If z is a point on the circle |z|=1 with oper | vedclass

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(C) Given $|z|=1$ and $\operatorname{Arg}(z)=\frac{\pi}{6}$.$z = |z|e^{i \operatorname{Arg}(z)} = 1 \cdot e^{i \frac{\pi}{6}} = e^{i \frac{\pi}{6}}$.T

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$3i$

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(C) Given $|z|=1$ and $\operatorname{Arg}(z)=\frac{\pi}{6}$.$z = |z|e^{i \operatorname{Arg}(z)} = 1 \cdot e^{i \frac{\pi}{6}} = e^{i \frac{\pi}{6}}$.Then $z^6 = (e^{i \frac{\pi}{6}})^6 = e^{i \pi} = -1$.And $z^{12} = (e^{i \frac{\pi}{6}})^{12} = e^{2 \pi i} = 1$.Substituting these values into the expression:$\frac{z^{12}+1-z^6}{z^{12}+i z^6-1} = \frac{1+1-(-1)}{1+i(-1)-1} = \frac{3}{-i} = \frac{3}{-i} \cdot \frac{i}{i} = \frac{3i}{-i^2} = \frac{3i}{1} = 3i$.

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

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