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

$z=\xi+i\eta = \rho e^{i\theta}$ 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.
A: 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 $i$.
$i\sqrt{(x-1)/(x+1)}$
