Question

Express ${\left(\frac{1-i}{1+i}\right)}^{1000}$ in the form of $a+ib$. Find $a+b$

Answer 1

$\to {\left(\frac{1-i}{1+i}\right)}^{1000}$

$\to {\left(\frac{(1-i)(1-i)}{(1+i)(1-i)}\right)}^{1000}={\left(\frac{1+i^2-2i}{1+i^2}\right)}^{1000}={\left(\frac{-2i}{2}\right)}^{1000}={(-i)}^{1000}$

$\to {(-i)}^{1000}=i^{1000}={\left(i^4\right)}^{250}=1=a+ib$

$a=1,b=0,a+b=1$