Question

$\frac{{(-1+i\sqrt{3})}^{12}}{{(1-i)}^{20}}$+$\frac{{(-1+i\sqrt{3})}^{12}}{{(1+i)}^{20}}$ is equal to

Answer 1

$\begin{array}{c}\frac{{\left(-1+i\sqrt{3}\right)}^{12}}{{(1-i)}^{20}}+\frac{{\left(-1+i\sqrt{3}\right)}^{12}}{{\left(1+i\right)}^{20}}\\ ={\left(-1+\sqrt{3}i\right)}^{12}\left[\frac{1}{{\left(2\right)}^{10}{\left(e^{-i\frac{\pi }{4}}\right)}^{20}}+\frac{1}{2^{10}{\left(e^{i\frac{\pi }{4}}\right)}^{20}}\right]\\ =\frac{{\left(-1+\sqrt{3}i\right)}^{12}}{2^{10}}\left[\frac{1}{e^{-i5\pi }}+\frac{1}{e^{i5\pi }}\right]\\ =\frac{{\left(-1+\sqrt{3}i\right)}^{12}}{2^9}\\ =-\frac{1}{2^9}\times 2^{12}{\left(-\frac{1}{2}+\frac{\sqrt{3}i}{2}\right)}^{12}\\ =-8\times {\left(e^{i2\frac{\pi }{3}}\right)}^{12}\\ =-8\times e^{i8\pi }\\ =-8\times 1=-8\end{array}$