How to construct Mobius transformation fixing real axis and mapping imaginary axis to circle

Question

I am working on an exercise from Fisher's Complex Variables text. In Exercise 7 part b on page 205 we are asked to find a fractional linear tranformation $T$ (Mobius Transformation) which maps the real axis onto itself and the imaginary axis onto the circle $C: |w-1/2| = 1/2$. In other words, $$ T(\mathbb{R}) = \mathbb{R} \qquad \& \qquad T(i\mathbb{R}) = C $$ Since I lack an intuition for how to do this I tried direct algebraic attack. We know there must be $A,B,C,D \in \mathbb{C}$ with $AD-BC \neq 0$ and $T(z) = \frac{Az+B}{Cz+D}$. If $z = x$ is real then $T(x)$ must also be real hence $T(x) = \overline{T(x)}$. Thus, for most real $x$, $$ \frac{Ax+B}{Cx+D} = \frac{\overline{A}x+\overline{B}}{\overline{C}x+\overline{D}} $$ and we can easily find: $$ B\overline{D} = \overline{B}D, \qquad A\overline{D}+B\overline{C} = \overline{B}C+\overline{A}D, \qquad A\overline{C}=\overline{A}C. $$ All of this said, I am not much closer to a solution. Next, the condition $T(i \mathbb{R}) = C$ amounts to the condition that $|T(iy)-1/2| = 1/2$ for $y \in \mathbb{R}$. That is: $$ \left|\frac{Aiy+B}{Ciy+D} - \frac{1}{2} \right| = \frac{1}{2} $$ multiply by $2$ and create a common denominator, $$ \left|\frac{2Aiy+2B-Ciy-D}{Ciy+D}\right| = 1 \ \ \Rightarrow \ \ \left|\frac{2B-D+i(2A-C)y}{Ciy+D}\right| = 1 $$ Setting $y=0$ provides $|2B-D| = |D|$ and in conjunction with the previous condition I find $B=D$. This is far as I have gotten.

The answer given in the back of the text is $T(z) = \frac{1}{z+1}$

this means $B=D=1$ which doesn't disagree with my analysis thus far. However, I am curious how to derive $A = 0$ and $C=1$.

Question: how to complete my analysis, or, better yet, how to replace my analysis with an nice clear solution.

Thanks!

Answer 1

Following up on OP's approach...

in conjunction with the previous condition I find $B=D$

Note that $\,B=D \ne 0\,$, otherwise the transformation would reduce to a constant. Then multiplying both the numerator and the denominator by $\,\dfrac{\bar B}{|B|^2}\,$ it can be assumed WLOG that $\,B=D=1\,$.

$A\overline{D}+B\overline{C} = \overline{B}C+\overline{A}D\,$

This reduces to $\;A-\bar A=C - \bar C \iff \operatorname{Im} A = \operatorname{Im} C\,$.

$A\overline{C}=\overline{A}C$

Using the above, this is equivalent to $\,\operatorname{Im} A\bar C = 0 \iff \operatorname{Im} A \cdot (\operatorname{Re} C - \operatorname{Re} A) = 0\,$. But $\,\operatorname{Re} A - \operatorname{Re} C \ne 0\,$, otherwise that would mean $\,A=C\,$ and the transformation would again be a constant, which leaves $\,\operatorname{Im} A = 0 \,$, and therefore $\,A, C \in \Bbb R\,$.

$\left|\dfrac{2B-D+i(2A-C)y}{Ciy+D}\right| = 1$

This now reduces to $\,|1 + i(2A-C)y| = |1+i Cy| \iff 1+(2A-C)^2y^2=1+C^2y^2\,$. Since $\,A\ne C\,$, it then follows that $\,A=0\,$.

So the general form of the tranformation is $\,T(z) = \dfrac{1}{C z + 1}\,$ with $\,C \in \Bbb R \setminus \{ 0 \}\,$.

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[ EDIT ]   A footnote about this line in the original post:

Setting $y=0$ provides $|2B-D| = |D|$ and in conjunction with the previous condition I find $B=D$.

This only follows if assuming that $\,B \ne 0\,$. But $\,B=0\,$ is a valid choice, and similar calculations would then lead to the alternative representation in @Lubin's answer $\,T(z)=\dfrac{z}{z+D}\,$. The two forms differ by an inversion $\,z \mapsto \dfrac{1}{z}\,$ which preserves both axes, so they are equivalent.

Answer 2

Here’s how I’d do it. I confess that I learned this stuff a long time ago, and I may have forgotten how much background material I’m using without being aware of it.

Sends real axis to real axis? Then should be represented as $z\mapsto\frac{Az+B}{Cz+D}$ with all constants real.

Might as well suppose $0\mapsto0$, so that means $B=0$; and we can try (real)$\infty\to1$, so that $A=C$. Since we can multiply all constants by the same number without changing the function, might as well take $A=C=1$, and that leaves only $D$ to be determined. Let’s see what the unknown function does, \begin{equation}f:\>z\mapsto\frac z{z+D}\,. \end{equation} Without calculating anything at all, but knowing that the image of the imaginary axis is a circle containing the origin and the unit point $1$, and further knowing that the image of the imaginary line is sent to itself by complex conjugation (because in this case $f(\overline z)=\overline{f(z)}\>$), I think that we can conclude that as long as $D\ne0$, this function already satisfies the demand.

Answer 3

Either $0$ is mapped to $0$ and $\infty$ is mapped to $1$ or $0$ is mapped to $1$ and $\infty$ is mapped to $0$. Therefore, the transform is either $z/(z + a)$ or $a/(z + a)$, $\,$for $a \in \mathbb R \backslash \{0\}$.

Answer 4

In the end, I think the answer my Maxim is important as a nice method to guide the solution. The key idea is that the Mobius transformation must map intersections to intersections for the given curves. For the problem given here, $T(\mathbb{R}) = \mathbb{R}$ and $T(i\mathbb{R}) = C$ we can see $\mathbb{R}$ and $i \mathbb{R}$ in the extended $z$-plane intersect at both $\infty$ and $z=0$. On the other hand, $C$ and $\mathbb{R}$ in the extended $w$-plane intersect at both $0$ and $1$. Therefore, we have two choices:

1. $T(0) = 0$ and $T(\infty) = 1$ which implies $T(z) = \frac{z}{z+A}$ for some $A$ which we can easily see must be real from the $T(\mathbb{R})=\mathbb{R}$ condition,
2. $T(0)=1$ and $T(\infty) = 0$ which implies $T(z) = \frac{1}{Az+1}$

These choices must conform to the other given data; $T(i \mathbb{R}) = C = \{ w \ | \ |w-1/2|=1/2 \}$. In particular, we can study an easy point on the imaginary axis and see if there is any choice of $A$ which allows us to meet the condition of $C$.

Working on case 1, Study $T(i)=w$, $$ T(i) = \frac{i}{i+A} = w $$ we need $|w-1/2|= 1/2$ or $|2w-1|=1$ so $$ \left| \frac{2i}{i+A} - 1\right| = 1 \ \ \Rightarrow \ \ \left| 2i - i-A \right| = |i+A| \ \ \Rightarrow \ \ |i-A| = |i+A|. $$ which is true. Apparently any choice of $A \neq 0$ will do as $T(i) = \frac{i}{i+A} \in C$ and we already arranged for two other points from $i\mathbb{R}$ to map to $C$. Recall, three points fix a Mobius transformation.

Likewise, work on case 2, Study $T(i) =w$, $$ T(i) = \frac{1}{Ai+1} = w $$ we need $|w-1/2|= 1/2$ or $|2w-1|=1$ so $$ \left| \frac{2}{Ai+1} - 1\right| = 1 \ \ \Rightarrow \ \ |2-Ai-1| = |Ai+1| \ \ \Rightarrow \ \ |1-iA| = |1+iA| $$ which is true, so again, any $A \neq 0$ will make $T(z) = \frac{1}{Az+1}$ a map which fixes the real axis while mapping the imaginary axis to $C$.