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(B) Let $z = \frac{\beta-\alpha}{1-\bar{\alpha} \beta}.$ We want to find $|z|.$ Consider $|z|^2 = z \cdot \bar{z} = \left( \frac{\beta-\alpha}{1-\bar{

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(B) Let $z = \frac{\beta-\alpha}{1-\bar{\alpha} \beta}.$ We want to find $|z|.$ Consider $|z|^2 = z \cdot \bar{z} = \left( \frac{\beta-\alpha}{1-\bar{\alpha} \beta} \right) \left( \frac{\bar{\beta}-\bar{\alpha}}{1-\alpha \bar{\beta}} \right).$ Since $|\beta|=1,$ we have $\beta \bar{\beta} = |\beta|^2 = 1,$ so $\bar{\beta} = \frac{1}{\beta}.$ Substituting this,$|z|^2 = \frac{(\beta-\alpha)(\bar{\beta}-\bar{\alpha})}{(1-\bar{\alpha} \beta)(1-\alpha \bar{\beta})} = \frac{\beta \bar{\beta} - \beta \bar{\alpha} - \alpha \bar{\beta} + \alpha \bar{\alpha}}{1 - \alpha \bar{\beta} - \bar{\alpha} \beta + \alpha \bar{\alpha} \beta \bar{\beta}}.$ Using $\beta \bar{\beta} = 1,$ the numerator becomes $1 - \beta \bar{\alpha} - \alpha \bar{\beta} + |\alpha|^2.$ The denominator becomes $1 - \alpha \bar{\beta} - \bar{\alpha} \beta + |\alpha|^2 (1) = 1 - \alpha \bar{\beta} - \bar{\alpha} \beta + |\alpha|^2.$ Since the numerator and denominator are equal,$|z|^2 = 1,$ which implies $|z| = 1.$

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

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