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

complex numberscomplex-analysisgraphing-functions

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?

Best Answer

Alternatively, you may use the polar form $z=R(\cos\theta+i\sin\theta)$. We then have $\frac 1z=\frac 1R(\cos\theta-i\sin\theta)$ and therefore $$f(z)=\left(R+\frac 1R\right)\cos\theta+i\left(R-\frac 1R\right)\sin\theta.$$ This is exactly the standard parametric expression for an ellipse with semi-major axis $R+\frac 1R$ and semi-minor axis $R-\frac 1R$.