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Prove that if  \(w,z\)are complex numbers such that \(|w|=|z|=1\) and \(wz\ne -1\)then  \(\frac{w+z}{1+wz}\)is a real number by proving that \(w,z\) and  are equal to their conjugates and that \(\overline z = 1/z\) and \(\overline w = 1/w\)

 May 6, 2020
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\(|a+bi|^2 =( \overline{a+bi}) \cdot (a+bi)\), and since \(|w| = 1\), we know \(|w|^2 = |w| \cdot |w| = 1 \cdot 1 = 1 =|w|\), giving us 

\(\begin{align*} |w| &= 1 \\ |w| &= |w|^2 \\ |w| &= \overline{w}w \\ \overline{w}w &= 1 \\ \overline{w} &= \frac1w. \end{align*}\)


we can get \(\overline{z} = \frac1z\) through the same process. we make \(x = \frac{w+z}{1+wz}\), which means \(\overline{x} = \frac{\overline{w}+\overline{z}}{1+\overline{w}\cdot\overline{z}}. \)we plug in our values for the congujates of \(w\) and \(z\), which gives us


\(\begin{align*} \overline{x} &= \frac{\frac{1}{w} + \frac{1}{z}}{1+ \frac{1}{w} \cdot \frac{1}{z}} \\ &= \frac{\frac{z}{wz} + \frac{w}{wz}}{\frac{wz}{wz}+\frac{1}{wz}} \\ &=\frac{\frac{w+z}{wz}}{\frac{1+wz}{wz}} \\ &= \frac{w+z}{1+wz} \\ &= x \end{align*}\)

 

we get \(\overline{x} = x\), which is only true when \(x\) is real.

 May 6, 2020
edited by chrissy  May 6, 2020

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