Question

The value of ${\left(\frac{1+i}{\sqrt{2}}\right)}^{8n}+{\left(\frac{1-i}{\sqrt{2}}\right)}^{8n}$ is, where $n\in N$

• A. $0$
• B. $1$
• C. $2$
• D. $-2$

Answer 1

Answer: (C)

$2$

We have,

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

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

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

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

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

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

$={\left(i^4\right)}^n+{(-i^4)}^n\therefore i^4=1$

$=1^n+1^n$

$=1+1=2$

Hence, this is the answer.