<s>[INST] Complex numbers lies on ellipse $ z $ is a variable complex number such that $ |z|=2$. Show that the point $ z+ \frac{1}{z} $ lies on an ellipse of eccentricity $\frac{4}{5}$ in the complex plane.
 [/INST] As you know $|z|=2$, you can write $z$ as a function of its argument $$z = 2\cos\varphi + 2i\sin\varphi $$
and then $$\begin{align}\frac 1z & = \frac 12\cos(-\varphi) + \frac i2\sin(-\varphi) \\ & = \frac 12\cos\varphi - \frac i2\sin\varphi\end{align}$$
so $$\begin{align}z+\frac 1z & = \left(2+\frac 12\right)\cos\varphi + i\left(2-\frac 12\right)\sin\varphi \\ & = \frac 52\cos\varphi + i\frac 32\sin\varphi \end{align}$$
Now you need only to show that the locus of $(x,y)=(\frac 52\cos\varphi, \frac 32\sin\varphi)$ is an ellipse of a given eccentricity.
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