The product of the four values of (1+i sqrt{3 | vedclass

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(D) Let $z = (1+i \sqrt{3})^{3/4}$.We know that for a complex number $w = z^n$,where $n = p/q$,the product of the $q$ values is given by $(-1)^{q-1} (

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Let $z = (1+i \sqrt{3})^{3/4}$.We know that for a complex number $w = z^n$,where $n = p/q$,the product of the $q$ values is given by $(-1)^{q-1} (z^p)$.Alternatively,let $z = r^{3/4} e^{i(3	heta/4 + 3k\pi/2)}$ for $k = 0, 1, 2, 3$.The product of these four values is $P = \prod_{k=0}^{3} r^{3/4} e^{i(3	heta/4 + 3k\pi/2)} = (r^{3/4})^4 e^{i \sum_{k=0}^{3} (3	heta/4 + 3k\pi/2)}$.Here,$r = |1+i\sqrt{3}| = \sqrt{1^2 + (\sqrt{3})^2} = 2$.So,$r^3 = 2^3 = 8$.The product is $8 	imes e^{i(3	heta + 3\pi/2(0+1+2+3))} = 8 	imes e^{i(3	heta + 9\pi)}$.Since $e^{i9\pi} = -1$,the product is $8 	imes e^{i3	heta} 	imes (-1) = -8(e^{i	heta})^3$.Given $1+i\sqrt{3} = 2e^{i\pi/3}$,we have $e^{i	heta} = e^{i\pi/3}$.Thus,$(e^{i	heta})^3 = e^{i\pi} = -1$.Therefore,the product is $-8 	imes (-1) = 8$.

The product of the four values of $(1+i \sqrt{3})^{3/4}$ is

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