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If $\alpha$ and $\beta$ are different complex numbers with $|\alpha|=1$, then
what is $\left|\frac{\alpha-\beta}{1-\alpha \beta}\right|$ equal to?
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what is $\left|\frac{\alpha-\beta}{1-\alpha \beta}\right|$ equal to?
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The correct answer is:
1
We know, it $|\alpha|=1$ $\Rightarrow \alpha \bar{\alpha}=1$
$\therefore\left|\frac{\alpha-\beta}{1-\alpha \bar{\beta}}\right|=\left|\frac{\alpha-\beta}{\alpha \cdot \bar{\alpha}-\alpha \bar{\beta}}\right| \quad($ from $(1))$
$=\left|\frac{\alpha-\beta}{\alpha(\bar{\alpha}-\bar{\beta})}\right|$
$=\frac{|\alpha-\beta|}{|\alpha||\alpha / \beta|} \quad(\because$ sin ce $|\bar{z}|=|z|)$
$=\frac{1}{|\alpha|}=\frac{1}{1}=1$
$\therefore\left|\frac{\alpha-\beta}{1-\alpha \bar{\beta}}\right|=\left|\frac{\alpha-\beta}{\alpha \cdot \bar{\alpha}-\alpha \bar{\beta}}\right| \quad($ from $(1))$
$=\left|\frac{\alpha-\beta}{\alpha(\bar{\alpha}-\bar{\beta})}\right|$
$=\frac{|\alpha-\beta|}{|\alpha||\alpha / \beta|} \quad(\because$ sin ce $|\bar{z}|=|z|)$
$=\frac{1}{|\alpha|}=\frac{1}{1}=1$
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