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Let $n \in \mathbb{N}^*$ and $k \in \{0, 1, 2, ..., n - 1\}$. $$z = \left( \cot \frac{(2k + 1)\pi}{2n} + i \right)^n$$

Find the real part of $z$.

I think we need to transform $z$ into something like $\cos \phi + i \sin \phi$, so we can compute its n-th power using De Moivre's rule. However, I haven't figured out any way to do this yet.

Thank you in advance!

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  • $\begingroup$ Try to get $$\dfrac {\cos \frac{(2k + 1)\pi}{2}}{\left(\sin \frac{(2k + 1)\pi}{2n}\right)^n}$$ It is not hard. $\endgroup$ – Piquito Nov 8 '16 at 15:33
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HINT.-$$\left(\frac{\cos X}{\sin X}+i\right)^n=\frac{(\cos X+i\sin X)^n}{(\sin X)^n}=\frac{\cos {nX}+i\sin {nX}}{(\sin X)^n}$$ In this case $\cos {nX}=\cos{\frac{(k+1)\pi}{2}}=0$ and $\sin{\frac{(k+1)\pi}{2}}\ne 0$ for all value of $k$.

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