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If $Z = \cos \theta+i\sin \theta$, $Z_1= \dfrac{z+1}{z-1}$, prove that $Z_1= -i\cot \dfrac{θ}{2}\\$

This was a proof that I ran into in a quiz. I couldn't really solve it. I only got as far as $Z_1 = z+1$, $Z_1 = \cos \theta+\sin \theta$ $-1$.

It's bugging me since then. A friend told me we'd prove it using a double angle formula.

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  • $\begingroup$ write $Z$ as $e^{i\theta}$ and factor by $e^{i\frac{\theta}{2}}$. $\endgroup$
    – hamam_Abdallah
    Oct 21, 2016 at 13:43

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\begin{align*} \frac{z+1}{z-1} &= \frac{(\cos \theta+1)+i\sin \theta}{(\cos \theta-1)+i\sin \theta} \\ &= \frac{2\cos^2 \frac{\theta}{2}+2i\sin \frac{\theta}{2} \cos \frac{\theta}{2}} {-2\sin^2 \frac{\theta}{2}+2i\sin \frac{\theta}{2} \cos \frac{\theta}{2}} \\ &= \cot \frac{\theta}{2} \left( \frac{\cos \frac{\theta}{2}+i\sin \frac{\theta}{2}} {-\sin \frac{\theta}{2}+i\cos \frac{\theta}{2}} \right) \\ &= \cot \frac{\theta}{2} \frac{\cos \frac{\theta}{2}+i\sin \frac{\theta}{2}} {i\left(\cos \frac{\theta}{2}+i\sin \frac{\theta}{2} \right)} \\ &= -i\cot \frac{\theta}{2} \end{align*}

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  • $\begingroup$ I put a negative in front of the $i$ in the OP $\endgroup$
    – imranfat
    Oct 21, 2016 at 15:24

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