Question

(1+iкорень из 3)^2003=?

Answer 1

$1+isqrt3=2left(sindfracpi6+icosdfracpi6right)=2e^{ipi/6}\ (2e^{ipi/6})^{2003}=2^{2003}cdot e^{icdot2003pi/6}=2^{2003}e^{334pi i-ipi/6}=2^{2003}(isqrt3-1)$