<s>[INST] $f(z)=\frac{(iz+2)}{(4z+i)}$ 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.
 [/INST] 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 
$$\left(\frac{9x}{16x^2+1}\right)^2 + \left(\frac{2(2x^2-1)}{16x^2+1}+\frac{7}{8}\right)^2 = \left(\frac{9}{8}\right)^2$$
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