QCE Specialist Mathematics - Unit 3 - Further complex numbers

Complex Arithmetic Using Polar Form | QCE Specialist Mathematics

Learn Specialist Mathematics polar complex arithmetic, modulus-argument identities and De Moivre's theorem for integral powers.

Updated 2026-05-18 - 6 min read

Cartesian form is useful for adding complex numbers. Polar form is useful for multiplying, dividing and taking powers because it separates a complex number into a size and a rotation.

If

$ z=x+yi $

then its modulus is

$ |z|=\sqrt{x^2+y^2} $

and its argument is the angle the vector from the origin to $z$ makes with the positive real axis. In polar form,

$ z=r\operatorname{cis}(\theta)=r(\cos\theta+i\sin\theta). $

Cartesian recap

Before polar form, make sure the Cartesian basics are secure. For:

$ z=a+bi, $

the real part is $\operatorname{Re}(z)=a$ and the imaginary part is $\operatorname{Im}(z)=b$. The imaginary part is the real coefficient of $i$, not $bi$ itself.

Two complex numbers are equal only when both parts match:

$ a+bi=c+di \quad\Longleftrightarrow\quad a=c\text{ and }b=d. $

The conjugate is:

$ \bar z=a-bi. $

Conjugates distribute over addition and multiplication:

$ \overline{z+w}=\bar z+\bar w,\quad \overline{zw}=\bar z\bar w. $

Division in Cartesian form uses the conjugate of the denominator:

$ \frac{3+2i}{1-i} =\frac{(3+2i)(1+i)}{(1-i)(1+i)} =\frac{1+5i}{2}. $

That same calculation is the Cartesian version of the polar fact that division subtracts arguments.

Why polar form helps

When you multiply complex numbers, the moduli multiply and the arguments add:

$ r_1\operatorname{cis}(\theta_1)\cdot r_2\operatorname{cis}(\theta_2) =r_1r_2\operatorname{cis}(\theta_1+\theta_2). $

That is why polar form feels so much cleaner for powers. Multiplication is just "scale, then rotate".

Division works the same way in reverse:

$ \frac{r_1\operatorname{cis}(\theta_1)}{r_2\operatorname{cis}(\theta_2)} =\frac{r_1}{r_2}\operatorname{cis}(\theta_1-\theta_2), \quad r_2\ne0. $

So if a question is full of products, quotients or powers, convert to polar form before doing the heavy lifting. If it is full of addition or subtraction, Cartesian form is usually easier because real parts and imaginary parts combine directly.

Modulus and argument identities

Specialist expects you to prove and use identities such as:

$ z\bar z=|z|^2 $

$ |z_1z_2|=|z_1||z_2| $

$ \arg(z_1z_2)=\arg(z_1)+\arg(z_2). $

For example, if $z=x+yi$, then $\bar z=x-yi$, so

$ z\bar z=(x+yi)(x-yi)=x^2+y^2=|z|^2. $

This is more than algebra. It tells you that multiplying a complex number by its conjugate removes direction and leaves the square of the length.

Arguments need a little care. The principal argument is usually chosen in a set interval such as $(-\pi,\pi]$, but the same complex number also has arguments $\theta+2k\pi$. When proving identities, write the geometric relationship first, then adjust the final angle into the required principal range if the question asks for it.

For example, if one argument is $\frac{3\pi}{4}$ and another is $\frac{2\pi}{3}$, the sum is $\frac{17\pi}{12}$. As a general argument that is fine, but as a principal argument it would be written as:

$ \frac{17\pi}{12}-2\pi=-\frac{7\pi}{12}. $

De Moivre's theorem

For integer $n$,

$ \left[r\operatorname{cis}(\theta)\right]^n=r^n\operatorname{cis}(n\theta). $

So powers multiply the angle by $n$. If $z=2\operatorname{cis}\left(\frac{\pi}{6}\right)$, then

$ z^3=8\operatorname{cis}\left(\frac{\pi}{2}\right)=8i. $

Choosing the form

The quickest method is usually decided before you start calculating. Use Cartesian form for $z_1+z_2$, $z_1-z_2$ and conjugates. Use polar form for $z_1z_2$, $\frac{z_1}{z_2}$, $z^n$ and roots. Many harder questions deliberately ask you to move between the forms: sketch the point in the Argand plane, find its modulus, choose the correct quadrant for the argument, and then apply the polar rule.

For a number $z=x+yi$, the quadrant matters because:

$ \tan(\theta)=\frac{y}{x} $

does not by itself tell you the correct angle. A negative real part and positive imaginary part places $z$ in quadrant II, not quadrant IV. This is one of the most common places where a correct calculation gets the wrong final angle.

Argand plane background

The Argand plane quietly supports almost every complex-number technique. The complex number $a+bi$ is plotted at $(a,b)$, the modulus is the distance from the origin, and the argument is the direction from the positive real axis. The origin has no argument, because there is no direction to measure.

Conditions involving modulus often describe circles or discs. For example:

$ |z-z_0|=r $

means all points at distance $r$ from $z_0$, so it is a circle centred at $z_0$. The inequality:

$ |z-z_0|\le r $

means the filled-in disc, including the boundary. A strict inequality such as $|z-z_0|<r$ excludes the circle itself.

Argument conditions describe rays or sectors. The condition:

$ \arg(z-z_0)=\theta $

is a ray starting at $z_0$ and heading at angle $\theta$. The starting point is excluded because $\arg(0)$ is undefined. A condition like $\alpha\le\arg(z-z_0)\le\beta$ describes the sector swept anticlockwise from $\alpha$ to $\beta$.

Useful identities beyond the headline

The modulus and argument rules have a few companion identities that are worth knowing:

$ |z|=|\bar z|,\quad \arg(\bar z)=-\arg(z),\quad z\bar z=|z|^2. $

The last one gives a compact way to find reciprocals:

$ \frac1z=\frac{\bar z}{|z|^2},\quad z\ne0. $

That is the same idea as multiplying by a conjugate in Cartesian division. In polar form, the reciprocal is even cleaner:

$ \left(r\operatorname{cis}\theta\right)^{-1}=\frac1r\operatorname{cis}(-\theta). $

The triangle inequality is another geometric result that occasionally appears:

$ |z_1+z_2|\le |z_1|+|z_2|. $

It says the direct route cannot be longer than travelling along two sides of a triangle. You usually will not need a full proof, but the picture is helpful when modulus expressions are being interpreted as distances.

De Moivre details and large powers

De Moivre's theorem works for every integer power. Negative powers are just reciprocals:

$ \left(r\operatorname{cis}\theta\right)^{-n} =r^{-n}\operatorname{cis}(-n\theta). $

For example:

$ \left(2\operatorname{cis}\frac{\pi}{6}\right)^{-3} =\frac18\operatorname{cis}\left(-\frac{\pi}{2}\right) =-\frac18 i. $

Do not use root logic for negative or fractional-looking powers unless the question is actually asking for roots. The expression $z^{1/3}$ is multi-valued in complex numbers unless a principal root is specified, while $z^{-3}$ is a single reciprocal power.

Large powers are usually easiest after converting to polar form and reducing the angle. If:

$ z=\sqrt2\operatorname{cis}\left(\frac{3\pi}{4}\right), $

then:

$ z^{20}=(\sqrt2)^{20}\operatorname{cis}(15\pi) =2^{10}\operatorname{cis}(\pi) =-1024. $

For powers of $i$, the cycle is even faster:

$ i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1. $

Reduce the exponent modulo $4$.

More Argand loci

Argand loci are just geometric conditions written in complex notation. The condition:

$ r_1<|z-z_0|\le r_2 $

describes an annulus centred at $z_0$: outside the circle of radius $r_1$, on or inside the circle of radius $r_2$. Strict inequalities exclude the boundary; non-strict inequalities include it.

The condition:

$ |z-z_1|=|z-z_2| $

means the point $z$ is equally far from $z_1$ and $z_2$, so the locus is the perpendicular bisector of the segment joining those two points.

The condition:

$ |z-z_1|=k|z-z_2| $

compares distances from two fixed points. When $k=1$, it is a line. When $k\ne1$, it is usually a circle after writing $z=x+iy$ and expanding.

For miscellaneous loci, set $z=x+iy$ and translate the complex condition into Cartesian algebra. For example:

$ \operatorname{Re}(z^2)=3 $

becomes:

$ \operatorname{Re}((x+iy)^2)=x^2-y^2=3. $

So the locus is a hyperbola. This method is slower than recognising circles and rays, but it works when the expression is not a standard distance or argument condition.

Quick check

Sources

• QCAA Specialist Mathematics subject page
• QCAA Specialist Mathematics 2025 syllabus
• OpenStax Algebra and Trigonometry 2e: Polar Form of Complex Numbers
• OpenStax Precalculus: Polar Form of Complex Numbers