Show that the center of the circle of the Mobius transformation $f(z)=i(1-z)/(1+z)$ of $z=k+iy$ is always on the $y$ axis.

Question

How can I show that the center of the circle of the Mobius transformation $f(z)=i(1-z)/(1+z)$ of $z=k+iy$ is always on the $y$ axis when $k$ is a constant? I know that this is a direct result of the Smith chart, but I tried to apply the function directly and expand but got stuck due to a possible mistake of long calculations. Is there a way to show this directly?

Answer 1

Hint: Use the fact that Möbius transformation is angle-preserving, namely the images of ${\bf Re}(z)=k$ and ${\bf Im}(z)=0$ should be orthogonal.

Details. Let $f(z)=i(1-z)/(1+z).$ One notes that the image under $f$ of ${\bf Im}(z)=0$ is the $y$-axis. Also the image of ${\bf Re}(z)=k~(k\neq -1)$ passes through circle $ABC,$ where $$A=f(k)=\frac {i(1-k)}{1+k},B=f(\infty)=-i,C=f(k+i)=\frac {1+(1-k)i}{(1+k)+i}.$$ Since $f$ is conformal, circle $ABC$ intersects the $y$-axis orthogonally at $A$ (being the image of $k$ that is the common intersection of the perpendicular lines ${\bf Re}(z)=k$ and ${\bf Im}(z)=0$).

It follows that the segment $AB$ (on the $y$-axis) must be the diagonal of circle $ABC,$ hence its center is the midpoint of $AB,$ i.e. $\frac{-k}{k+1}i$ or $(0,-k/(k+1)).$ Alternatively, one may check that the circumcenter of triangle $ABC$ is indeed $\frac{-k}{k+1}i.$

Remark 1. When $k=-1,$ the image of $z=-1+yi$ under $f$ is $\{z~|~{\bf Im}(z)=-1\}.$ In this case, one may still regard the center $\infty$ as a point on the $y$-axis.

Remark 2. One can also use algebraic manipulations as follows. The line ${\bf Re}(z)=k$ can be written as $$z+{\bar z}=2k.\qquad (1)$$ Let $u+iv=w=f(z)=i(1-z)/(1+z).$ Then $$z=\frac {i-w}{i+w}.\qquad (2)$$ Substituting (2) in (1), one has $$\frac{i-w}{i+w}+\frac{-i-{\bar w}}{-i+{\bar w}}=2k$$ $$\Leftrightarrow (i-w)(-i+{\bar w})+(-i-{\bar w})(i+w)=2k(i+w)(-i+{\bar w})$$ $$\Leftrightarrow 2-2w{\bar w}=2k(1+i{\bar w}-iw +w{\bar w})$$ $$\Leftrightarrow (k+1)w{\bar w}+ki({\bar w}-w)+k-1=0$$ $$\Leftrightarrow (k+1)(u^2+v^2)+2kv+k-1=0~(w=u+iv)$$ $$\Leftrightarrow u^2+\left(v+\frac k{k+1}\right)^2=\frac 1{(k+1)^2},$$ as required.

Answer 2

Assuming $\,k\,$ is a real constant, $\,L = \{k + i y \mid y \in \mathbb R\} = \{ z \in \mathbb C \mid \text{Re}(z) = k\}\,$ is a vertical line parallel with the imaginary axis. Let the circle $\,\Gamma = f(L)\,$ be its image, then:

• $\displaystyle L \ni z \implies \Gamma \ni f(z) = i \frac{1-z}{1+z}$

• $\displaystyle L \ni z \implies L \ni \overline z \implies \Gamma \ni f(\overline{z}) = i \frac{1 - \overline{z}}{1+\overline{z}}= - \overline{f(z)}$

With $\,w = f(z)\,$, it follows that for each $\,z \in L\,$ both $\,w, -\overline{w} \in \Gamma\,$. But points $\,w, -\overline{w}\,$ are symmetric across the imaginary axis, so the imaginary axis must be an axis of symmetry of $\,\Gamma\,$, thus a diameter of circle $\,\Gamma\,$, thus contains its center.