I need to find conformal mapping from area outside the two circles $|z-1|=1$,$|z+1|=1$ onto a half plane. We want to find trans’ that take $Z=0→W=∞$. such trans’ is $t(z)=1/z$ Now we find images of points from the domains according to $t(z)$. On the right circle $z_1=1+i$→$t(z_1 )=1/(1+i)=1/2 (1-i)$ $z_2=2$ →$t(z_2 )=1/2$
On the left circle $z_3=-2$ →$t(z_3 )=-1/2$ $z_4=-1+i$ → $t(z_4 )=1/(-1+i)=-1/2 (1+i)$ lets check point z=1 which is out of the source domain, and we get w=1,which is right to the right boundery. Our current image domain is strip $C_t:-1/2<\mathrm{Re}\,z<1/2$
How can i advance from here?
The first thing is to check that such a conformal map exists. Let us call $U$ your domain (including the $\infty$ point). Since $U$ is simply connected and different from the whole plane, we are good.
Now, the boundary of your domain is the reunion of two circles both passing through the origin point (and tangent here - I admit I had a few seconds of horror, since if these circles had been either disjoint or intersecting, the problem would have been insoluble).
If you remember a bit of elementary geometry, you might know that an inversion maps a circle through its center to a line. So using an inversion centered at zero was indeed the good choice. However, I don't see any use to all your computations of the images of some points of the circles -- even though your result is good, you should be able to give a direct proof that the image of $U$ is a band between two parallel lines.
Now, the tangent function maps a (suitable) band to an open disk: for $x \in ]-\pi, +\pi[$ and $y \in \mathbb R$, $$ \tan (x + i y) = \frac{\sin(2x) + i \mathrm{sh} (2y)}{\cos (2x) + i \mathrm{ch} (2y)},$$ so that any band $]-R, R[$ is mapped to an open disk of radius $\tan R$ (check it!).
Finally, changing this open disk back to the half-plane is standard via $z \mapsto i \frac{z-1}{z+1}$.
To sum it up, to find the full transform, you must in order:
- inverse your domain to a band;
- change your band by an affine transform to the vertical band $]-\pi/4, +\pi/4[$;
- map this to the unit disk with the tangent map;
- map this to the half-plane by a homography;
- since everything will be a small formula involving homographies and exponentials, it should simplify back to a quite manageable result.