b) Write $w = 4e^{-i\pi/3}$ in Cartesian form.

Solution.

(a) $r = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$. The point is in the second quadrant (see the figure above). The reference angle is $\arctan\dfrac{2}{2} = \dfrac{\pi}{4}$, so $\theta = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}$.

$$z = 2\sqrt{2}\,\text{cis}\,\frac{3\pi}{4} = 2\sqrt{2}\,e^{3\pi i/4}$$

(b) Use $a = r\cos\theta$ and $b = r\sin\theta$:

$$\begin{aligned} w &= 4\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right) \\ &= 4\left(\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) \\ &= 2 - 2\sqrt{3}\,i \end{aligned}$$

Example 2: Products and quotients

Let $z = 4\,\text{cis}\,\dfrac{3\pi}{4}$ and $w = 2\,\text{cis}\,\dfrac{\pi}{2}$. Find $zw$ and $\dfrac{z}{w}$ in polar form (principal argument), then in Cartesian form.

Solution. Product: multiply the moduli, add the arguments.

$$zw = 8\,\text{cis}\left(\frac{3\pi}{4} + \frac{\pi}{2}\right) = 8\,\text{cis}\,\frac{5\pi}{4}$$

$\dfrac{5\pi}{4}$ is bigger than $\pi$, so subtract $2\pi$: $\dfrac{5\pi}{4} - 2\pi = -\dfrac{3\pi}{4}$.

$$zw = 8\,\text{cis}\left(-\frac{3\pi}{4}\right) = 8\left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\right) = -4\sqrt{2} - 4\sqrt{2}\,i$$

Quotient: divide the moduli, subtract the arguments.

$$\frac{z}{w} = 2\,\text{cis}\left(\frac{3\pi}{4} - \frac{\pi}{2}\right) = 2\,\text{cis}\,\frac{\pi}{4} = \sqrt{2} + \sqrt{2}\,i$$

Check the product in Cartesian form: $z = -2\sqrt{2} + 2\sqrt{2}\,i$ and $w = 2i$, so $zw = -4\sqrt{2}\,i + 4\sqrt{2}\,i^2 = -4\sqrt{2} - 4\sqrt{2}\,i$. ✓

Example 3: Multiplication as a rotation and a stretch

Let $w = 2 + i$ and $u = 1 + i$. Describe geometrically what multiplying by $u$ does to points on the Argand diagram. Then find $uw$ and check that it fits your description.

Solution. In polar form, $u = \sqrt{2}\,\text{cis}\,\dfrac{\pi}{4}$. So multiplying by $u$ rotates every point $\dfrac{\pi}{4}$ anticlockwise about the origin and enlarges its distance from the origin by a factor of $\sqrt{2}$.

$$uw = (1 + i)(2 + i) = 2 + i + 2i + i^2 = 1 + 3i$$

Check the stretch: $|w| = \sqrt{5}$ and $|uw| = \sqrt{10} = \sqrt{2}\cdot\sqrt{5}$. ✓

Check the rotation: $\arg w = \arctan\dfrac{1}{2} = 0.4636\ldots$ and $\arg(uw) = \arctan 3 = 1.2490\ldots$. The difference is $0.7854\ldots = \dfrac{\pi}{4}$. ✓

Multiplying $w = 2 + i$ by $1 + i = \sqrt{2}\,\text{cis}\,\dfrac{\pi}{4}$ rotates it by $\dfrac{\pi}{4}$ and stretches it by $\sqrt{2}$.

Example 4: Adding two voltages (AI HL)

Two AC voltage sources in a circuit give $V_1 = 5\cos(50t)$ and $V_2 = 8\cos\left(50t + \dfrac{\pi}{3}\right)$ volts, where $t$ is time in seconds. Write the total voltage $V = V_1 + V_2$ in the form $V = R\cos(50t + \beta)$, with $R$ and $\beta$ to 3 s.f.

Solution. Both waves have the same angular frequency, $50$, so the sum is one wave of that frequency. Write each as a complex number (amplitude, phase), and add in Cartesian form:

$$\begin{aligned} 5e^{i\cdot 0} + 8e^{i\pi/3} &= 5 + 8\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) \\ &= 5 + 4 + 4\sqrt{3}\,i \\ &= 9 + 4\sqrt{3}\,i \end{aligned}$$

Convert to Euler form. It’s in the first quadrant, so

$$R = \sqrt{9^2 + (4\sqrt{3})^2} = \sqrt{81 + 48} = \sqrt{129} = 11.357\ldots$$ $$\beta = \arctan\frac{4\sqrt{3}}{9} = 0.65605\ldots$$ $$V = 11.4\cos(50t + 0.656) \text{ volts (3 s.f.)}$$

Check at $t = 0$: $V_1 + V_2 = 5 + 8\cos\dfrac{\pi}{3} = 9$, and $\sqrt{129}\cos(0.65605\ldots) = 9.00$. ✓ On a GDC, you can enter $5 + 8e^{i\pi/3}$ and ask for the answer in polar form to get $R$ and $\beta$ in one step.

Common mistakes

Using arctan without checking the quadrant. For $-2 + 2i$, $\arctan\dfrac{2}{-2} = -\dfrac{\pi}{4}$ points into the fourth quadrant, but the number is in the second. Sketch first, then adjust: $\theta = \dfrac{3\pi}{4}$.

Leaving the argument outside the principal range. After adding arguments, $8\,\text{cis}\,\dfrac{5\pi}{4}$ is a correct number but not the principal form. Subtract $2\pi$ to get $8\,\text{cis}\left(-\dfrac{3\pi}{4}\right)$.

Multiplying the arguments or adding the moduli. For products, the moduli multiply and the arguments add. It’s the exponent laws: $e^{i\theta_1}e^{i\theta_2} = e^{i(\theta_1 + \theta_2)}$.

Adding complex numbers in polar form directly. There’s no rule for $r_1\,\text{cis}\,\theta_1 + r_2\,\text{cis}\,\theta_2$. Convert to Cartesian form, add, then convert back, just as in Example 4.

Calculator in degree mode. Arguments here are in radians. If your GDC is in degree mode, $e^{i\theta}$ and the conversions will give wrong answers. Check the mode before you start.

Mixing up the phase sign when adding waves. A wave $\cos\left(\omega t - \dfrac{2\pi}{3}\right)$ has phase $-\dfrac{2\pi}{3}$, so its complex number is $e^{-2\pi i/3}$, not $e^{2\pi i/3}$.

Practice

1. (Warm-up) Write each number in Cartesian form.

• (a) $6\,\text{cis}\,\dfrac{\pi}{2}$
• (b) $4e^{i\pi}$
• (c) $2\,\text{cis}\left(-\dfrac{\pi}{6}\right)$
• (d) $\sqrt{2}\,e^{3\pi i/4}$

Solution

(a) $6(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}) = 6(0 + i) = 6i$.

(b) $4(\cos\pi + i\sin\pi) = 4(-1 + 0i) = -4$.

(c) $2\left(\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}i\right) = \sqrt{3} - i$.

(d) $\sqrt{2}\left(-\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}i\right) = -1 + i$.

2. (Warm-up) Write each number in polar form $r\,\text{cis}\,\theta$ and Euler form $re^{i\theta}$, with $-\pi \lt \theta \le \pi$.

• (a) $5i$
• (b) $-3$
• (c) $1 - i$

Solution

(a) On the positive imaginary axis: $5i = 5\,\text{cis}\,\dfrac{\pi}{2} = 5e^{i\pi/2}$.

(b) On the negative real axis: $-3 = 3\,\text{cis}\,\pi = 3e^{i\pi}$.

(c) $r = \sqrt{1 + 1} = \sqrt{2}$, fourth quadrant, reference angle $\dfrac{\pi}{4}$: $1 - i = \sqrt{2}\,\text{cis}\left(-\dfrac{\pi}{4}\right) = \sqrt{2}\,e^{-i\pi/4}$.

3. (Core) Write $z = -1 - \sqrt{3}\,i$ in Euler form, then write down $z^*$ in Euler form.

Solution

$r = \sqrt{1 + 3} = 2$. The point is in the third quadrant, with reference angle $\arctan\sqrt{3} = \dfrac{\pi}{3}$. So $\theta = -\left(\pi - \dfrac{\pi}{3}\right) = -\dfrac{2\pi}{3}$.

$$z = 2e^{-2\pi i/3}$$

The conjugate has the same modulus and the opposite argument: $z^* = 2e^{2\pi i/3}$ (which is $-1 + \sqrt{3}\,i$ ✓).

4. (Core) Let $z_1 = 6e^{2\pi i/3}$ and $z_2 = 3e^{-i\pi/6}$. Find $z_1 z_2$ and $\dfrac{z_1}{z_2}$, in Euler form and in Cartesian form.

Solution

$$z_1 z_2 = 18\,e^{i(2\pi/3 - \pi/6)} = 18\,e^{i\pi/2} = 18i$$$$\frac{z_1}{z_2} = 2\,e^{i(2\pi/3 + \pi/6)} = 2\,e^{5\pi i/6} = 2\left(-\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = -\sqrt{3} + i$$

5. (Core) Let $z = 1 - i$. Use polar form to find $z^8$ and $z^5$ in Cartesian form, without a calculator.

Solution

$z = \sqrt{2}\,\text{cis}\left(-\dfrac{\pi}{4}\right)$.

$$z^8 = (\sqrt{2})^8\,\text{cis}(-2\pi) = 16(1 + 0i) = 16$$$$z^5 = (\sqrt{2})^5\,\text{cis}\left(-\frac{5\pi}{4}\right) = 4\sqrt{2}\,\text{cis}\,\frac{3\pi}{4} = 4\sqrt{2}\left(-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\right) = -4 + 4i$$

Check $z^5$: $(1 - i)^2 = -2i$, so $(1 - i)^4 = (-2i)^2 = -4$ and $(1 - i)^5 = -4(1 - i) = -4 + 4i$. ✓

6. (Core) The point $A$ on an Argand diagram represents $z = 3 + i$. Find the complex number representing the image of $A$ after:

• (a) a rotation of $\dfrac{\pi}{2}$ anticlockwise about the origin;
• (b) a rotation of $\dfrac{\pi}{3}$ anticlockwise about the origin followed by an enlargement with scale factor $2$, centre the origin.

Solution

(a) Multiply by $i$: $i(3 + i) = 3i + i^2 = -1 + 3i$.

(b) Multiply by $2\,\text{cis}\,\dfrac{\pi}{3} = 2\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right) = 1 + \sqrt{3}\,i$:

$$(1 + \sqrt{3}\,i)(3 + i) = 3 + i + 3\sqrt{3}\,i + \sqrt{3}\,i^2 = (3 - \sqrt{3}) + (1 + 3\sqrt{3})i$$

7. (Core) Give answers to 3 s.f.

• (a) Write $z = -4 + 3i$ in Euler form.
• (b) Write $w = 3e^{-2i}$ in Cartesian form.

Solution

(a) $r = \sqrt{16 + 9} = 5$. The point is in the second quadrant, with reference angle $\arctan\dfrac{3}{4} = 0.6435\ldots$, so $\theta = \pi - 0.6435\ldots = 2.4980\ldots$

$$z \approx 5e^{2.50i}$$

(b) The argument is $-2$ radians (third quadrant, since $-\pi \lt -2 \lt -\dfrac{\pi}{2}$):

$$w = 3\cos(-2) + 3i\sin(-2) = -1.248\ldots - 2.727\ldots i \approx -1.25 - 2.73i$$

8. (Challenge) (AI HL) Two voltages in a circuit are $V_1 = 4\cos t$ and $V_2 = 6\cos\left(t - \dfrac{2\pi}{3}\right)$ volts. Write $V_1 + V_2$ in the form $R\cos(t + \beta)$, giving $R$ exactly and $\beta$ to 3 s.f.

Solution

Add the complex numbers for the two waves:

$$\begin{aligned} 4 + 6e^{-2\pi i/3} &= 4 + 6\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) \\ &= 4 - 3 - 3\sqrt{3}\,i \\ &= 1 - 3\sqrt{3}\,i \end{aligned}$$

$R = \sqrt{1 + 27} = \sqrt{28} = 2\sqrt{7}$. The number is in the fourth quadrant, so

$$\beta = -\arctan\frac{3\sqrt{3}}{1} = -1.38 \text{ (3 s.f.)}$$$$V_1 + V_2 = 2\sqrt{7}\cos(t - 1.38) \text{ volts}$$

Check at $t = 0$: $4 + 6\cos\left(-\dfrac{2\pi}{3}\right) = 4 - 3 = 1$, and $2\sqrt{7}\cos(-1.3806\ldots) = 1.00$. ✓

9. (Challenge) By writing $1 + i$ and $\sqrt{3} + i$ in polar form and multiplying them in two ways, find the exact value of $\cos\dfrac{5\pi}{12}$.

Solution

$1 + i = \sqrt{2}\,\text{cis}\,\dfrac{\pi}{4}$ and $\sqrt{3} + i = 2\,\text{cis}\,\dfrac{\pi}{6}$. In polar form,

$$(1 + i)(\sqrt{3} + i) = 2\sqrt{2}\,\text{cis}\left(\frac{\pi}{4} + \frac{\pi}{6}\right) = 2\sqrt{2}\,\text{cis}\,\frac{5\pi}{12}$$

In Cartesian form,

$$(1 + i)(\sqrt{3} + i) = \sqrt{3} + i + \sqrt{3}\,i + i^2 = (\sqrt{3} - 1) + (\sqrt{3} + 1)i$$

The real parts must be equal: $2\sqrt{2}\cos\dfrac{5\pi}{12} = \sqrt{3} - 1$, so

$$\cos\frac{5\pi}{12} = \frac{\sqrt{3} - 1}{2\sqrt{2}} = \frac{\sqrt{6} - \sqrt{2}}{4}$$

(Equating imaginary parts gives $\sin\dfrac{5\pi}{12} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$ as a bonus.)