$f(z)=\frac{(iz+2)}{(4z+i)}$ 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.
 A: Let $f(z)=w=x+iy$ and show that $(x,y)$ is on a circle, provided $z \in \mathbb{R}$.  First, solve for $z$ so we can get a look at its real and imaginary parts:  $$w=\frac{(iz+2)}{(4z+i)} \\
(4z+i)w= iz+2 \\
z(4w-i)=2-iw \\
z=\frac{2-iw}{4w-i} = \frac{2-iw}{4w-i}\frac{4\overline{w}+i}{4\overline{w}+i}=\frac{8\overline{w}+2i-4i|w|^2+w}{|4w-i|^2}$$
Since $z$ is a real number, the imaginary part of the numerator of that last expression must be zero.  So
$$-8y+2-4(x^2+y^2)+y=0 \\
x^2 + y^2 +\frac{7}{4}y=\frac{1}{2} \\
x^2 + y^2 +\frac{7}{4}y + \left(\frac{7}{8}\right)^2=\frac{1}{2}+ \left(\frac{7}{8}\right)^2 = \frac{81}{64} \\
x^2 +\left(y+ \frac{7}{8}\right)^2=\left(\frac{9}{8}\right)^2 $$
Now claim the point $f(z)=(x,y)$ does lie on a circle, the circle has center $(0,-7/8)$ and the radius of the circle is $\frac{9}{8}$.
Verified by plotting some points.
To find the point $z_c$ that $f$ maps to the center of the circle $w_c$, we can use our simplest expression for $z$ in terms of $w$ from above. 
$$z_c=\frac{2-iw_c}{4w_c-i}=\frac{2-i\left( \frac{-7}{8}i \right)}{4\left( \frac{-7}{8}i \right)-i}=\frac{2-\frac{7}{8}}{-\frac{28}{8}i-i}=\frac{i}{4}$$
A: 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 $-\frac{7}{8}i$
A: For a real $x$, we have 
$$f(x) = \frac{9x}{16x^2+1} + \frac{2(2x^2-1)}{16x^2+1}i $$
When $x=0$, we have $f(0) = -2i$ and when $x \rightarrow \infty$, $f(x) \rightarrow \frac{1}{4}i$. Clearly, changing $x$ to $-x$ does not change the imaginary part of $f(x)$ and hence $f(\mathbb{R})$ is symmetrical about the imaginary axis. Thus $(0,-2)$ and $(0,\frac{1}{4})$ are ends of diameters of the image circle. Hence center is $\left(0, -\frac{7}{8}\right)$ and radius is $\frac{9}{8}$.
It is easy to see that $\frac{i}{4}$ maps to the center of the circle.
A simple computation shows that 
$$\left(\frac{9x}{16x^2+1}\right)^2 + \left(\frac{2(2x^2-1)}{16x^2+1}+\frac{7}{8}\right)^2 = \left(\frac{9}{8}\right)^2$$
