Question

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

• A. $2i$
• B. $-2i$
• C. $-2$
• D. $2$

Answer 1

Answer: (C)

$-2$

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

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

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