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# HELP PLZ I NEED HELP ASAP

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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