The product of the distinct (2n)^{text{th}} r | vedclass

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Question

(B) Let $z = 1+i\sqrt{3}$. We want to find the product of the $(2n)^{	ext{th}}$ roots of $z$.Let the roots be $w_0, w_1, \dots, w_{2n-1}$.The product

Vedclass · Accepted Answer

$-1-i\sqrt{3}$

Vedclass · Answer

(B) Let $z = 1+i\sqrt{3}$. We want to find the product of the $(2n)^{	ext{th}}$ roots of $z$.Let the roots be $w_0, w_1, \dots, w_{2n-1}$.The product of the roots of the equation $w^{2n} - z = 0$ is given by $(-1)^{2n-1} 	imes (	ext{constant term})$.Here,the constant term is $-z$.So,the product $P = (-1)^{2n-1} (-z) = (-1) 	imes (-z) = z$.Wait,let us re-evaluate: The equation is $w^{2n} - z = 0$.The product of the roots is $(-1)^{2n} 	imes (-z) = 1 	imes (-z) = -z$.Thus,the product is $-(1+i\sqrt{3}) = -1-i\sqrt{3}$.

The product of the distinct $(2n)^{\text{th}}$ roots of $1+i\sqrt{3}$ is equal to:

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