The polar form

Length and direction

Mark $z = x + iy$ as the point $(x, y)$ of the Argand plane. You can describe the point in a second way: by its distance from the origin and by the angle that its arrow makes with the positive real axis.

• The distance is the modulus, $r = |z| = \sqrt{x^2 + y^2}$.
• The angle is an argument of $z$, written $\arg z$. Then $x = r\cos\theta$ and $y = r\sin\theta$.

So every $z \ne 0$ can be written in polar form:

The number $0$ has modulus $0$ and no argument at all.

The principal argument

Turning a whole turn more, or less, gives the same point. So $\theta$, $\theta + 2\pi$ and $\theta - 2\pi$ are all arguments of the same $z$. To make the answer unique, we choose the principal argument, the one with $-\pi < \theta \le \pi$.

• Above the real axis it is between $0$ and $\pi$.
• Below the real axis it is between $0$ and $-\pi$.
• On the negative real axis it is $\pi$, not $-\pi$.

Finding the principal argument

A calculator gives $\tan^{-1}\dfrac{y}{x}$ between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$, so it cannot tell quadrant II from quadrant IV. Work in three steps.

1. Find the acute angle $\alpha = \tan^{-1}\left|\dfrac{y}{x}\right|$.
2. Find the quadrant from the signs of $x$ and $y$.
3. Place the angle. Quadrant I: $\alpha$. Quadrant II: $\pi - \alpha$. Quadrant III: $-(\pi - \alpha)$. Quadrant IV: $-\alpha$.

Example 1. Write $z = -1 - i\sqrt{3}$ in polar form.

1. Modulus. $r = \sqrt{1 + 3} = 2$.
2. Acute angle. $\tan\alpha = \sqrt{3}$, so $\alpha = \dfrac{\pi}{3}$.
3. Quadrant. Both parts are negative: quadrant III.
4. Principal argument. $\theta = -\left(\pi - \dfrac{\pi}{3}\right) = -\dfrac{2\pi}{3}$.

So $z = 2\left(\cos\left(-\dfrac{2\pi}{3}\right) + i\sin\left(-\dfrac{2\pi}{3}\right)\right)$. The angle $\dfrac{4\pi}{3}$ is also an argument, but it is not the principal one.

Move the point across the negative real axis and watch the readout jump from close to $\pi$ to close to $-\pi$. The point moved a little; only the label we chose for its direction jumped.

Back from polar form

To go back to $a + ib$, work out $a = r\cos\theta$ and $b = r\sin\theta$.

Example 2. $z = 6\left(\cos\dfrac{5\pi}{6} + i\sin\dfrac{5\pi}{6}\right)$ gives $a = 6 \times \left(-\dfrac{\sqrt{3}}{2}\right) = -3\sqrt{3}$ and $b = 6 \times \dfrac{1}{2} = 3$. So $z = -3\sqrt{3} + 3i$.

Practise with new numbers

Try it in Python

Python's cmath.polar returns the modulus and the principal argument, and cmath.rect goes back. The program finds the polar form of a point in each quadrant.

Common mistakes

• Trusting $\tan^{-1}\dfrac{y}{x}$. For $-1 + i$ it gives $-\dfrac{\pi}{4}$, but the point is in quadrant II, so the principal argument is $\dfrac{3\pi}{4}$.
• Giving $\dfrac{4\pi}{3}$ as the principal argument. It is an argument, but it lies outside $(-\pi, \pi]$. The principal one is $-\dfrac{2\pi}{3}$.
• A negative modulus. $r$ is a length, so it is never negative. Write $-2(\cos\theta + i\sin\theta)$ as $2(\cos(\theta + \pi) + i\sin(\theta + \pi))$.
• Mixed units. Keep the angle in radians, unless the question uses degrees throughout.

Remember these

• $z = r(\cos\theta + i\sin\theta)$, with $r = |z|$ and $\theta$ an argument.
• The principal argument lies in $(-\pi, \pi]$.
• Find the acute angle, then place it by the quadrant.
• $a = r\cos\theta$ and $b = r\sin\theta$ take you back to $a + ib$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What is the principal argument of $-5i$?

2. Kabir says the principal argument of $-2 + 2i$ is $-\dfrac{\pi}{4}$, because $\tan^{-1}(-1) = -\dfrac{\pi}{4}$. Correct him.

3. Write $4\left(\cos\left(-\dfrac{\pi}{3}\right) + i\sin\left(-\dfrac{\pi}{3}\right)\right)$ as $a + ib$.

4. Two complex numbers have the same modulus and arguments that differ by $2\pi$. Are they equal?