Question

If $a=\frac{1}{\sqrt 2}+\frac{i}{\sqrt 2}$, find the value of a^5 + conjugate of a^5=?

(a) $cos⁡ \frac{3\pi}{4}$

(b) $2 sin ⁡\frac{5\pi}{4}$

(c) $2 cos \frac{5\pi}{4}$

(d) $cos \frac{5\pi}{4}$

I had been asked this question in an interview.

This question is from DeMoivre’s Theorem in portion Complex Numbers of Engineering Mathematics

Answer 1

Correct option is (c) $2 cos \frac{5\pi}{4}$

Easy explanation: We know that,

$a=(\frac{1}{\sqrt 2}+\frac{i}{\sqrt 2})=(cos \frac{\pi}{4}+isin \frac{\pi}{4})$

Conjugate of a = $(\frac{1}{\sqrt 2}-\frac{i}{\sqrt 2})=(cos \frac{\pi}{4}-isin \frac{\pi}{4})$

$a^5$ + (conj of a)^5=$(cos \frac{\pi}{4}+isin \frac{\pi}{4})^5+(cos \frac{\pi}{4}-isin \frac{\pi}{4})^5$

By Applying the DeMoivre’s Theorem

a^5 + (conj of a)^5 = $cos⁡ \frac{5\pi}{4} +isin \frac{5\pi}{4}+cos⁡ \frac{5\pi}{4} -isin \frac{5\pi}{4}$

a^5 + (conj of a)^5 = $2 cos ⁡\frac{5\pi}{4}$.