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I've been self studying complex analysis, and I read that there is a canonical conformal bijection $$\varphi(z) = \frac{z-i}{z+i}$$ from the open unit disc to the open half-plane.

I see why $\varphi$ is analytic, and I was able to show that it is one to one by taking differences and counting zeros. I cannot see why $\varphi$ is onto, however, why is this true?

Martin R
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user5583511
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    Which open half-plane? – PrincessEev Apr 07 '21 at 04:41
  • Negative real I believe – user5583511 Apr 07 '21 at 05:11
  • Are you familiar with Möbius transformations? That are bijective analytic mappings of the extended complex plane (Riemann sphere) onto itself, they map “generalized circles” (i.e. circles or lines) to generalized circles, and are uniquely determined by the image of three distinct points. – Often these three properties are sufficient to understand the behavior of a Möbius transformation. – Martin R Apr 07 '21 at 05:26
  • I'm not familiar with these transformations, very cool. Is the theory behind them needed to attack the simple case above? – user5583511 Apr 07 '21 at 05:47
  • Maybe it helps to write $\frac{(z-i)}{(z+i)}=1-\frac{2i}{(z+i)}.$ So $\psi$ is the composition of the functions $z\mapsto z+i$, $z \mapsto \frac 1 z$, $z \mapsto i \cdot z$, $z \mapsto 2z$, $z \mapsto -z$, $z \mapsto 1+z$ – miracle173 Apr 07 '21 at 06:50
  • @user5583511: It is not required, but allows to determine the image easily, without much explicit computation. – Martin R Apr 07 '21 at 06:55

3 Answers3

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You can use the fact that the inverse function is $\varphi^{-1}(z)=-i\frac{z+1}{z-1}$, and that for $z$ in the left half plane, $|z-1|^2$ is always greater than $|z|^2+1$ (expand $(z-1)\overline{(z-1)}$, and see how $\text{Re}(z)$ appears). Meanwhile, $|z+1|^2 < |z|^2 + 1$ for similar reasons. So, the norm-square of $\varphi^{-1}(z)$ on the left half plane always has a larger denominator than numerator; thus its norm-square is always less than 1, and therefore so is its norm. So, the image of $\varphi^{-1}$ on the left half plane is contained in the unit disc, and we’re done.

Here’s a more geometric way of looking at it: $\varphi(z) = \frac{i-z}{-i-z}$, i.e. the ratio of the two “vectors” from $z$ to $\pm i$. This ratio is the complex number $s$ that makes a similar triangle with $1$ in the complex plane, upon identifying $i$ with $s$, $-i$ with $1$, and $z$ with the origin. We know that for any $s$ in the left open half plane, the angle $s$ makes with $1$ through the origin is between $\pi /2$ and $3\pi /2$. As such, since the triangles are similar, the angle through $0$ is the same as the angle through $z$ and so the point $z$ must lie within the unit circle. So, we’ve constructed a point in the unit disc that yields an arbitrary point in the left half plane.

If any step is unclear here let me know! :)

thorimur
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Here is how you can solve the problem using basic properties of Möbius transformations:

  • Möbius transformations map circles to circles or extended lines. Here $\varphi(-i) = \infty$ and $\varphi(i) = 0$. It follows that $\varphi$ maps the unit circle $S = \{ z : |z| = 1 \}$ onto a line through the origin.

  • Möbius transformations preserve angles. Here $\varphi(i\Bbb R) \subset i \Bbb R$ and the unit circle $S$ intersects the imaginary axis at a right angle. It follows that $\varphi(S)$ intersects the imaginary axis at a right angle as well.

  • Combining the previous two results, we see that $\varphi(S) = \Bbb R \cup \{ \infty \}$.

  • Möbius transformations are bijective conformal mappings from the extended complex plane (aka Riemann sphere) onto itself. The boundary of the unit disk $\Bbb D$ is mapped to the extended real line. It follows that the unit disk itself is conformally mapped onto either the left or the right half-plane.

  • $\varphi(0) = -1$, therefore $\varphi$ maps the unit disk conformally onto the left half-plane.

Martin R
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It is onto because you can write its inverse by inverting the matrix

$$\begin{pmatrix} 1 & -i \\ 1 & i\end{pmatrix}.$$

Igor Rivin
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