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(B) Let $t = z^4$. The equation becomes $t^2 - 20t + 64 = 0$. Factoring the quadratic equation,we get $(t-16)(t-4) = 0$. So,$z^4 = 16$ or $z^4 = 4$. F

Vedclass · Accepted Answer

$-12$

Vedclass · Answer

(B) Let $t = z^4$. The equation becomes $t^2 - 20t + 64 = 0$. Factoring the quadratic equation,we get $(t-16)(t-4) = 0$. So,$z^4 = 16$ or $z^4 = 4$. For $z^4 = 16$,the roots are $z = 2, -2, 2i, -2i$. The imaginary roots are $2i$ and $-2i$. For $z^4 = 4$,the roots are $z = \sqrt{2}, -\sqrt{2}, \sqrt{2}i, -\sqrt{2}i$. The imaginary roots are $\sqrt{2}i$ and $-\sqrt{2}i$. The squares of the imaginary roots are $(2i)^2 = -4$,$(-2i)^2 = -4$,$(\sqrt{2}i)^2 = -2$,and $(-\sqrt{2}i)^2 = -2$. The sum of the squares of the imaginary roots is $(-4) + (-4) + (-2) + (-2) = -12$.

Sum of the squares of the imaginary roots of the equation $z^8-20z^4+64=0$ is

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