If $f(z)= z + \frac{1}{z}$ and $z$-points move on the circle $|z|=R$, then what is the locus of $f(z)$ points Here's what I did to graph 
$$f(z) = z + \frac{1}{z}$$
where $|z|=R$. Given $|z|=R$ (all points being in a circle of radius R), I can get y depending on x and vice-versa, so $y=\pm \sqrt{R^2-x^2}$.
After some rearrangements 
$$f(x,y) = x\frac{R^2+1}{R} + i*y\frac{R^2-1}{R^2}$$
From where I got the real and imaginary parts of the graph.
Real: $$u(x) = x\frac{R^2+1}{R^2}$$
And 2 imaginary part functions: 
$$v_1(x) = \sqrt{R^2-x^2} \frac{R^2-1}{R^2}$$
$$v_2(x) = -\sqrt{R^2-x^2} \frac{R^2-1}{R^2}$$
Also some helpful inequations I deduced: $$\frac{R^2+1}{R^2} > 1$$
$$\frac{R^2-1}{R^2} < 1$$
$$|x|\leq R$$
$$|y|\leq R$$
By calculating a few values of the functions I got some points on the Argand plane, and a graph might look like an ellipse centered at the origin, but I'm not sure about that.
The question is: what is a graph of that function and how I can obtain it analytically?

Continue with what you obtained and let $f(x,y)=u+iv$,
$$u+iv=  x\frac{R^2+1}{R^2} + iy\frac{R^2-1}{R^2}$$
Then, equate both real and imaginary parts,
$$x=\frac{R^2}{R^2+1}u,\>\>\>\>\>y=\frac{R^2}{R^2-1}v\tag 1$$
Given that $|z|=R$, a circle of radius $R$ center at the origin, we have 
$$x^2+y^2 = R^2$$
Substitute (1) into above equation to obtain 
$$\frac{u^2}{\frac{(R^2+1)^2}{R^2} }+ \frac{v^2}{\frac{(R^2-1)^2}{R^2} }=1$$
which represents an ellipse centered at origin, with the major and the minor axes of $\frac{R^2+1}{R}$ and $\frac{R^2-1}{R}$, respectively.
