Question

If n is be a positive integer, then $(1+i{)}^n+(1+i{)}^n=2^{k}cos\frac{n\pi }{4},$ where k is equald to

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$(cos\theta +isin\theta {)}^2=cosn\theta +isin4\theta$

Solution : $(1+i{)}^n+(1-i{)}^n=2^{k}cos\frac{n\pi }{4}\mid$

L.H.S.$=(1+i{)}^n+(1-i{)}^n$ $(\sqrt{2}{)}^n\left(\right(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}{)}^n+(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}{)}^n)$

$(\sqrt{2}{)}^n(cos\frac{n\pi }{4}+isin\frac{n\pi }{4}+cos\frac{n\pi }{4}-isin\frac{n\pi }{4})$

$=2^{\frac{n}{2}+1}cos\frac{n\pi }{4}$

$\Rightarrow k=\frac{n}{2}+1$