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I want to find the Möbius transformation from the disk $\{|z-1|\leq2\}$ to the half-plane $\{Re(z)\geq3\}$ that moves the point $0$ to $4+4i$.

I know that by specifying the values at 3 points, the Möbius transformation is determined.

If I impose that $$T(0)=4+4i, T(3)=3+i, T(-1)=\infty$$

I get the relations $c=d$, $b=(4+4i)c$, $c=\frac{3}{8}a$ and therefore my transformation is $$T(z)=\frac{az+\frac{3(4+4i)}{8}a}{3az/8+3/8}=\frac{8z+12+12i}{3z+3}$$

This transformation does fulfill my 3 initial conditions, however the are points outside the disk $\{|z-1|\leq2\}$ that, with this transformation, also move to the half-plane $\{Re(z)\geq3\}$. Does this mean I am wrong? If I am correct why can I ensure that all the points inside my disk go to the half-plane $\{Re(z)\geq3\}$?

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Regarding your guess about the images under $T$ of the three points $0$, $3$, and $-1$, since those three points all lie on a line, with $3$ and $-1$ on the boundary of the disc, you may conclude that their three images $T(0)=4+4i$, $T(3)$, and $T(-1)$ must all lie on a line (which includes $\infty$) or a circle, with $T(3)$ and $T(-1)$ all on the boundary of the region $Re(z)\ge 3$. You've chosen $T(-1)=\infty$, and so the choice of $T(3)$ is still open. But there are still infinitely many points to choose for $T(3)$ on the boundary line $Re(z)=3$.

You chose $T(3) = 3+3i$.

But that cannot be correct, because the $0$, $3$, and $-1$ lie on a diameter that hits the boundary circle at a right angle at the point $3$, whereas the line through $\infty$, $4+4i$, and $3+3i$ definitely does not hit the boundary line $Re(z)=3$ at a right angle, which would violate conformality of $T$.

On the other hand, the line through $\infty$, $4+4i$, and $3+4i$ does hit the boundary line $Re(z)=3$ at a right angle.....

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  • $\begingroup$ I am sorry there was a typo I meant $T(3)=3+i$ and not $T(3)=3+3i$. Anyway, if I understood you answer it would still be wrong since the line through $\infty$, $4+4i$, and $3+i$ is not even in the same line and that violates the conformality of $T$. Now using $T(0)=4+4i, T(3)=3+4i, T(-1)=\infty$ gives $T(z)=\frac{z(8+12i)+12+12i}{3z+3}$ How can I ensure this is the transformation I was looking for, will any points outside of the intial disk move outside of the plane $\{Re\geq 3\}$? $\endgroup$ – John Keeper Mar 23 at 20:33
  • $\begingroup$ You can probably make use of the fact that a Möbius transformation is a homeomorphism of the Riemann sphere which takes circles (including lines union $\infty$) to circles. $\endgroup$ – Lee Mosher Mar 23 at 20:57
  • $\begingroup$ You say "since those three points all lie on a line, with $3$ and $-1$ on the boundary of the disc, you may conclude that their three images $T(0)=4+4i$, $T(3)$, and $T(-1)$ must all lie on a line (which includes $\infty$) or a circle, with $T(3)$ and $T(-1)$ all on the boundary of the region $Re(z)\ge 3$". Specifically $T(-1)=\infty$, $\infty$ does not lie on the boundary of the region $Re(z)\ge 3$, right? $\endgroup$ – John Keeper Mar 24 at 17:06
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    $\begingroup$ You should think about how $\mathbb C$ sits inside the Riemann sphere. Evern line in $\mathbb C$, after being extended to the Riemann sphere, does contain $\infty$. So in that sense $\infty$ does lie on the line $Re(z)=3$ which forms the boundary of the region $Re(z) \ge 3$, and furthermore $\infty$ does line on any line determined by any two points of $\mathbb C$. The salient issue is about the angles of intersection, and the fact that a Möbius transformation is a conformal map hence it preserves angles of intersection. $\endgroup$ – Lee Mosher Mar 24 at 17:23
  • $\begingroup$ Just one more question: since Möbius transformations are open maps, they send boundaries to boundaries. $-i$ is on the boundary of the disk, however $T(-i)=\frac{(8+12i)(-i)+12+12i}{3i+3}=\frac{10}{3}+\frac{14i}{3}$ whis does not lie on the boundary of the plane $\{Re(z)\geq 3\}$. Doesn't this mean this is not the right Möbius transformation I was looking for? $\endgroup$ – John Keeper Mar 25 at 7:59

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