The following example is taken from Conway's book, Functions of One Complex Variable I.

Find an analytic function $f:G\to\mathbb C$, where $G=\{z:Re(z)>0\}$, such that $f(G)=D=\{z:|z|<1\}$.

Conway's explanation towards a solution is as follows:

enter image description here enter image description here

However, I have the following questions regarding the explanation above:

  • Why $\{z:\Im(z,-i,0,i)>0\}=\{z:\Im(iz)>0\} ?$
  • How does $T:=R^{-1}\circ S$ map the imaginary axis onto the circle $\Gamma\subset\mathbb C_\infty ?$
  • How did he arrive at the expression $g(z)=\dfrac{e^z -1}{e^z+1}$ from the expression for $Tz$ ?

I understood his calculations other than these three conclusions.


1 Answer 1


I don't quite understand Conway's notation, but, this should be pretty straightforward using Cayley's transform and a rotation.

That is, $$f(z)=\frac{z-i}{z+i}$$ maps the upper half plane onto the unit circle.

So you just need to compose with a rotation by $\pi/2$. That's multiplication by $i$.

The result is $$\frac {z-1}{z+1}$$.

  • $\begingroup$ Thanks, but I am not used to Cayley's Transform (at least not yet). Could you show this in line with Conway's arguments? $\endgroup$ Jul 15, 2022 at 15:37
  • $\begingroup$ Well, I'm not exactly sure what he means by an "orientation". I don't have the book. Why he's concerned with that horizontal strip I don't know. But it results in $e^z$. $\endgroup$
    – calc ll
    Jul 15, 2022 at 15:40
  • $\begingroup$ His definition of an orientation is as follows:\\ "If $\Gamma$ is a circle then an orientation for $\Gamma$ is an ordered triple of points $(z_1,z_2,z_3)$ such that each $z_j$ is in $\Gamma$." $\endgroup$ Jul 15, 2022 at 15:43
  • 1
    $\begingroup$ Hmm, well, he's done this from scratch I guess. But Cayley's transform makes this easy. To learn it, just check the images of $-1,0,1$. You'll find they're on the unit circle. Then use a test point, like $i$, to see that the upper half plane indeed maps onto the unit disk. That's the way I would do it. $\endgroup$
    – calc ll
    Jul 15, 2022 at 16:04
  • 1
    $\begingroup$ Okay, understood your argument now. $\endgroup$ Jul 15, 2022 at 16:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .