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If \( \alpha \) and \( \beta \) are two different comlex numbers with \( |\beta|=1 \), then \( \left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right| \) is equal to
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\( 11 \)
We have, \( |\beta|=1 \)
We know that \( |z|^{2}=(z)(\bar{z}) \)
Thus, \[ \begin{array}{l}\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|^{2}=\left(\frac{\beta-\alpha}{1-\alpha \beta}\right)\left(\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right) \\ =\left(\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right)\left(\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right) \\ =\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|^{2}\left(\frac{\bar{\beta}-\bar{\alpha}}{\overline{1}-\overline{\bar{\alpha}} \bar{\beta}}\right) \\ \text { since, } \bar{\alpha}=\alpha\end{array} \]
and conjugate of \( 1 \) is \( 1 \). So,
\[
\begin{aligned}
&\left(\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right)\left(\frac{\bar{\beta}-\bar{\alpha}}{1-\alpha \bar{\beta}}\right)=\frac{(\beta-\alpha)(\bar{\beta}-\bar{\alpha})}{(1-\bar{\alpha} \beta)(1-\alpha \bar{\beta})} \\
=& \frac{\beta(\bar{\beta}-\bar{\alpha})-\alpha(\beta-\bar{\alpha})}{1(1-\alpha \bar{\beta}-\bar{\alpha} \beta(1-\alpha \bar{\beta})} \\
=& \frac{\beta \bar{\beta}-\beta \alpha-\alpha \bar{\beta}+\alpha \bar{\alpha}}{1-\alpha \bar{\beta}-\bar{\alpha} \beta+\alpha \bar{\alpha} \beta \bar{\beta}} \\
\text { As } &|z|^{2}=(z)(\bar{z}) \\
=& \frac{|\beta|^{2}-\beta \bar{\alpha}-\bar{\beta} \alpha+|\alpha|^{2}}{1-\alpha \bar{\beta}-\bar{\alpha} \beta+|\alpha|^{2}|\beta|^{2}} \\
\text { Given }|\beta|=1 \\
=& \frac{1-\beta \bar{\alpha}-\bar{\beta} \alpha+|\alpha|^{2}}{1-\alpha \bar{\beta}-\bar{\alpha} \beta+|\alpha|^{2} \cdot 1}=\frac{1-\beta \bar{\alpha}-\bar{\beta} \alpha+|\alpha|^{2}}{1-\alpha \bar{\beta}-\bar{\alpha} \beta+|\alpha|^{2}}=1
\end{aligned}
\]
We know that \( |z|^{2}=(z)(\bar{z}) \)
Thus, \[ \begin{array}{l}\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|^{2}=\left(\frac{\beta-\alpha}{1-\alpha \beta}\right)\left(\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right) \\ =\left(\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right)\left(\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right) \\ =\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|^{2}\left(\frac{\bar{\beta}-\bar{\alpha}}{\overline{1}-\overline{\bar{\alpha}} \bar{\beta}}\right) \\ \text { since, } \bar{\alpha}=\alpha\end{array} \]
and conjugate of \( 1 \) is \( 1 \). So,
\[
\begin{aligned}
&\left(\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right)\left(\frac{\bar{\beta}-\bar{\alpha}}{1-\alpha \bar{\beta}}\right)=\frac{(\beta-\alpha)(\bar{\beta}-\bar{\alpha})}{(1-\bar{\alpha} \beta)(1-\alpha \bar{\beta})} \\
=& \frac{\beta(\bar{\beta}-\bar{\alpha})-\alpha(\beta-\bar{\alpha})}{1(1-\alpha \bar{\beta}-\bar{\alpha} \beta(1-\alpha \bar{\beta})} \\
=& \frac{\beta \bar{\beta}-\beta \alpha-\alpha \bar{\beta}+\alpha \bar{\alpha}}{1-\alpha \bar{\beta}-\bar{\alpha} \beta+\alpha \bar{\alpha} \beta \bar{\beta}} \\
\text { As } &|z|^{2}=(z)(\bar{z}) \\
=& \frac{|\beta|^{2}-\beta \bar{\alpha}-\bar{\beta} \alpha+|\alpha|^{2}}{1-\alpha \bar{\beta}-\bar{\alpha} \beta+|\alpha|^{2}|\beta|^{2}} \\
\text { Given }|\beta|=1 \\
=& \frac{1-\beta \bar{\alpha}-\bar{\beta} \alpha+|\alpha|^{2}}{1-\alpha \bar{\beta}-\bar{\alpha} \beta+|\alpha|^{2} \cdot 1}=\frac{1-\beta \bar{\alpha}-\bar{\beta} \alpha+|\alpha|^{2}}{1-\alpha \bar{\beta}-\bar{\alpha} \beta+|\alpha|^{2}}=1
\end{aligned}
\]
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