frac{{{{( - 1 + isqrt 3 )}^{15}}}}{{{{(1 - i) | 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 becomes $

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 becomes $\frac{(2\omega)^{15}}{(1-i)^{20}} + \frac{(2\omega^2)^{15}}{(1+i)^{20}}$.Since $\omega^3 = 1$,$\omega^{15} = 1$ and $\omega^{30} = 1$.So,the expression is $\frac{2^{15}}{(1-i)^{20}} + \frac{2^{15}}{(1+i)^{20}} = 2^{15} \left[ \frac{(1+i)^{20} + (1-i)^{20}}{(1-i^2)^{20}} ight]$.Note that $(1-i^2) = (1 - (-1)) = 2$,so $(1-i^2)^{20} = 2^{20}$.Thus,$\frac{2^{15}}{2^{20}} [(1+i)^{20} + (1-i)^{20}] = \frac{1}{2^5} [(1+i)^2]^{10} + [(1-i)^2]^{10}$.$(1+i)^2 = 1 + i^2 + 2i = 2i$ and $(1-i)^2 = 1 + i^2 - 2i = -2i$.Substituting these,we get $\frac{1}{32} [(2i)^{10} + (-2i)^{10}] = \frac{1}{32} [2^{10} i^{10} + 2^{10} i^{10}] = \frac{1}{32} [2 	imes 2^{10} 	imes (-1)]$.$= \frac{1}{32} [-2^{11}] = -\frac{2048}{32} = -64$.

$\frac{{{{( - 1 + i\sqrt 3 )}^{15}}}}{{{{(1 - i)}^{20}}}} + \frac{{{{( - 1 - i\sqrt 3 )}^{15}}}}{{{{(1 + i)}^{20}}}} = \dots$

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