Show that the function $f(z) = z^2 + z^{-2}$ maps the unit circle onto the interval $[-2, 2]$.
Okay so far, doing previous questions I firstly try and find the inverse mapping. Here I considered the unit circle as $|z| = 1$. I then converted this to $x^2 + y^2 = 1$.
However with this question I don't know how to find the inverse mapping of the above function so I'm stuck unfortunately.
On
You want to show two things: First, if $z$ is on the unit circle, i.e., if $z=e^{i\theta}$, then the given function maps $z$ to the interval $[-2,2]$. Also, if $x$ is a point on that interval, you want to produce a $z$ on the unit circle that maps to that $x$. Once you figure out the first part, the second part will be easier, because you'll have a function of one variable to invert.
Does that help?
Since our domain is the unit circle, $z$ is the form $e^{i\theta}=\cos(\theta)+i\sin(\theta)$
$$z^2=\cos(2\theta)+i\sin(2\theta)$$
$$z^{-2}=\cos(2\theta)-i\sin(2\theta)$$
$$z^2+\frac1{z^2}=2\cos(2\theta)$$
And we are done.