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9231 · 2.5

Complex numbers — common mistakes

Common exam mistakes on 9231 Complex numbers. Learn what loses marks, then practise the topic with Examiner’s Ink.

Exam tip 1

Always add 2kπ2k\pi to the argument before dividing by nn. When stating your final answers, double-check the required range for the argument, typically (π,π](-\pi, \pi], and adjust your angles if necessary by adding or subtracting multiples of 2π2\pi.

What's the difference between finding the roots of unity and the roots of any other complex number?

The roots of unity are the solutions to zn=1z^n = 1. Since 1=1|1|=1 and arg(1)=0\arg(1)=0, the roots of unity all have a modulus of 11/n=11^{1/n}=1 and are located on the unit circle. The method for finding roots of any complex number w=reiθw = re^{i\theta} is identical, but the roots will lie on a circle of radius r1/nr^{1/n} and will be rotated by an angle of θ/n\theta/n compared to the roots of unity.

How do I know whether to use the real or imaginary part when summing a series?

If your target series involves cosines (e.g., cos(kθ)\sum \cos(k\theta)), you will take the real part of the complex sum. If your target series involves sines (e.g., sin(kθ)\sum \sin(k\theta)), you will take the imaginary part. Always define your target series as CC or SS at the start and form the complex sum C+iSC+iS.

When expressing $\sin^n\theta$, what happens if $n$ is odd or even?

If nn is even, the binomial expansion of (zz1)n(z-z^{-1})^n will have an odd number of terms. The terms will pair up to form cosine terms, e.g., zk+zk=2cos(kθ)z^k+z^{-k}=2\cos(k\theta). If nn is odd, the expansion will have an even number of terms, and they will pair up to form sine terms, e.g., zkzk=2isin(kθ)z^k-z^{-k}=2i\sin(k\theta). Pay close attention to the powers of ii from (2isinθ)n(2i\sin\theta)^n.