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Find an analytic function that maps the disk $\{|z|<1\}$ onto the disk $\{|w-1|<1\}$ so that $w(0)=1/2$ and $w(1)=0$

The 3 points theorem: Given 3 point $z_1, z_2, z_3 $ always map into 3 distinct point $w_1,w_2,w_3$ and only one linearly tranformation map $w=f(z)$ then 

<p>$$\frac{w-w_1}{w-w_3} .\frac{w_2-w_3}{w_2-w_1}= \frac{z-z_1}{z-z_3} .\frac{z_2-z_3}{z_2-z_1}$$</p>

I'm not sure I really know how to do this problem, all info I can get from the question is the domain is a unit circle with center at the origin and the image is also a unit circle with center at $(1.0)$. I also know that this function map 2 points $0 \to 1/2$ and $1 \to 0$.

How can I use the 3 points theorem with only 2 points? I can't just make up th ethird point, can I ?

Diane Vanderwaif
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2 Answers2

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You can't "just make up" the third point, but Möbius transformations have a very helpful property that allow you to determine where a specific other point is mapped:

If $T$ is a Möbius transformation mapping the circle (or straight line) $C$ to the circle (or straight line) $K$, and $z,\, z^\ast$ are two points symmetric with respect to $C$, then $T(z)$ and $T(z^\ast)$ are symmetric with respect to $K$.

Now, for your desired map, you know $C$ and $K$, and you know one point $z$ (namely $0$) not lying on $C$ and where it is mapped (namely $T(z) = \frac{1}{2}$). So you need only find the point symmetric to $z$ with respect to $C$ and the point symmetric to $T(z)$ with respect to $K$, and then you have your three points.

Daniel Fischer
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  • oh, so if I pick point $z=1$ which map to zero, then the symmetric of $z=1$ respect to $C$ is $z*=-1$ will map to the symmetric point of zero respect to $K$ is $w(-1)=2$. Am I correct? – Diane Vanderwaif Feb 22 '15 at 15:14
  • No, $z = 1$ lies on $C$, so its symmetric point with respect to $C$ is itself. The point not lying on the circle is $z = 0$. Thus the point symmetric to $0$ with respect to $C$ is the third point. – Daniel Fischer Feb 22 '15 at 15:16
  • I'm sorry, I'm not sure I understand, zero is the center of the disk $C$ right?, what do you mean by saying symmetric respect to $C$? – Diane Vanderwaif Feb 22 '15 at 15:26
  • Two points are symmetric with respect to a circle if they are mapped to each other by reflection in the circle. You can get the reflection in a circle by first mapping the circle to $\mathbb{R}\cup {\infty}$ by a Möbius transformation, then taking the complex conjugate, and finally mapping $\mathbb{R}\cup {\infty}$ back to the circle with the inverse of the previous Möbius transformation. If the circle is $\lvert z-a\rvert = r$, then two points $z,z^\ast$ that are neither $a$ nor $\infty$ are symmetric if and only if $(z-a)\overline{(z^\ast-a)} = r^2$. The centre is symmetric to $\infty$. – Daniel Fischer Feb 22 '15 at 15:33
  • so if I do the same thing to the image, I will have $(1/2-1) \overline{(T^(z)-1)}=1$ so $T^(z)=-1$, but the $-1$ is on the exterior region of the disk – Diane Vanderwaif Feb 22 '15 at 15:41
  • Yes, that's right. $T(z^\ast)$ must be outside the disk, since $z^\ast$ is outside of the other disk. – Daniel Fischer Feb 22 '15 at 15:47
  • oh, I thought I did something wrong again. I think I understand it now, thank you very much. – Diane Vanderwaif Feb 22 '15 at 15:50
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We just need a holomorphic function $f:\mathbb{D}\to\mathbb{D}$ that is onto and maps $0$ to $-\frac{1}{2}$ and $1$ to $-1$, so that $1+f$ fulfills the given constraints. Now, from this related question, we know that: $$ f(z) = e^{i\theta}\frac{z-a}{1-\bar{a} z} $$ for some $a\in\mathbb{D}$, hence: $$ \frac{1}{2}=ae^{i\theta},\qquad -1=e^{i\theta}\frac{1-a}{1-\bar{a}},$$ so $\theta=\pi$ and $a=-\frac{1}{2}$ leads to: $$ f(z) = - \frac{z+\frac{1}{2}}{1+\frac{z}{2}} = -\frac{2z+1}{2+z}$$ and the wanted analytic function is $\color{red}{-1+\frac{3}{2+z}}.$

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Jack D'Aurizio
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