Question

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

• A. $32$
• B. $12$
• C. $43$
• D. $10$

Answer 1

The correct option is A $32$

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

$=\frac{1}{{\left(\sqrt{8}\right)}^8}[{(1+i)}^8+{(1-i)}^8]$

$=\frac{1}{{\left(2^{\frac{1}{2}}\right)}^8}[{\left({(1+i)}^2\right)}^4+{\left({(1-i)}^2\right)}^4]$

$=\frac{1}{16}[{(1-1+2i)}^4+{(1-1-2i)}^4]$

$=\frac{1}{16}[{\left(2i\right)}^4+{(-2i)}^4]$

$=\frac{16}{16}\left(2{\left(2i\right)}^4\right)$

$=32\times {(-1)}^2=32$