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

Question

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?

Answer 1

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.

Answer 2

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 $R-\frac 1R$.

Answer 3

Also when $R=1$, then $w=u+iv=f(z)=z+\frac{\bar z}{z \bar z}=z+\bar z=2x$ $$\implies u+iv=2x \implies u=2x, v=0,$$ then the locus $w=f(z)$ is a line segment on the x-axis in [-2,2].