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{\bf Question}

By the means of the Argand diagram,or otherwise, show that if
$|z|=1$ the real part of $\ds \frac{1+z}{1-z}$ is zero.

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{\bf Answer}

\begin{eqnarray*} \frac{1+e^{i\theta}}{1-e^{i\theta}} & = &
\frac{e^{-i\frac{\theta}{2}} +
e^{i\frac{\theta}{2}}}{e^{-i\frac{\theta}{2}} -
e^{i\frac{\theta}{2}}}\\ & = &
\frac{2\cos\frac{\theta}{2}}{-2i\sin\frac{\theta}{2}} \\ & = &
i\cot\frac{\theta}{2} \end{eqnarray*}

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Therefore $\arg(1+z) - \arg(1-z) = \frac{\pi}{2}$
($-\frac{\pi}{2}$ if z is replaced by -z)  In either case the
ra=eal part is zero.


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