Question: For $|z_0|<R$, I want to show that the mapping $$T(z)=\frac {R(z-z_0)} {R^2-\bar{z_0}z}$$ takes the open disc of radius $R$ $1-1$ and onto the unit disc and $z_0\rightarrow 0$.

Hint: Use the max modulus theorem and verify that $z_0\rightarrow0$ and $|z|=R\implies |T(z)|=1$.

this is how the question was posed with the hint.

Approach: I covered the automorphism $\phi_\alpha(z)=\frac{z-\alpha}{1-\bar{\alpha}z}$ of unit disc mapping onto a unit disc in class. But since $|z_0|<R$ inside the disc of radius R, I took $\frac{|z_0|}{R}<1$ and plugged it in $\phi$ to get $\phi_{\frac{z_0}{R}}=\frac{Rz-z_0}{R-\bar z_0z}$. Which is not what i want.

Also i tried the straight forward approach to show its 1-1 and onto, but the onto part looks like this: $$z=\frac{R^2w+Rz_0}{w\bar z_0+R}$$ Is this right? Also I'm so confused as to how to use the max modulus theorem in this case.

Any help would be greatly appreciated.


  • 1
    $\begingroup$ Did youtry scaling both $z$ and $z_0$ by $R$? $\endgroup$ May 14, 2013 at 22:54
  • $\begingroup$ how do i do that? $\frac{|z-z_0|}{R}<1$? i don't think that would work. $\endgroup$
    – d13
    May 15, 2013 at 3:31
  • $\begingroup$ Please any help would be greatly appreciated. Thankyou $\endgroup$
    – d13
    May 15, 2013 at 16:41

1 Answer 1


d13: If you had divided both $z_0$ and $z$ by $R$, you'd map the disk of radius $R$ to the disk of radius $1$: Letting $\alpha = z_0/R$ and $Z=z/R$, $$T(z) = \frac{R(z-z_0)}{R^2-\bar z_0z} = \frac{R^2(Z-\alpha)}{R^2(1-\bar\alpha z)} = \frac{Z-\alpha}{1-\bar\alpha Z}\,,$$ with $|Z|<1$. Now we apply the standard knowledge about the automorphisms of the unit disk.

  • $\begingroup$ yeah i did just that after giving it some thought. thanx by the way for answering! I appreciate your help. $\endgroup$
    – d13
    May 20, 2013 at 6:46

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