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This transformation question was asked in an assignment question and I was unable to solve it.

Prove that the transformation f(z) =$\frac { z-z_{0}} { \bar z_{0} z -1} $ maps unit disc to itself.

Attempt : I tried to assume that x+iy lies in U and then prove that f(z) also lies in U. But that's a very lenghty and primitive approach.

Can you please tell any better approach for this problem.

Thanks!!

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  • $\begingroup$ Show that $|z-z_0|^{2} <|\overline {z_0}z-1|^{2}$ using the fact that $|a+b|^{2}=|a|^{2}+|b|^{2}+2 \Re a \overline {b}$. $\endgroup$ – Kavi Rama Murthy Oct 1 '20 at 6:22
  • $\begingroup$ You can use the maximum modulus principle: it then suffices to prove that $|f(e^{i\vartheta})|\le 1$ $\endgroup$ – Caffeine Oct 1 '20 at 6:23
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Let $|z|=1$ be the unit cfircle. The transformation is $z=\frac{w-a}{\bar aw-1}$ Then $$|z|=1 \implies \left|\frac{w-a}{\bar a w-1} \right|=1 \implies |w-a|=|\bar a w-1|\implies (w-a)(\bar w-\bar a)=(\bar a w-1)(a \bar w-1). $$ $$\implies (a\bar a-1)(w\bar w-1)=0 \implies |w|=1, ~\text{if}~ |a|\ne 1.$$ So $z$ points move in a unit circle so will be $w$ points, provided $|a| \ne 1.$

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  • $\begingroup$ @Z Ahmed Thanks for the nice proof. If $|a| =1 $ then how should I approach it? $\endgroup$ – No -One Oct 1 '20 at 13:35

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