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 = \(2 cos ⁡\frac{5\pi}{4}\).