Let’s consider $w = z + \frac1z$
and I want to find out the effect of this transformation on the circle $| z | = 1$
A method that seems effective
if $z = x + iy$ and $|z| = 1$ then $1/z = z^*$
So $w = z + \frac1z = z + z^* = x + iy + x – iy = 2x$
So the image, $w$, runs along the real axis and further, since $-1 \leq x \leq +1$ then $-2 \leq w \leq +2$. That is the image is constrained to run between $-2$ and $+2$ on the real axis.
The very useful Geogebra visualisation allowed me to input the transformation then drag $z_1$ roughly around the path of the circle $|z| = 1$. The image point $z_2$ in the Geogebra, which plays the role of w, traced a path along the real axis $(-2,+2)$ as the algebra above showed.
Now my challenge is to perform the transformation on $|z| = 2$.
Using the Geogebra visualisation, the image $z_2$ appears as if it could be an ellipse.
If I do similar algebra to above starting with $|z|= 2$ and $z = x + iy$
I get $w = \frac{5}{4}x + i\frac{3}{4}y$
And I can’t work out how to interpret / represent this in coordinate or polar form in order to create an “equation” for the image.
In general, the image of circle $\,|z|=r \gt 0\,$ is an ellipse with foci at $\,\pm2\,$, except for the case $r=1$ when the ellipse degenerates to segment $\,[-2,2]\,$ on the real axis.
Let $u$ be a square root of $z\,$, thus $\,u^2=z\,$ and $\,|u|^2=|z|=r\,$. Then:
$$ w+2=u^2+\frac{1}{u^2}+2=\left(u+\frac{1}{u}\right)^2 $$
Therefore:
$$ |w+2|=\left|u+\frac{1}{u}\right|^2 = \left(u+\frac{1}{u}\right)\left(\bar u+\frac{1}{\bar u}\right) = |u|^2+\frac{1}{|u|^2} + \frac{u}{\bar u} + \frac{\bar u}{u}=r+\frac{1}{r} + \frac{u}{\bar u} + \frac{\bar u}{u} \tag{1} $$
Redoing the same steps for $\,w-2\,$ gives:
$$ |w-2| =r+\frac{1}{r} - \frac{u}{\bar u} - \frac{\bar u}{u} \tag{2} $$
Adding $\,(1)+(2)\,$:
$$ |w+2|+|w-2|=2\left(r+\frac{1}{r}\right) $$
The latter is the equation of an ellipse with foci at $\,\pm 2\,$ and semi-major axis $\,r+\cfrac{1}{r} \ge 2\,$. The equality holds (only) for $r=1\,$, in which case the ellipse degenerates to the real segment $[-2,2]\,$.