Question

calcule : (1+i/1-i)^102

Answer 1

$\left(\dfrac{1+i}{1-i}\right)^{102}=\\\\\left(\dfrac{1+i}{1-i}\cdot\dfrac{1+i}{1+i}\right)^{102}=\\\\\left(\dfrac{(1+i)^2}{1^2-i^2}\right)^{102}=\\\\\left(\dfrac{1+2i-1}{1+1}\right)^{102}=\\\\\left(\dfrac{2i}{2}\right)^{102}=\\\\i^{102}\\\\102\div4=25\ R:2\\\\i^{102}=i^2=-1$