2. Polar Form & Loci

Video coming soon4 worked examples

Theory

A complex number can be written in polar (modulus–argument) form:

z = r(\cos\theta + i\sin\theta), \qquad r = |z| = \sqrt{a^2 + b^2}, \quad \theta = \arg z

The modulus $r$ is the distance from the origin on an Argand diagram, and the argument $\theta$ is the angle measured from the positive real axis (taken in the range $-\pi < \theta \leq \pi$ ).

Polar form is worth the effort because multiplication and division become simple. Writing $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$ :

\begin{aligned} z_1z_2 &= r_1r_2\bigl[\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2)\bigr] \\ \frac{z_1}{z_2} &= \frac{r_1}{r_2}\bigl[\cos(\theta_1-\theta_2) + i\sin(\theta_1-\theta_2)\bigr] \end{aligned}

In words: multiply the moduli and add the arguments. Geometrically, multiplying by a complex number of modulus $r$ and argument $\theta$ scales by $r$ and rotates by $\theta$ . This gives a set of results worth knowing:

\begin{aligned} |z_1z_2| &= |z_1||z_2| & \arg(z_1z_2) &= \arg z_1 + \arg z_2 \\ \left|\frac{z_1}{z_2}\right| &= \frac{|z_1|}{|z_2|} & \arg\!\left(\frac{z_1}{z_2}\right) &= \arg z_1 - \arg z_2 \\ |\bar{z}| &= |z| & \arg(\bar{z}) &= -\arg z \\ z\bar{z} &= |z|^2 && \end{aligned}

Loci describe sets of points: $|z - a| = k$ is a circle of radius $k$ centred at the point $a$ ; $|z - a| = |z - b|$ is the perpendicular bisector of the segment joining $a$ and $b$ .

The Golden Rule: always identify which quadrant $z$ lies in before stating the argument — $\arctan\frac{b}{a}$ alone cannot tell the second quadrant from the fourth.

⚠️ Common Examiner Traps

Argument quadrant: $\arctan\frac{b}{a}$ gives a principal value — adjust by $\pm\pi$ for the second and third quadrants.
Circle centre: $|z - a| = k$ is centred at the point $a$ , not at the origin.
Argument range: stick to $-\pi < \theta \leq \pi$ unless told otherwise.

Worked examples

Example 1

Express $z = 1 + i\sqrt{3}$ in polar form.

Step 1: Find the modulus:

r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2

Step 2: The point $(1, \sqrt{3})$ is in the first quadrant, so the argument is:

\theta = \arctan\!\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}

Step 3: Write in polar form:

z = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)

Example 2

Express $z = -1 + i$ in polar form.

Step 1: Find the modulus:

r = \sqrt{(-1)^2 + 1^2} = \sqrt{2}

Step 2: The point $(-1, 1)$ is in the second quadrant. The related acute angle is $\frac{\pi}{4}$ , so measuring from the positive real axis:

\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}

Step 3: Write in polar form:

z = \sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)

Example 3

Describe, and sketch mentally, the locus of points satisfying $|z - 2 - i| = 3$ .

Step 1: Rewrite the condition to identify the fixed point:

|z - (2 + i)| = 3

Step 2: This is the set of points whose distance from $2 + i$ is $3$ .

Step 3: The locus is a circle of radius $3$ centred at the point $(2, 1)$ on the Argand diagram.

Example 4

Given $z_1 = -1 + \sqrt{3}\,i$ and $z_2 = 1 + i$ , find $|z_1z_2|$ and $\arg(z_1z_2)$ without multiplying the two numbers out.

Step 1: Find the modulus and argument of $z_1$ . The point $(-1, \sqrt{3})$ lies in the second quadrant, so the argument is $\pi$ minus the acute reference angle:

|z_1| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{4} = 2, \qquad \arg z_1 = \pi - \tan^{-1}\!\left(\frac{\sqrt{3}}{1}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}

Step 2: Do the same for $z_2$ . The point $(1,1)$ is in the first quadrant, so no adjustment is needed:

|z_2| = \sqrt{1^2 + 1^2} = \sqrt{2}, \qquad \arg z_2 = \tan^{-1}(1) = \frac{\pi}{4}

Step 3: Multiply the moduli and add the arguments:

|z_1z_2| = 2\sqrt{2}, \qquad \arg(z_1z_2) = \frac{2\pi}{3} + \frac{\pi}{4} = \frac{8\pi}{12} + \frac{3\pi}{12} = \frac{11\pi}{12}

Step 4: Check the argument lies in the required range $-\pi < \theta \le \pi$ . Since $\frac{11\pi}{12} < \pi$ , no adjustment is needed.