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Question

(B) Given that $|\beta|=1$,we have $|\beta|^2 = \beta \bar{\beta} = 1$.Consider the expression $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|

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Vedclass · Answer

(B) Given that $|\beta|=1$,we have $|\beta|^2 = \beta \bar{\beta} = 1$.Consider the expression $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|$.Substituting $1 = \beta \bar{\beta}$ in the denominator,we get:$\left|\frac{\beta-\alpha}{\beta \bar{\beta}-\bar{\alpha} \beta}\right| = \left|\frac{\beta-\alpha}{\beta(\bar{\beta}-\bar{\alpha})}\right|$.Using the property of modulus $|z_1/z_2| = |z_1|/|z_2|$ and $|z_1 z_2| = |z_1||z_2|$,we have:$= \frac{|\beta-\alpha|}{ |\beta| |\bar{\beta}-\bar{\alpha}| }$.Since $|\beta|=1$ and $|\bar{z}| = |z|$,we know $|\bar{\beta}-\bar{\alpha}| = |\overline{\beta-\alpha}| = |\beta-\alpha|$.Therefore,the expression becomes $\frac{|\beta-\alpha|}{1 \cdot |\beta-\alpha|} = 1$.

If $\alpha$ and $\beta$ are different complex numbers with $|\beta|=1$,then $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|$ is equal to

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