Find an analytic function that maps the plane with the slit $[-1,1]$ onto the upper half plane

Normaly I can find the function from $3$ points easily, but this problem doesn't give me any point. So I know that $z$ is in the plane that delete the line segment on the real aaxis from $-1$ to $1$ and I know that the image is only cover the upper half plane, so I guess $f(z)$ must have $z^2$ in it? or the whole expression is square?

However, these infor is not enough for me to get the 3 points, I wonder if anyone would please give me a hint.


A useful conformal map to know is $$ w = \frac{1}{2}\left(z + \frac{1}{z}\right) \tag{*}$$ It maps the inside of the unit circle onto the Riemann sphere except the interval $[-1,1]$; and also maps the outside of the unit circle to the same thing.

Here is a picture from the web: conformal
$z=\xi+i\eta = \rho e^{i\theta}$ is on the right, $w = u+iv$ is on the left. The outsde of the disk on the right maps to the outside of the slit on the left. Circles of constant $\rho$ on the right map to ellipses (with foci $1,-1$) on the left. Radial lines with constant $\theta$ on the right map to hyperbolas (with foci $1,-1$) on the left.

Of course the inside (or outside) of the unit clrcle is easily mapped to the upper half-plane by a linear fractional transformation.

So, use the inverse of the map (*) to map the plane without the slit onto the exterior of the disk, then a linear fractional transformation to map the exterior of the disk onto the upper half plane.

  • $\begingroup$ what if you have a negative $z$, say $z=-2-i$ the $w= \frac{-6-2i}{5}$, the image of this point is on the lower half plane, right? $\endgroup$ – Diane Vanderwaif Feb 28 '15 at 14:45
  • $\begingroup$ I'm sorry, but I still can't see it, I checked around and know this is a Joukowski map, I also find the inverse $z=2w \pm \sqrt{4w^-1}$, nad tried to let $w=u+iv$ but it still doesn't go anywhere, can you explain a little bit more please. $\endgroup$ – Diane Vanderwaif Feb 28 '15 at 17:23
  • $\begingroup$ from your picture, the slit $[-1,1]$ is on the $w$-plane, but I think in my question, the slit is on $z$-plane, right? $\endgroup$ – Diane Vanderwaif Mar 1 '15 at 14:57
  • $\begingroup$ So that is why you need to use the inverse of the map (*). $\endgroup$ – GEdgar Mar 1 '15 at 15:00
  • $\begingroup$ I'm sorry, just want to make sure that I understand this correctly, by saying "Of course the inside (or outside) of the unit clrcle is easily mapped to the upper half-plane by a linear fractional transformation." you mean take $w^2$? $\endgroup$ – XiaoXiao Zhen Mar 19 '15 at 0:43

The 360 degree angle at $1$ has to be reduced to $180$, so put in a $\sqrt{x-1}$.
The other key point, at $-1$, has to be moved to $\infty$, so put in a $1/(x+1)$.
Going in a full circle around the slit must turn into an ordinary loop in the upper half plane. So $\sqrt{(x-1)/(x+1)}$ does not change sign.
Lastly, the boundary is the imaginary axis, so multiply by $i$.

  • $\begingroup$ can you explain it a little bit further please? For example , why $-1$ is move to $\infty$, so $f(-1) = \infty$? $\endgroup$ – Diane Vanderwaif Feb 28 '15 at 14:27
  • $\begingroup$ The double line from $-1$ to $1$ has to become the whole real line. So at one point it becomes infinite. Also, I have to deal with the $360^{\circ}$ at $-1$, so it is convenient to deal with both as I did. It is tricky to make, for example, $0$ in the upper half plane of the slit go to infinity without making $0$ in the lower half-plane of the slit go to infinity. $\endgroup$ – Empy2 Feb 28 '15 at 14:34
  • $\begingroup$ Why your answer does not match with the answer of the book: $w=\sqrt{\frac{z+1}{1-z}}$? $\endgroup$ – Bellatrix Aug 12 '18 at 10:12
  • $\begingroup$ The reciprocal of the upper half plane is the lower half plane $\endgroup$ – Empy2 Aug 12 '18 at 12:07
  • $\begingroup$ what ? ${}{}{}$ $\endgroup$ – Bellatrix Aug 12 '18 at 16:23

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