# Image of real and imaginary axis under $f(z)= \frac {2z-1} {2-z}$

I have to find the image of real and imaginary axis under $f(z)= \frac {2z-1} {2-z}$, I have proved that it maps unit disk and upper half to itself.In some book it is written that only looking at the image of $0$ and $\infty$ we can tell what is the image of real and imaginary axis, I know Möbius transformation maps lines and circles to lines or circles. Please help me

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Linear fractional transformations, i.e. mappings of the form $\displaystyle z\mapsto\frac{az+b}{cz+d}$, map circles in $\mathbb C$ to circles, provided one construes straight lines as circles of infinite radius.
This mapping takes $0$, $1$, and $-1$ to real numbers; therefore it takes $\mathbb R\cup\{\infty\}$ to itself. It takes $0$, $i$, and $-i$ to $-1/2$, $-(4/5)+(3/5)i$ and $-(4/5)-(3/5)i$ respectively. Therfore it takes the imaginary axis to the circle that passes through those two points. It takes $\infty$ to $-2$, which is also on that circle. The diameter is $3/2$. Draw the picture and you'll see the center at $-5/4$ and the Pythagorean theorem will confirm that the distance from there to either of $-(4/5)\pm(3/5)i$ is $3/4$.
@Mathematician : Yes. The fact that the coefficients are real means there's symmetry about the real axis. So look at where $0$ is mapped to ($-1/2$) and where $\infty$ is mapped to ($-2$), and notice that both are real, and the center is half-way between those points. –  Michael Hardy Apr 1 '13 at 1:57