Lets analyse this Complex Numbers question from JPJC 2017 JC H2 Maths Prelim Paper 2.
Increasingly asked in many different prelim questions, this type of question, which usually contains the keywords “purely real” or “purely imaginary”.
The technique to solve this type of question is generic, as we will discuss in detail below.
Q4 (ii). The complex number \(z\) is given by $$z=-1+\text{i}c$$, where \(c\) is a non-zero real number.
Given that \(\frac{z^n}{z^*}\) is purely real, find the three smallest positive integer values of \(n\) when \(c=\sqrt{3}\).
Solution with explanation:
Step 1: To solve this type of question, we first find the argument and modulus of the complex number. This is to facilitate changing to exponential form.
\(z=-1+\text{i}\sqrt{3}\)\(|z|=2\), \(\arg(z)=\frac{2\pi}{3}\)
Step 2: Convert to exponential form, then polar form.
\(\frac{z^n}{z^*}=\frac{\left(2e^{\text{i}\frac{2\pi}{3}}\right)^n}{2e^{\text{i}\left(-\frac{2\pi}{3}\right)}}=2^{n-1}e^{\text{i}\left(\frac{2n\pi}{3}+\frac{2\pi}{3}\right)} =2^{n-1}e^{\text{i}\left(n+1\right)\frac{2\pi}{3}} = 2^{n-1}\left(\cos\left(n+1\right)\frac{2\pi}{3} + i \sin \left(n+1\right)\frac{2\pi}{3} \right)\)Notice that by doing this, the power \(n\) is now “stuck” inside the trigo function, instead of being in the power. This is exactly what we want.
Step 3: For purely real, we set imaginary part = 0.
Since\(\frac{z^n}{z^*}\) is purely real, Im(\(\frac{z^n}{z^*}\))\(=0\),
\(\sin\left[\left(n+1\right)\frac{2\pi}{3}\right]=0\)
**\((n+1)\frac{2\pi}{3}=k\pi, k \in \mathbb{Z}\)
\(n+1 = \frac{3k}{2}\)
We want the smallest 3 positive integers of \(n\), which corresponds to \(k=2,4,6\)
\(n+1=3, 6, 9\)
\(n = 2, 5, 8\)
**Note: We need to learn the generic solution for \(\sin \theta = 0\) and \(\cos \theta = 0\). To remember the following results, you just have to recall the graph of sine or cosine, and note down the positions of the zeros!
\(\sin \theta = 0\)
\(\theta =k\pi, k \in \mathbb{Z}\)
\(\cos \theta = 0\)
\(\theta =\frac{\pi}{2} + k\pi, k \in \mathbb{Z}\)
