Question

The value of ${\left(\frac{1-i}{1+i}\right)}^{10}+{\left(\frac{1+i}{1-i}\right)}^8=$

• A. $0$
• B. $2i$
• C. $i$
• D. $-i$

Answer 1

The correct option is A $0$

$\frac{1-i}{1+i}=\frac{(1-i{)}^2}{1^2-i^2}=\frac{1-1-2i}{2}=-i$

$\frac{1+i}{1-i}=\frac{(1+i{)}^2}{1^2-i^2}=\frac{1-1+2i}{2}=i$

$\therefore {\left(\frac{1-i}{1+i}\right)}^{10}+{\left(\frac{1+i}{1-i}\right)}^8$

$=(-i{)}^{10}+(i{)}^8$

$=i^{10}+i^8$

$=-1+1$ - ($i^{4n-2}=-1$ & $i^{4n}=8$)

$=0$