Exponential Form

A2 · P3 · 2 min

Euler's relation $e^{i\theta} = \cos\theta + i\sin\theta$ compresses the polar form to $z = re^{i\theta}$ . Multiplying and dividing then follow the ordinary index laws, which is the cleanest way to see why arguments add.

The form

Key result

z = re^{i\theta} \quad\text{means}\quad z = r(\cos\theta + i\sin\theta), \qquad r = |z|,\ \theta = \arg z.

Index laws give:

r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} = r_1 r_2 e^{i(\theta_1 + \theta_2)}, \qquad \frac{r_1 e^{i\theta_1}}{r_2 e^{i\theta_2}} = \frac{r_1}{r_2}e^{i(\theta_1 - \theta_2)}, \qquad \left(re^{i\theta}\right)^n = r^n e^{in\theta}.

Special values: $e^{i\pi} = -1$ , $e^{i\pi/2} = i$ , $e^{2\pi i} = 1$ . The conjugate of $re^{i\theta}$ is $re^{-i\theta}$ .

Converting to exponential form

Write $1 - i$ and $-4$ in the form $re^{i\theta}$ .

Solution

$|1 - i| = \sqrt{2}$ , $\arg(1 - i) = -\tfrac{\pi}{4}$ : $\sqrt{2}e^{-i\pi/4}$ .

$-4 = 4e^{i\pi}$ .

From exponential to Cartesian

Express $6e^{i\pi/3}$ and $2e^{-5i\pi/6}$ in the form $x + iy$ .

Solution

$6e^{i\pi/3} = 6\left(\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i\right) = 3 + 3\sqrt{3}i$ .

$2e^{-5i\pi/6} = 2\left(-\tfrac{\sqrt{3}}{2} - \tfrac{1}{2}i\right) = -\sqrt{3} - i$ .

Multiplying and dividing

$z = 3e^{i\pi/4}$ and $w = 2e^{i\pi/3}$ . Find $zw$ , $\dfrac{z}{w}$ and $z^4$ in exponential form, and $z^4$ in Cartesian form.

Solution

$zw = 6e^{7i\pi/12}$ . $\dfrac{z}{w} = \tfrac{3}{2}e^{-i\pi/12}$ . $z^4 = 81e^{i\pi} = -81$ .

Using the conjugate

Show that $z + z^*$ is real and $z - z^*$ is purely imaginary using exponential form, for $z = re^{i\theta}$ .

Solution

$z + z^* = re^{i\theta} + re^{-i\theta} = r(\cos\theta + i\sin\theta) + r(\cos\theta - i\sin\theta) = 2r\cos\theta$ , real.

$z - z^* = 2ir\sin\theta$ , purely imaginary.

Solving an equation in exponential form

Find both square roots of $9e^{i\pi/3}$ .

Solution

If $z = re^{i\phi}$ then $r^2 e^{2i\phi} = 9e^{i\pi/3}$ , so $r = 3$ and $2\phi = \tfrac{\pi}{3} + 2k\pi$ , giving $\phi = \tfrac{\pi}{6}$ or $\phi = \tfrac{\pi}{6} + \pi = \tfrac{7\pi}{6}$ (principal value $-\tfrac{5\pi}{6}$ ).

$z = 3e^{i\pi/6}$ or $z = 3e^{-5i\pi/6}$ ; in Cartesian form, $\pm\left(\tfrac{3\sqrt{3}}{2} + \tfrac{3}{2}i\right)$ .

Watch out

$e^{i\theta}$ has modulus $1$ for every real $\theta$ : it is a point on the unit circle. Do not confuse $e^{i\theta}$ (a rotation) with $e^{\theta}$ (a real number). And the argument in the exponent must be in radians.

Exam tip

Any of the three forms is accepted unless the question specifies. "In the form $re^{i\theta}$ " means give exact $r$ and $\theta$ where possible, e.g. $\sqrt{2}e^{-i\pi/4}$ , and otherwise 3 significant figures for $\theta$ .

Practice

Question

Write $-1 - i\sqrt{3}$ in the form $re^{i\theta}$ .
Express $4e^{3i\pi/4}$ in Cartesian form.
$z = 2e^{2i\pi/3}$ . Find $z^3$ and $\dfrac{1}{z}$ in exponential form with principal arguments.
Find both square roots of $16e^{i\pi/2}$ in Cartesian form.

Answers

$2e^{-2i\pi/3}$ .
$-2\sqrt{2} + 2\sqrt{2}i$ .
$z^3 = 8e^{2i\pi} = 8e^{0} = 8$ ; $\dfrac{1}{z} = \tfrac{1}{2}e^{-2i\pi/3}$ .
$4e^{i\pi/4}$ and $4e^{-3i\pi/4}$ : $\pm(2\sqrt{2} + 2\sqrt{2}i)$ .

The form

Practice

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