# How to show that $\Re(\frac{e^{i \theta}+z}{e^{i \theta}-z})=\frac{1+|z|^2}{|z-e^{i \theta}|^2}$?

How to show that $\Re(\frac{e^{i \theta}+z}{e^{i \theta}-z})=\frac{1+|z|^2}{|z-e^{i \theta}|^2}$?

I have got $\Re(\frac{e^{i \theta}+z}{e^{i \theta}-z})=\frac{1+|z|}{1-|z|}$. Is this right?

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Does $|w+z|=|w|+|z|$ hold? Try $z = -w$. –  Lord_Farin Apr 4 '13 at 8:51
Both statements are wrong. Try $\theta = 0$, $z = -1$ to see this. It should be $1-|z|^2$ in the numerator. Show us your work so that we can tell you what went wrong. –  Ayman Hourieh Apr 4 '13 at 11:39

Hint: $\Re z= \frac{1}{2}(z+z^*)$.