$f(z)=\frac{(iz+2)}{(4z+i)}$ maps the real axis in the $\mathbb{C}$-plane into a circle Find the center and radius of the circle. Also find the points on the complex plane which is mapped onto the center of the circle.

The inverse $z'$ of a point $z$ with respect to the circle centred at $a$ with radius $r$ is given by $$z' = \frac{r^2}{\overline{z}-\overline{a}}+a$$
We have $$\frac{iz+2}{4z+i} = \frac{9}{4(4z+i)}+ \frac{i}{4} = \frac{9}{16 \left(z - \frac{-i}{4} \right)} + \frac{i}{4} =\frac{9}{16 \left(\overline{z} - \frac{-i}{4} \right)} + \frac{i}{4} $$
($\because z \in \mathbb{R} \Rightarrow \overline{z} = z$)
With $a = \frac{i}{4}$ and $r = \frac{3}{4}$ its now evident that the real line is being inverted with respect to the circle $ \left|z - \frac{i}{4} \right| = \frac{3}{4}$ Since the real axis does not pass through the centre of the circle it is inverted into a circle under the given transformation.
The projection of $\frac{i}{4}$ on the real axis i.e. $z=0$ will transform into one end of a diameter of this circle ($z=0$ transforms to $-2i$), while $\frac{i}{4}$ itself will be the other end of the diameter. Hence the radius is $\frac{9}{8}$ and the centre is $-\frac{7}{8}i$
