Search any question & find its solution
Question:
Answered & Verified by Expert
If $\mathrm{z}$ is a point on the circle $|\mathrm{z}|=1$ with $\operatorname{Arg}(\mathrm{z})=\frac{\pi}{6}$, then
$\frac{z^{12}+1-z^6}{z^{12}+i z^6-1}=$
Options:
$\frac{z^{12}+1-z^6}{z^{12}+i z^6-1}=$
Solution:
2889 Upvotes
Verified Answer
The correct answer is:
$1 / 2$
$|z|=1$ and $\operatorname{Arg}(z)=\frac{\pi}{6}$
$\begin{aligned} & \because z=|z| e^{i \operatorname{Arg}(z)}=i \cdot e^{i \frac{\pi}{6}}=e^{\frac{\pi}{6} i} \\ & \Rightarrow z^6=e^{i \pi}=-1 \text { and } z^{12}=e^{2 \pi i}=1\end{aligned}$
Now, $\frac{z^{12}+1-z^6}{z^{12}+i z^6-1}=\frac{1+1+1}{1+i(-1)-1}=\frac{3}{-i}=3 i$
$\begin{aligned} & \because z=|z| e^{i \operatorname{Arg}(z)}=i \cdot e^{i \frac{\pi}{6}}=e^{\frac{\pi}{6} i} \\ & \Rightarrow z^6=e^{i \pi}=-1 \text { and } z^{12}=e^{2 \pi i}=1\end{aligned}$
Now, $\frac{z^{12}+1-z^6}{z^{12}+i z^6-1}=\frac{1+1+1}{1+i(-1)-1}=\frac{3}{-i}=3 i$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.