Question

$\frac{(-1+i\sqrt{3}{)}^{15}}{(1-i{)}^{20}}+\frac{(-1-i\sqrt{3}{)}^{15}}{(1+i{)}^{20}}$ is equal to

• A. $132$
• B. $64$
• C. $-64$
• D. $-132$

Answer 1

The correct option is C $-64$

We have,
$\frac{(-1+i\sqrt{3}{)}^{15}}{(1-i{)}^{20}}+\frac{(-1-i\sqrt{3}{)}^{15}}{(1+i{)}^{20}}=[\frac{{\left(\frac{-1+i\sqrt{3}}{2}\right)}^{15}}{{\left(\frac{1-i}{\sqrt{2}}\right)}^{20}}+\frac{{\left(\frac{-1-i\sqrt{3}}{2}\right)}^{15}}{{\left(\frac{1+i}{\sqrt{2}}\right)}^{20}}]\times 2^5$

[Dividing and multiplying both terms by $2^{15}$]

$=\frac{e^{i\frac{2\pi }{3}\times 15}}{e^{i(2\pi -\pi /4)\times 20}}+\frac{e^{i(\pi +\pi /3)\times 15}}{e^{i\pi /4\times 20}}$

$=\frac{e^{i\left(10\right)\pi }}{e^{i\left(35\right)\pi }}+\frac{e^{i(4\pi \times 5)}}{e^{i5\pi }}{e}^{ik\pi }=\{\begin{array}{cc}1;& k=even\\ -1;& k=odd\end{array}$

$=(-1-1){.2}^5=-2^6$
$=(-64)$