Evaluate (1 + i)^12

First convert to mod-arg form in order to use de Moivre's theorem. |1 + i| = (1^2 + 1^2)^1/2 = 2^1/2 arg(z) is the angle made by the vector of the complex number and the positive real axis. I world recommend by always drawing a diagram arg(1+i) = tan^-1(1/1) = tan^-1(1) = π /4Therefor 1+i = 2^1/2(cos(π /4) +isin(π /4))Now use de Maivre's theorem to raise to the power of 12(1+i)^12 = (2^1/2)^12(cos(π /4) + isin(π /4))^12 =2^6(cos(12π /4) + isin(12π /4))= 64 (cos(3π ) + isin(3π )) As 2π is a full rotation is can be equated to zero.= 64(cosπ + isinπ ) = 64(-1)=-64