Ultimate Guide to Complex Arithmetic (Part 3): Modulus, Division, and Square Roots

Introduction

In the previous parts of our guide, we learned how to add, subtract, multiply, and conjugate complex numbers. But two critical algebraic operations remain: division and square rooting. Complex Arithmetic (Part 3) equips you with the tools to handle fractions containing imaginary numbers and introduces a powerful algebraic technique for finding the square root of a complex number $z = x + iy$ . We will also formalize the "size" of a complex number through the concept of the modulus.

Executive Summary

This guide finalizes the algebraic toolkit for complex numbers:

The Modulus ( $|z|$ ): Defining the magnitude of a complex number as $\sqrt{x^2 + y^2}$ .
Modulus Relationships: Using the vital identity $z\bar{z} = |z|^2$ to solve algebraic problems.
Complex Division: Dividing non-zero complex numbers by "realising" the denominator.
Square Roots: Setting up simultaneous equations to find the two square roots of any complex number algebraically.

What is this about?

When you face a fraction like $\frac{2+3i}{1-4i}$ , you can't leave it in this form. Just as mathematicians prefer not to leave surds in the denominator ( $\frac{1}{\sqrt{2}}$ is converted to $\frac{\sqrt{2}}{2}$ ), we do not leave imaginary numbers in the denominator. You will use the complex conjugate (learned in Part 2) to force the denominator to become a real number.

Furthermore, while taking the square root of a positive real number is simple on a calculator, taking the square root of a complex number (like $\sqrt{3+4i}$ ) requires a clever algebraic setup. Because a complex number is two-dimensional, taking its square root requires equating real and imaginary parts to solve simultaneous equations.

Main Content

1. The Modulus of a Complex Number

The modulus (or absolute value) of a complex number $z = x + iy$ is a measure of its "size" or distance from the origin on the complex plane. It is defined using Pythagoras' theorem: $|z| = \sqrt{x^2 + y^2}$

Because $x$ and $y$ are real numbers, the modulus $|z|$ is always a non-negative real number.

2. Modulus Relationships

In Part 2, we saw that multiplying a complex number by its conjugate results in $x^2 + y^2$ . We can now link this directly to the modulus: $z\bar{z} = (x+iy)(x-iy) = x^2 + y^2 = |z|^2$

This relationship, $z\bar{z} = |z|^2$ , is one of the most frequently used algebraic tricks in Extension 2. It allows us to express the reciprocal of a complex number easily: $\frac{1}{z} = \frac{\bar{z}}{z\bar{z}} = \frac{\bar{z}}{|z|^2}$

Other important properties include:

$|z_1 z_2| = |z_1| |z_2|$
$\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$

3. Complex Division

To divide one complex number by another (e.g., $\frac{z_1}{z_2}$ ), we multiply both the numerator and the denominator by the conjugate of the denominator. This process is called "realising the denominator".

Example layout: $\frac{a+ib}{c+id} = \frac{(a+ib)}{(c+id)} \times \frac{(c-id)}{(c-id)}$ The denominator becomes $c^2 + d^2$ (a real number), and you simply expand the numerator using FOIL. Finally, you split the fraction into the standard $X + iY$ Cartesian form.

4. Square Roots of a Complex Number

To find the square roots of a complex number $a + ib$ , we cannot just push a button on a calculator. We must use an algebraic method. Let the square root be $x + iy$ (where $x, y \in \mathbb{R}$ ).

Set up the equation: $(x + iy)^2 = a + ib$
Expand the left side: $x^2 + 2xyi + i^2y^2 = a + ib$
Group real and imaginary parts: $(x^2 - y^2) + i(2xy) = a + ib$
Equate the real parts: $x^2 - y^2 = a$ (Equation 1)
Equate the imaginary parts: $2xy = b$ (Equation 2)

By solving these two simultaneous equations, you will find two pairs of $(x, y)$ values, representing the two square roots.

Simple Worked Example

Question: (a) Express $\frac{7 + i}{2 - i}$ in the form $x + iy$ . (b) Find the two square roots of $3 - 4i$ .

Solution: (a) Division Multiply numerator and denominator by the conjugate of the denominator ( $2 + i$ ): $\frac{7 + i}{2 - i} \times \frac{2 + i}{2 + i}$ Expand the numerator: $= \frac{14 + 7i + 2i + i^2}{2^2 + 1^2}$ $= \frac{14 + 9i - 1}{4 + 1}$ $= \frac{13 + 9i}{5}$ $= \frac{13}{5} + \frac{9}{5}i$

(b) Square Roots Let $(x + iy)^2 = 3 - 4i$ . Expanding: $x^2 - y^2 + 2xyi = 3 - 4i$ Equating parts:

$x^2 - y^2 = 3$
$2xy = -4 \implies y = -\frac{2}{x}$

Substitute (2) into (1): $x^2 - \left(-\frac{2}{x}\right)^2 = 3$ $x^2 - \frac{4}{x^2} = 3$ Multiply by $x^2$ : $x^4 - 3x^2 - 4 = 0$ This is a quadratic in $x^2$ . Factorizing: $(x^2 - 4)(x^2 + 1) = 0$ Since $x$ must be a real number, $x^2 = -1$ has no valid solutions. Therefore, $x^2 = 4 \implies x = 2$ or $x = -2$ .

Find corresponding $y$ values using $y = -\frac{2}{x}$ : If $x = 2$ , $y = -1$ . Root 1 is $2 - i$ . If $x = -2$ , $y = 1$ . Root 2 is $-2 + i$ .

The two square roots are $\pm(2 - i)$ .

mini-FAQ page

Q: Can I use De Moivre's theorem to find square roots instead of simultaneous equations? A: Yes! When we cover polar form in the next topic, you will learn a geometric way to find roots. However, if the complex number isn't easily convertible to an exact polar angle (like $3 - 4i$ ), the algebraic simultaneous equation method shown above is much faster and more accurate.

Q: Why do complex numbers always have two square roots? A: The Fundamental Theorem of Algebra states that a polynomial of degree $n$ has $n$ roots in the complex number system. The equation $z^2 = w$ is a polynomial of degree 2, so it must have exactly 2 roots.

Common mistakes to avoid

Forgetting the denominator in division: When expanding $\frac{(c-id)}{(c-id)}$ , students sometimes forget that $c^2 + d^2$ is the divisor for both the real and imaginary parts. Don't write $13 + 9i$ when the answer is $\frac{13}{5} + \frac{9}{5}i$ .
Accepting imaginary values for $x$ in square roots: In the equation $x^4 - 3x^2 - 4 = 0$ , you must discard the $x^2 = -1$ solution. The setup specifically defines $x$ and $y$ as real numbers. If you allow $x$ to be imaginary, the whole logical structure collapses.

Practice on Vu's Maths Hub

Division and square root calculations are guaranteed to appear in your HSC exams. Build your speed and accuracy with our resources:

Practice realising denominators and solving simultaneous root equations in the HSC Complex Numbers Booklet.
Try solving quadratic equations that have complex coefficients (requiring the square root method) in the HSC Polynomials Booklet.

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