If $\mathrm{z}$ is a point on the circle $|\mathrm{z}|=1$ with $\operatorname{Arg}(\mathrm{z})=\frac{\pi}{6}$, then $\frac{z^{12}+1-z^6}{z^{12}+i z^6-1}=$

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Aicharya · Accepted Answer

$|z|=1$ and $\operatorname{Arg}(z)=\frac{\pi}{6}$ $\begin{aligned} & \because z=|z| e^{i \operatorname{Arg}(z)}=i \cdot e^{i \frac{\pi}{6}}=e^{\frac{\pi}{6} i} \ & \Rightarrow z^6=e^{i \pi}=-1 	ext { and } z^{12}=e^{2 \pi i}=1\end{aligned}$ Now, $\frac{z^{12}+1-z^6}{z^{12}+i z^6-1}=\frac{1+1+1}{1+i(-1)-1}=\frac{3}{-i}=3 i$

If $\mathrm{z}$ is a point on the circle $|\mathrm{z}|=1$ with…

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