I have found that the curve $z(x)$ on the complex plane with $$ z=\frac{ax+b}{x^2+1} $$ at real $x$ spanning from $-\infty$ to $+\infty$ looks very similar to an ellipse at any complex parameters $a,b$ (see the example on the picture).
Is there any easy proof that this is indeed the ellipse?
On
$z=\frac{ax+b}{x^2+1}$
where x is just a parameter, so substitute it as t
let a and b be $a_1+ia_2$ and $b_1+ib_2$,
Clearly, Re(z)= $\frac{a_1t+b_1}{t^2+1}$ and Im(z)= $\frac{a_2t+b_2}{t^2+1}$
Re(z) is plotted on x axis and Im(z) is plotted on y, so simply remove t from both equations simultaneously and make it in cartesian form. Following that, check the eccentricity (there's a general formula as well) and it indeed comes out less than or equal to 1.
https://www.desmos.com/calculator/xjyxtjfyio
Another reason for it being elliptical is that, the function z is bounded in real domain i.e. $(ax+b)/(x^2+1)$ never goes above a finite value.
Do the variable substitution $x=\tan \cfrac{t}{2},\,\text{where }t\in[-\pi,\pi]$. This equation has a unique solution respect $t$. Further simplify and get $z=\cfrac{b}{2}+\cfrac{b}{2}\cos t +\cfrac{a}{2}\sin t $ The latter is obviously the equation of an ellipse.