Question

$n\in N,{\left(\frac{1+i}{\sqrt{2}}\right)}^{8n}+{\left(\frac{1-i}{\sqrt{2}}\right)}^{8n}=$

• A. $0$
• B. $1$
• C. $2$
• D. $-2$

Answer 1

Answer: (C)

$2$

We have,

${\left(\frac{1+i}{\sqrt{2}}\right)}^{8n}+{\left(\frac{1-i}{\sqrt{2}}\right)}^{8n}$

${\left[{\left(\frac{1+i}{\sqrt{2}}\right)}^2\right]}^{4n}+{\left[{\left(\frac{1-i}{\sqrt{2}}\right)}^2\right]}^{4n}$

$={\left[\frac{1^2+i^2+2i}{2}\right]}^{4n}+{\left[\frac{1^2+i^2-2i}{2}\right]}^{4n}$

$={\left[\frac{1-1+2i}{2}\right]}^{4n}+{\left[\frac{1-1-2i}{2}\right]}^{4n}$

$={\left(i^4\right)}^n+{\left({(-i)}^4\right)}^n$

$=1^n+1^n$

$=1+1$

$=2$

Hence, this is the answer.