(-1+i 3)^ 15 (1-i)^ 20 + (-1-i 3)^ 15 (1+i)^ 20 is equal to

MCQ

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

✓
$-64$
B
$-32$
C
$-16$
D
$\frac{1}{16}$

✓

Answer

Correct option: A.

$-64$

(A)
$2^{15}\left[\frac{\left(-\frac{1}{2}+\frac{ i \sqrt{3}}{2}\right)^{15}}{(1- i )^{20}}+\frac{\left(\frac{-1}{2}-\frac{ i \sqrt{3}}{2}\right)^{15}}{(1+ i )^{20}}\right]$
$=2^{15}\left[\frac{\omega^{15}}{(1- i )^{20}}+\frac{\omega^{30}}{(1+ i )^{20}}\right]$
$=2^{15}\left[\frac{1}{(1-i)^{20}}+\frac{1}{(1+i)^{20}}\right]$
$=2^{15}\left[\frac{(1+i)^{20}+(1-i)^{20}}{\left(1-i^2\right)^{20}}\right]$
$=\frac{2^{15}}{2^{20}}\left[(1+i)^{20}+(1-i)^{20}\right]$
$=\frac{1}{2^5}\left[\left\{(1+i)^2\right\}^{10}+\left\{(1-i)^2\right\}^{10}\right]$
$=\frac{1}{2^5}\left[(2 i )^{10}+(-2 i )^{10}\right]=\frac{2^{11} \cdot 1^{10}}{2^5}$
$=-2^6 \quad \ldots\left[\because i ^{10}=\left( i ^4\right)^2 \cdot i ^2= i ^2=-1\right]$
$=-64$

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