Image of line under Mobius transformation The question states to calculate the image of the line $\{\text{Re}(z)=c\}$ under the Mobius transformation $z\mapsto \frac{z+1}{z-1}$. For $c=1$, it can be shown that the line is invariant under the transformation (in the extended real sense, i.e. $1\mapsto \infty$ and $\infty\mapsto 1$. For $c\neq1$, I have a hint saying that the image is a circle, but I am having trouble understanding what is going on. We know that 
$$
c+iv\mapsto\frac{c+1+iv}{c-1+iv}=\frac{(c+1+iv)(c-1-vi)}{(c-1)^2+v^2}
$$
but I do not see how the result can be shown to lie on a circle of some sort. Is there a better way of approaching this? Thanks.

Every non-singular Möbius transformation maps circles and lines to circles and lines. Equivalently, circles on the Riemann sphere are maps to other such circles. That does not mean that circles are mapped to circles or lines are mapped to lines.
For $c\ne 1$, the line where $-\infty < v < \infty$ is mapped under
$$
             c+iv \mapsto \frac{c+1+iv}{c-1+iv}
$$
to a circle in the complex plane. It cannot be mapped to a line because the point at $\infty$ is not in the image. The image is symmetric about the real axis, $c+i0$ is mapped to $\frac{c+1}{c-1}$, and $c\pm i\infty$ is mapped to $1$. So the center of the circle should be $\frac{1}{2}(1+\frac{c+1}{c-1})=\frac{c}{c-1}$. To verify,
\begin{align}
   \left|\frac{c+1+iv}{c-1+iv}-\frac{c}{c-1}\right|&=\left|\frac{c^2-1+iv(c-1)-c(c-1)-icv}{(c-1)(c-1+iv)}\right| \\
     &=\left|\frac{c-1-iv}{(c-1)(c-1+iv)}\right| \\
     &=\left|\frac{1}{c-1}\right|.
\end{align}
So the radius of the circle is given on the right, and you can see the above does define a circle in the complex plane.
