frac{(-1 + isqrt{3})^{15}}{(1 - i)^{20}} + fr | vedclass

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(A) Let $z_1 = -1 + i\sqrt{3} = 2\omega$ and $z_2 = -1 - i\sqrt{3} = 2\omega^2$,where $\omega$ is the cube root of unity. Then the expression is $E =

Vedclass · Accepted Answer

$-64$

Vedclass · Answer

(A) Let $z_1 = -1 + i\sqrt{3} = 2\omega$ and $z_2 = -1 - i\sqrt{3} = 2\omega^2$,where $\omega$ is the cube root of unity. Then the expression is $E = \frac{(2\omega)^{15}}{(1 - i)^{20}} + \frac{(2\omega^2)^{15}}{(1 + i)^{20}}$. Since $\omega^3 = 1$,$\omega^{15} = 1$ and $\omega^{30} = 1$. $E = \frac{2^{15}}{(1 - i)^{20}} + \frac{2^{15}}{(1 + i)^{20}} = 2^{15} \left[ \frac{(1 + i)^{20} + (1 - i)^{20}}{(1 - i^2)^{20}} ight]$. Since $1 - i^2 = 2$,the denominator is $2^{20}$. $E = \frac{2^{15}}{2^{20}} [(1 + i)^{20} + (1 - i)^{20}]$. Note that $(1 + i)^2 = 1 + i^2 + 2i = 2i$ and $(1 - i)^2 = 1 + i^2 - 2i = -2i$. $E = \frac{1}{2^5} [(2i)^{10} + (-2i)^{10}] = \frac{1}{32} [2^{10} i^{10} + 2^{10} i^{10}] = \frac{2 	imes 2^{10} 	imes (-1)}{32} = \frac{-2^{11}}{2^5} = -2^6 = -64$.

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

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