Mobius transformations between two sets

I am doing some revision of complex analysis, and am stuck on this question. I am looking for A mobius mapping sending the set {z: |z+1|<$$\sqrt{2}$$}, |z-1|<$$\sqrt2$$} onto the sector {z:3pi/4< argz< 5pi<4}.

I thought that a Mobius mapping is uniquely determined by three points, so I was going to consider the points i,0 and -i in the first set, and send those to -1+i, 0, -1-i respectively? But if this is right (and I'm not sure if it is), I can't find values for a Mobius transformation to work.

Any help appreciated.

• Just to clarify, I decided on i, 0 and -i as my points as these were the points of intersection of the two circles, so the corner points of the sets. – jessg12345 Apr 17 at 17:45
• Do you mean $|z+1| < rt_2$ or $|z+1| < r \cdot t \cdot 2$? Consider using MathJax for better readability. What is $r$? What is $t$? – Strichcoder Apr 17 at 20:15
• No sorry, I meant the square root of 2 for both of these. – jessg12345 Apr 18 at 7:45

$$i \mapsto 0, ~-i \mapsto \infty, ~0 \mapsto -1$$
Afterwards, check that that this Möbius transform is indeed what you want (e.g. compute the values of $$1+\sqrt{2}$$ and $$1-\sqrt{2}$$ and remember that it maps circles and lines to circles and lines)
• I get that $f(z)=z-i/z+i$. From here, how could I find a conformal mapping sending this region to the upper half plane? I guess I want to rotate by 3pi/4 and square? Is this correct-how would I do this> – jessg12345 Apr 18 at 13:55
• Rotation by $\frac{3\pi}{4}$ is just multiplication by $e^{-\frac{3\pi}{4}i}$ ($-$ in the exponent because you want to rotate clockwise). And as you said, afterwards you square everything. – agb Apr 23 at 6:44