Q: Find a conformal map f(z) which maps which maps the unit disc D bijectively onto the whole complex plane minus the non-negative real axis, and which in addition satisfies $\lim_{z\to -1}f(z)=0$ and $f(0)=-7$.

My Attempt:

I know that $\frac{z+1}{z-1}$ goes to the left half of imaginary axis.

Now, I take the expression raised by 2, $\frac{(z+1)^2}{(z-1)^2}$, and get the whole complex-plane except the non-negative real axis.

Am I in the right direction? And what must I do to satisfy

$\lim_{z\to -1}f(z)=0$ and $f(0)=-7$?

  • $\begingroup$ It goes to 0. But how about $f(0)=-7$? $\endgroup$ – Aerdennis Mar 9 '17 at 18:05
  • $\begingroup$ $z \rightarrow i \cdot \frac{1+z}{1-z} \rightarrow \left ( i \cdot \frac{1+z}{1-z} \right )^2=-\left (\frac{1+z}{1-z} \right )^2\rightarrow -7\left (\frac{1+z}{1-z} \right )^2$ $\endgroup$ – Daniil Mar 9 '17 at 18:41
  • $\begingroup$ Why did you rotate it by pi? Is it because it fullfill f(0)=-7? $\endgroup$ – Aerdennis Mar 10 '17 at 9:50
  • $\begingroup$ 1) map unit disk to the upper half-plane 2) map upper half-plane to the whole plane minus nonnegative real axis 3) is made in order to satisfy initial condition $\endgroup$ – Daniil Mar 11 '17 at 22:11

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