If Z = r(cos theta + i sin theta), (theta neq | vedclass

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(C) Given that $Z = r(\cos 	heta + i \sin 	heta)$ is a solution of $x^3 = i$. By De Moivre's Theorem,$Z^3 = r^3(\cos 3	heta + i \sin 3	heta)$. Sin

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$-i$

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(C) Given that $Z = r(\cos 	heta + i \sin 	heta)$ is a solution of $x^3 = i$. By De Moivre's Theorem,$Z^3 = r^3(\cos 3	heta + i \sin 3	heta)$. Since $Z^3 = i$,we have $r^3(\cos 3	heta + i \sin 3	heta) = i$. We know that $i = \cos(\pi / 2) + i \sin(\pi / 2)$. Thus,$r^9(\cos 	heta + i \sin 	heta)^9 = (r^3(\cos 	heta + i \sin 	heta)^3)^3$. Since $Z^3 = i$,this expression becomes $(i)^3$. Calculating the value: $i^3 = i^2 	imes i = -1 	imes i = -i$.

If $Z = r(\cos \theta + i \sin \theta), (\theta \neq -\pi / 2)$ is a solution of $x^3 = i$,then $r^9(\cos \theta + i \sin \theta)^9 = $

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