Let alpha be a fixed non-zero complex number | vedclass

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(C) We are given $w = \frac{z-\alpha}{1-\bar{\alpha}z}$ with $|\alpha| < 1$.Consider the condition $|z| < 1$.We want to check the value of $|w|^2 = w\

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$|w| < 1$ for all $z$ such that $|z| < 1$

Vedclass · Answer

(C) We are given $w = \frac{z-\alpha}{1-\bar{\alpha}z}$ with $|\alpha| Consider the condition $|z| We want to check the value of $|w|^2 = w\bar{w}$.$|w|^2 = \left(\frac{z-\alpha}{1-\bar{\alpha}z}\right) \left(\frac{\bar{z}-\bar{\alpha}}{1-\alpha\bar{z}}\right) = \frac{z\bar{z} - z\bar{\alpha} - \alpha\bar{z} + \alpha\bar{\alpha}}{1 - \alpha\bar{z} - \bar{\alpha}z + \alpha\bar{\alpha}z\bar{z}}$.Since $|z|^2 = z\bar{z}$ and $|\alpha|^2 = \alpha\bar{\alpha}$,we have $|w|^2 = \frac{|z|^2 - z\bar{\alpha} - \bar{z}\alpha + |\alpha|^2}{1 - \bar{z}\alpha - z\bar{\alpha} + |\alpha|^2|z|^2}$.Now,consider $|w|^2 - 1 = \frac{|z|^2 - z\bar{\alpha} - \bar{z}\alpha + |\alpha|^2 - (1 - \bar{z}\alpha - z\bar{\alpha} + |\alpha|^2|z|^2)}{1 - \bar{z}\alpha - z\bar{\alpha} + |\alpha|^2|z|^2}$.$|w|^2 - 1 = \frac{|z|^2 + |\alpha|^2 - 1 - |\alpha|^2|z|^2}{1 - \bar{z}\alpha - z\bar{\alpha} + |\alpha|^2|z|^2} = \frac{-(1 - |z|^2)(1 - |\alpha|^2)}{|1 - \bar{\alpha}z|^2}$.Since $|z|  0$ and $(1 - |\alpha|^2) > 0$.Thus,$|w|^2 - 1 < 0$,which implies $|w| < 1$ for all $|z| < 1$.

Let $\alpha$ be a fixed non-zero complex number with $|\alpha| < 1$ and $w = \frac{z-\alpha}{1-\bar{\alpha}z}$,where $z$ is a complex number. Then,

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