Complex Number Question 9
Question 9 - 31 January - Shift 2
The complex number $z=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$ is equal
to:
(1) $\sqrt{2}(\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12})$
(2) $\cos \frac{\pi}{12}-i \sin \frac{\pi}{12}$
(3) $\sqrt{2}(\cos \frac{\pi}{12}+i \sin \frac{\pi}{12})$
(4) $\sqrt{2} i(\cos \frac{5 \pi}{12}-i \sin \frac{5 \pi}{12})$
Show Answer
Answer: (1)
Solution:
Formula: Polar form, Properties of conjugate
$ \begin{aligned} & z=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}=\frac{i-1}{\frac{1}{2}+\frac{\sqrt{3}}{2} i} \\ & =\frac{i-1}{\frac{1}{2}+\frac{\sqrt{3}}{2} i} \times \frac{\frac{1}{2}-\sqrt{\frac{3}{2}} i}{\frac{1}{2}-\sqrt{3 / 2} i}=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{2} i \end{aligned} $
Apply polar form,
$r \cos \theta=\frac{\sqrt{3}-1}{2}$
$r \sin \theta=\frac{\sqrt{3}+1}{2}$
Now, $\tan \theta=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
$ \theta=\frac{5 \pi}{12} $
