Polar form and converting between forms

The second way to describe a complex number

• Rectangular form $a + bi$ locates a point by how far right and how far up.
• Polar form locates the same point by how far from the origin and in what direction.
• Both describe exactly the same number. Which one you use is a matter of which operation you are about to do.

Building polar form from the triangle

• From the right-angled triangle on the Argand diagram, with $r = |z|$ and $\theta = \arg(z)$:

$$a = r\cos\theta \qquad b = r\sin\theta$$

• Substituting into $z = a + bi$:

$$z = r\cos\theta + (r\sin\theta)i = r(\cos\theta + i\sin\theta)$$

• NZQA abbreviates $\cos\theta + i\sin\theta$ to $\text{cis}\,\theta$, so:

$$z = r\,\text{cis}\,\theta$$

• Each variable:
  • $r$ — the modulus, $\sqrt{a^2 + b^2}$; always positive
  • $\theta$ — the argument, the anticlockwise angle from the positive real axis
  • "cis" — short for cosine i sine; it is notation, not a function you can split up

Converting rectangular to polar

• Find the modulus: $r = \sqrt{a^2 + b^2}$.

• Sketch the point to identify the quadrant.

• Find the reference angle $\alpha = \tan^{-1}\left|\frac{b}{a}\right|$ and adjust for the quadrant.

• Write the answer as $r\,\text{cis}\,\theta$.

• Worked through for $z = 1 + i$:

  • $$r = \sqrt{1 + 1} = \sqrt{2}$$
  • First quadrant, $\alpha = \tan^{-1}(1) = \frac{\pi}{4}$, so $\theta = \frac{\pi}{4}$
  • $$z = \sqrt{2}\,\text{cis}\,\frac{\pi}{4}$$

Converting polar to rectangular

• Evaluate $\cos\theta$ and $\sin\theta$, then multiply each by $r$.

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

• Worked through for $z = 8\,\text{cis}\,\frac{2\pi}{3}$:

  • $\cos\frac{2\pi}{3} = -\frac{1}{2}$ and $\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$
  • $$z = 8\left(-\frac{1}{2}\right) + 8\left(\frac{\sqrt{3}}{2}\right)i = -4 + 4\sqrt{3}\,i$$

• Note that this direction is much easier than the other — there is no quadrant decision to make, because the angle already tells you where the point is.

The exact values worth knowing

• These come up constantly, and knowing them lets you give exact answers rather than decimals:

$\theta$	$\cos\theta$	$\sin\theta$
$0$	$1$	$0$
$\frac{\pi}{6}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$
$\frac{\pi}{4}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$
$\frac{\pi}{3}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$
$\frac{\pi}{2}$	$0$	$1$
$\pi$	$-1$	$0$

Equal complex numbers in polar form

• Two numbers in polar form are equal when their moduli are equal and their arguments differ by a whole number of full turns:

$$r_1\,\text{cis}\,\theta_1 = r_2\,\text{cis}\,\theta_2 \quad \Longleftrightarrow \quad r_1 = r_2 \;\text{ and }\; \theta_1 = \theta_2 + 2k\pi$$

• That $2k\pi$ is not a technicality. It is exactly what generates multiple roots when you solve $z^n = w$, which is covered later in this standard.

Fixing an "illegal" polar form

• A polar form is only standard when $r$ is positive. Two situations need fixing:

  • Negative modulus: $-r\,\text{cis}\,\theta = r\,\text{cis}(\theta + \pi)$. Multiplying by $-1$ rotates the point half a turn.
  • Argument out of range: add or subtract $2\pi$ until $\theta$ lies in $(-\pi, \pi]$.

• So $-2\,\text{cis}\,\frac{\pi}{6}$ becomes $2\,\text{cis}\,\frac{7\pi}{6}$, and then, bringing the argument into the principal range, $2\,\text{cis}\left(-\frac{5\pi}{6}\right)$.

Choosing which form to work in

• Rectangular for: adding, subtracting, equating real and imaginary parts, checking whether something is purely real or imaginary.
• Polar for: multiplying, dividing, powers, roots, and anything involving rotation.
• Many questions are a conversion sandwich: convert to polar, do the multiplication or the power, convert back to rectangular because that is the form the answer was requested in.