The value of $\left(\cos 45^{\circ}+\mathrm{i} \sin 45^{\circ}\right)^2=$

Question

Professor PC Thomas Classes · Answer

By De Moivre's theorem, $(\cos 	heta + i \sin 	heta)^n = \cos(n	heta) + i \sin(n	heta)$. So, $\cos(90^{\circ}) + i \sin(90^{\circ}) = 0 + i(1) = i$.