1.13 Modulus–argument and Euler form (IB Maths: Analysis and Approaches HL): Revision notes

Section 1

Modulus–argument (polar) form

A complex number with modulus $r$ and argument $\theta$ can be written in modulus–argument form: $z = r(\cos\theta + \mathrm{i}\sin\theta) = r\,\mathrm{cis}\,\theta.$ To convert from Cartesian form $a + b\mathrm{i}$ : $r = \sqrt{a^2 + b^2}$ , and find $\theta$ from the quadrant (usually $-\pi < \theta \le \pi$ ).

To convert back: $a = r\cos\theta$ , $b = r\sin\theta$ . For example $2\,\mathrm{cis}\frac{\pi}{3} = 1 + \sqrt3\,\mathrm{i}$ and $-2 - 2\mathrm{i} = 2\sqrt2\,\mathrm{cis}\left(-\frac{3\pi}{4}\right)$ .

Key termsmodulus–argument formcis

Common mistake

Taking $\arctan\frac{b}{a}$ blindly. For $-2 - 2\mathrm{i}$ it gives $\frac{\pi}{4}$ , but the point is in the third quadrant, so $\theta = -\frac{3\pi}{4}$ .

Section 2

Euler form

Euler's formula states $\mathrm{e}^{\mathrm{i}\theta} = \cos\theta + \mathrm{i}\sin\theta$ , so $z = r\,\mathrm{cis}\,\theta = r\mathrm{e}^{\mathrm{i}\theta}.$ This is the Euler (exponential) form. It makes the rules for products and quotients follow from the laws of exponents. Two useful consequences: $\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta} = 2\cos\theta, \qquad \mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{-\mathrm{i}\theta} = 2\mathrm{i}\sin\theta.$ Also $|\mathrm{e}^{\mathrm{i}\theta}| = 1$ for every real $\theta$ , and the conjugate of $r\mathrm{e}^{\mathrm{i}\theta}$ is $r\mathrm{e}^{-\mathrm{i}\theta}$ .

Key termsEuler's formulaEuler form

Exam tip

For $1 \pm \mathrm{e}^{\mathrm{i}\theta}$ , take out the 'half-angle' factor $\mathrm{e}^{\frac{\mathrm{i}\theta}{2}}$ : $1 + \mathrm{e}^{\mathrm{i}\theta} = \mathrm{e}^{\frac{\mathrm{i}\theta}{2}}\left(\mathrm{e}^{-\frac{\mathrm{i}\theta}{2}} + \mathrm{e}^{\frac{\mathrm{i}\theta}{2}}\right) = 2\cos\frac{\theta}{2}\,\mathrm{e}^{\frac{\mathrm{i}\theta}{2}}$ .

Section 3

Products and quotients

In polar or Euler form: $r_1\mathrm{e}^{\mathrm{i}\theta_1}\times r_2\mathrm{e}^{\mathrm{i}\theta_2} = r_1r_2\,\mathrm{e}^{\mathrm{i}(\theta_1 + \theta_2)}, \qquad \frac{r_1\mathrm{e}^{\mathrm{i}\theta_1}}{r_2\mathrm{e}^{\mathrm{i}\theta_2}} = \frac{r_1}{r_2}\,\mathrm{e}^{\mathrm{i}(\theta_1 - \theta_2)}.$ So $|zw| = |z||w|$ and $\arg(zw) = \arg z + \arg w$ ; $\left|\frac{z}{w}\right| = \frac{|z|}{|w|}$ and $\arg\frac{z}{w} = \arg z - \arg w$ .

Example: $2\,\mathrm{cis}\frac{\pi}{3}\times4\,\mathrm{cis}\frac{\pi}{6} = 8\,\mathrm{cis}\frac{\pi}{2} = 8\mathrm{i}$ . Adjust the final argument into $-\pi < \theta \le \pi$ by adding or subtracting $2\pi$ if needed.

Sums are easiest in Cartesian form: convert first, add, then convert back if required.

Key termsproduct rulequotient rule

Common mistake

Adding the moduli or multiplying the arguments. Moduli multiply (or divide); arguments add (or subtract).

Section 4

Geometric interpretation

In the complex plane:

Multiplying $z$ by $r\mathrm{e}^{\mathrm{i}\alpha}$ enlarges by scale factor $r$ and rotates anticlockwise by $\alpha$ about the origin. In particular, multiplying by $\mathrm{i} = \mathrm{e}^{\frac{\pi}{2}\mathrm{i}}$ rotates by $\frac{\pi}{2}$ .
Dividing by $r\mathrm{e}^{\mathrm{i}\alpha}$ scales by $\frac1r$ and rotates by $-\alpha$ .
Adding $z + w$ follows the parallelogram rule: the point $z + w$ is the fourth vertex of the parallelogram with sides $Oz$ and $Ow$ .
If $\frac{z}{w}$ is purely imaginary, the lines from $O$ to $z$ and $w$ are perpendicular; if it is real, they are collinear with $O$ .

Application: in AC circuits $V = IZ$ , so the voltage's argument is the current's argument plus $\arg Z$ . With $Z = 8\mathrm{e}^{\frac{\pi}{6}\mathrm{i}}$ , the voltage leads the current by $\frac{\pi}{6}$ .

Key termsrotationenlargementparallelogram rule

Example

$1 + \mathrm{e}^{\mathrm{i}\theta}$ : both have modulus 1, so the parallelogram is a rhombus and the sum bisects the angle, giving argument $\frac{\theta}{2}$ .

Must know

$z = r(\cos\theta + \mathrm{i}\sin\theta) = r\,\mathrm{cis}\,\theta = r\mathrm{e}^{\mathrm{i}\theta}$ .
Convert with $r = \sqrt{a^2 + b^2}$ , $\theta$ from the correct quadrant, $a = r\cos\theta$ , $b = r\sin\theta$ .
Products: multiply moduli, add arguments. Quotients: divide moduli, subtract arguments.
Multiplication by $r\mathrm{e}^{\mathrm{i}\alpha}$ = enlargement by $r$ and rotation by $\alpha$ .
$\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta} = 2\cos\theta$ and $\mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{-\mathrm{i}\theta} = 2\mathrm{i}\sin\theta$ .

That's the notes covered.

Carry on to the next subtopic.