If z is a non-real complex number, then the m | vedclass

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(B) Let $z = r(\cos 	heta + i \sin 	heta) = r e^{i 	heta}$, where $\operatorname{Im}(z) = r \sin 	heta$.Then $z^5 = r^5(\cos 5	heta + i \sin 5	h

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$-4$

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(B) Let $z = r(\cos 	heta + i \sin 	heta) = r e^{i 	heta}$, where $\operatorname{Im}(z) = r \sin 	heta$.Then $z^5 = r^5(\cos 5	heta + i \sin 5	heta)$, so $\operatorname{Im}(z^5) = r^5 \sin 5	heta$.The expression becomes $\frac{\operatorname{Im}(z^5)}{(\operatorname{Im} z)^5} = \frac{r^5 \sin 5	heta}{(r \sin 	heta)^5} = \frac{\sin 5	heta}{\sin^5 	heta}$.Using the identity $\sin 5	heta = 16 \sin^5 	heta - 20 \sin^3 	heta + 5 \sin 	heta$, we get:$\frac{\sin 5	heta}{\sin^5 	heta} = \frac{16 \sin^5 	heta - 20 \sin^3 	heta + 5 \sin 	heta}{\sin^5 	heta} = 16 - 20 \csc^2 	heta + 5 \csc^4 	heta$.Let $x = \csc^2 	heta$. Since $z$ is non-real, $\sin 	heta 
eq 0$, so $x \geq 1$.The expression is $f(x) = 5x^2 - 20x + 16$.This is a parabola opening upwards with vertex at $x = -\frac{b}{2a} = -\frac{-20}{2(5)} = 2$.Since $x=2$ is in the domain $[1, \infty)$, the minimum value is $f(2) = 5(2)^2 - 20(2) + 16 = 20 - 40 + 16 = -4$.

If $z$ is a non-real complex number, then the minimum value of $\frac{\operatorname{Im}(z^5)}{(\operatorname{Im} z)^5}$ is

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