If $z_1$ and $z_2$ are two different complex numbers and $\lvert z_1\rvert=1 $ then find $$ \frac{\lvert z_1-z_2 \rvert}{\lvert 1-z_1\bar{z_2} \rvert} $$
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https://math.stackexchange.com/questions/506058/show-that-left-cfrac-alpha-beta1-bar-alpha-beta-right-1-when – Guy Fsone Nov 23 '17 at 15:31
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$$\left|\dfrac{z_1-z_2}{1-z_1\bar{z_2}}\right|= \left|\dfrac{\overline{z_1-z_2}}{1-z_1\bar{z_2}}\right|= \left|\dfrac{\overline{z_1}-\overline{z_2}}{1-z_1\bar{z_2}}\right|=\dfrac1{|z_1|}\cdot\left|\dfrac{z_1\overline{z_1}-z_1\overline{z_2}}{1-z_1\bar{z_2}}\right|$$
Now $z_1\cdot\overline{z_1}=|z_1|^2$
lab bhattacharjee
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Can you explain the last step pls? I am not able to understand it – Vankshu Bansal Jul 13 '16 at 06:15
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@VankshuBansal, Multiplied the numerator & teh denominator by $|z_1|$ – lab bhattacharjee Jul 13 '16 at 06:25
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@VankshuBansal Note that $$;|z_1\overline{z_1}-z_1\overline{z_2}|=||z_1|^2-z_1\overline{z_2}|=|1-z_1 \overline{z_2}|$$ – DonAntonio Jul 13 '16 at 06:56
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Since $|z_1| = |\bar{z_1}| = 1$, we have \begin{align*} \frac{\lvert z_1-z_2 \rvert}{\lvert 1-z_1\bar{z_2} \rvert} &= \frac{|\bar{z_1}|\lvert z_1-z_2 \rvert}{\lvert 1-z_1\bar{z_2} \rvert} \\ &= \frac{\lvert \bar{z_1}z_1-\bar{z_1}z_2 \rvert}{\lvert 1-z_1\bar{z_2} \rvert}\\ &=\frac{\lvert 1-\bar{z_1}z_2 \rvert}{\lvert 1-z_1\bar{z_2} \rvert}\\ &=\frac{\overline{\lvert 1-\bar{z_1}z_2 }\rvert}{\lvert 1-z_1\bar{z_2} \rvert}\\ &=\frac{\lvert 1-{z_1}\bar{z_2} \rvert}{\lvert 1-z_1\bar{z_2} \rvert}\\ &=1 \end{align*}