Ultimate Guide to Complex Geometry (Part 2): Polar Operations and Identities

Introduction

In our last guide, we introduced the polar (modulus-argument) form of complex numbers, $z = r(\cos\theta + i\sin\theta)$ . But why go through the effort of converting from Cartesian form? The answer lies in multiplication and division. In Complex Geometry (Part 2), we reveal how polar form transforms messy algebraic expansions into simple arithmetic. We will also explore the geometric identities that govern the complex plane, concluding with the crucial Triangle Inequality.

Executive Summary

This guide explores the operational advantages of polar form and key geometric proofs:

Polar Operations: Multiplying and dividing complex numbers in polar form (multiply moduli, add arguments).
Geometric Interpretation: Understanding multiplication as a combined rotation and dilation.
Complex Identities: Proving standard modulus, argument, and conjugate identities.
The Triangle Inequality: Proving and applying $|z_1 + z_2| \le |z_1| + |z_2|$ in the complex plane.

What is this about?

If you try to multiply $z_1 = 3+4i$ and $z_2 = 1-i$ , you have to use FOIL, group terms, and simplify $i^2$ . But what if you had to multiply ten complex numbers together? The algebraic expansion would be a nightmare.

Polar form provides a shortcut. When you multiply complex numbers, their lengths (moduli) simply multiply, and their angles (arguments) simply add. This geometric interpretation—that multiplication is just scaling and rotating—is one of the most elegant and useful concepts in Mathematics Extension 2.

Main Content

1. Multiplication and Division in Polar Form

Let $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$ .

By expanding the product $z_1 z_2$ and applying the compound angle trigonometric identities ( $\cos(A+B)$ and $\sin(A+B)$ ), a beautiful result emerges: $z_1 z_2 = r_1 r_2 \left(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\right)$

Using the cis shorthand, this rule is very easy to memorize: Multiplication: $r_1 \text{cis}(\theta_1) \times r_2 \text{cis}(\theta_2) = (r_1 r_2) \text{cis}(\theta_1 + \theta_2)$ (Multiply the moduli, add the arguments).

Similarly, by dividing and realizing the denominator, we get: Division: $\frac{r_1 \text{cis}(\theta_1)}{r_2 \text{cis}(\theta_2)} = \left(\frac{r_1}{r_2}\right) \text{cis}(\theta_1 - \theta_2)$ (Divide the moduli, subtract the arguments).

2. The Geometric Interpretation

What does it mean to multiply $z_1$ by $z_2$ geometrically? If you start with a vector $z_1$ on the Argand diagram and multiply it by $z_2 = r_2 \text{cis}(\theta_2)$ :

The length of the vector is scaled (dilated) by a factor of $r_2$ .
The vector is rotated anticlockwise by an angle of $\theta_2$ .

For example, multiplying by $i$ (which has a modulus of $1$ and an argument of $\frac{\pi}{2}$ ) results in a pure $90^\circ$ anticlockwise rotation with no change in size!

3. Key Complex Identities

The multiplication rule above immediately proves two vital identities:

$|z_1 z_2| = |z_1||z_2|$
$\arg(z_1 z_2) = \arg(z_1) + \arg(z_2) + 2k\pi$ (The $2k\pi$ is required if you are forcing the result back into the Principal Argument range $(-\pi, \pi]$ ).

You must also be able to prove conjugate identities:

$\overline{z_1 + z_2} = \bar{z_1} + \bar{z_2}$ (The conjugate of a sum is the sum of the conjugates).
$\overline{z_1 z_2} = \bar{z_1} \cdot \bar{z_2}$ (The conjugate of a product is the product of the conjugates).
$|z| = |\bar{z}|$ (A number and its conjugate have the same length).
$\arg(\bar{z}) = -\arg(z)$ (A number and its conjugate are reflected across the real axis).

4. The Triangle Inequality

If we plot $z_1$ , $z_2$ , and their sum $z_1 + z_2$ on the Argand diagram, they form a triangle with the origin. The lengths of the sides of this triangle are $|z_1|$ , $|z_2|$ , and $|z_1 + z_2|$ .

A fundamental theorem of Euclidean geometry states that the length of any side of a triangle must be less than or equal to the sum of the lengths of the other two sides. Therefore: $|z_1 + z_2| \le |z_1| + |z_2|$

This is the Triangle Inequality. Equality only holds when $z_1$ and $z_2$ lie on the exact same ray from the origin (i.e., their arguments are identical).

There is also a reverse triangle inequality for subtraction: $|z_1 - z_2| \ge ||z_1| - |z_2||$

Simple Worked Example

Question: Let $z = 2\text{cis}(\frac{\pi}{6})$ and $w = 3\text{cis}(\frac{3\pi}{4})$ . Evaluate $zw$ and express the result in exact Cartesian form ( $x+iy$ ).

Solution: Step 1: Multiply in polar form Using the rule: multiply moduli, add arguments. $zw = (2 \times 3) \text{cis}\left(\frac{\pi}{6} + \frac{3\pi}{4}\right)$ To add the fractions, find a common denominator (12): $\frac{\pi}{6} = \frac{2\pi}{12}$ and $\frac{3\pi}{4} = \frac{9\pi}{12}$ $\frac{2\pi}{12} + \frac{9\pi}{12} = \frac{11\pi}{12}$

So, $zw = 6\text{cis}\left(\frac{11\pi}{12}\right)$

Step 2: Convert to Cartesian form $zw = 6\left(\cos\left(\frac{11\pi}{12}\right) + i\sin\left(\frac{11\pi}{12}\right)\right)$

To find the exact values of $\cos(\frac{11\pi}{12})$ , we use compound angle formulas, since $\frac{11\pi}{12} = \frac{3\pi}{4} + \frac{\pi}{6}$ . $\cos\left(\frac{3\pi}{4} + \frac{\pi}{6}\right) = \cos\frac{3\pi}{4}\cos\frac{\pi}{6} - \sin\frac{3\pi}{4}\sin\frac{\pi}{6}$ $= \left(-\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) = \frac{-\sqrt{3}-1}{2\sqrt{2}}$

$\sin\left(\frac{3\pi}{4} + \frac{\pi}{6}\right) = \sin\frac{3\pi}{4}\cos\frac{\pi}{6} + \cos\frac{3\pi}{4}\sin\frac{\pi}{6}$ $= \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(-\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{3}-1}{2\sqrt{2}}$

Now substitute back into $zw$ : $zw = 6 \left( \frac{-\sqrt{3}-1}{2\sqrt{2}} + i \frac{\sqrt{3}-1}{2\sqrt{2}} \right)$ Multiply by $\frac{\sqrt{2}}{\sqrt{2}}$ to rationalize the denominator: $= 6 \left( \frac{-\sqrt{6}-\sqrt{2}}{4} + i \frac{\sqrt{6}-\sqrt{2}}{4} \right)$ $= \frac{3}{2}(-\sqrt{6}-\sqrt{2}) + i\frac{3}{2}(\sqrt{6}-\sqrt{2})$

mini-FAQ page

Q: Do I have to use the compound angle formula in exams for Cartesian conversion? A: If the angle is a standard exact value (like $\frac{\pi}{3}, \frac{\pi}{4}, \frac{\pi}{6}$ ), you just read it from the exact value triangles. But for angles like $\frac{\pi}{12}$ or $\frac{5\pi}{12}$ , yes, you will likely need to use compound angle trigonometric identities if the question asks for exact form.

Q: Why does the Triangle Inequality matter? A: HSC questions frequently ask you to find the maximum or minimum possible value of $|z_1 + z_2|$ given constraints on the moduli. The triangle inequality provides the algebraic bounds needed to solve these optimization problems without drawing complicated graphs.

Common mistakes to avoid

Forgetting to check the Principal Argument after multiplying: If you multiply $z_1 = \text{cis}(\frac{3\pi}{4})$ and $z_2 = \text{cis}(\frac{3\pi}{4})$ , you add arguments to get $\frac{6\pi}{4} = \frac{3\pi}{2}$ . This is outside the principal range $(-\pi, \pi]$ . You must subtract $2\pi$ to get the final principal argument: $-\frac{\pi}{2}$ .
Confusing addition and multiplication: You can only multiply moduli and add arguments when multiplying complex numbers. You cannot add $z_1 + z_2$ by adding their moduli and arguments! Addition must be done in Cartesian form.

Practice on Vu's Maths Hub

Polar multiplication unlocks the hardest questions in Extension 2 complex geometry.

Practice vector rotations and triangle inequality proofs in the HSC Complex Numbers Booklet.
Revise your compound angle identities in the HSC Trigonometry Booklet.

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