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My Attempt:- Let $z_1=z_0,z_2=\overline{z_0},z_3=0$ and $w_1=0,w_2=\infty, w_3=\frac{z_0}{\overline{z_0}}$. Applying this in Result in the box, we get $$\frac{w.(1-\frac{w_3}{w_2})}{(w-\frac{z_0}{\overline{z_0}})(1-\frac{w_3}{w_2})}=\frac{(z-z_0).\overline{z_0}}{z.(\overline{z_0}-z_{0})}.$$ Simplication, $$\frac{w.\overline{z_0}}{(\overline{z_0}w-{z_0)}{}}=\frac{(z-z_0).\overline{z_0}}{z.(\overline{z_0}-z_{0})}$$ $\implies$ $$\frac{w}{(\overline{z_0}w-{z_0)}{}}=\frac{(z-z_0)}{z.(\overline{z_0}-z_{0})}$$ I am not able to simplify and get the answer in the picture.

Reference:- Complex Variables with Applications, Ponnusamy S. and Silverman H.

Please help me.

1 Answers1

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It only remains to simplify the expression by multiplying everything out and collecting common factors.

$$ \begin{align} & & \frac{w}{\overline{z_0}w - z_0} &= \frac{z-z_0}{z\cdot (\overline{z_0}-z_0)}\\ &\iff & \color{red}{wz \overline{z_0}} - wzz_0 &= \color{red}{wz\overline{z_0}} - wz_0\overline{z_0} - zz_0 + z_0^2\\ &\iff & wz_0(\overline{z_0} - z) &= z_0(z_0 - z) \\ &\iff & w &= \frac{z - z_0}{z - \overline{z_0}} \end{align} $$

The cancellation of the $z_0$'s is allowed at each instance because $z_0 \neq 0$ is assumed.

A minor typo in the question details: the substitution is incorrectly made in the first line. The LHS should read: $$ \frac{w \cdot \left(1 - \frac{w_3}{w_2}\right)}{\left(w - \frac{z_0}{\overline{z_0}}\right)\left(1-\frac{\color{red}{w_1}}{w_2}\right)}. $$ This doesn't affect the other steps because $w_2 = \infty$ and $w_1 = 0\ (\neq \infty)$.