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  1. $|z|=|w|=1$

  2. $1 + zw \neq 0$

Then $\dfrac{z+w}{1+zw}$ is real. How can prove that.

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One thing, if you want to see whether $\frac{a}{b}$ is real (or purely imaginary) is to multiply with $\frac{\overline{b}}{\overline{b}}$. – Daniel Fischer Jun 19 '14 at 17:56
See… . – alexqwx Jun 19 '14 at 17:57
Thanks. This is a repeated question. I'm sorry. – Jorge Jun 19 '14 at 18:03
up vote 5 down vote accepted

Hint: Take the complex conjugate then multiply top and bottom by $zw$.

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Let $z=e^{ia}$ and $w=e^{ib}$ for some $a,b \in [0,2\pi)$. This takes care of condition 1. Then $1+zw \neq 0$ means $1+e^{i(a+b)} \neq 0$, which is the same as saying $a+b$ is not an odd multiple of $\pi$.

Now consider \begin{align*} \dfrac{z+w}{1+zw} & = \frac{e^{ia}+e^{ib}}{1+e^{i(a+b)}} \end{align*} Rationalize the denominator and see what happens.

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