<s>[INST] Find z s.t. $z+\frac{1}{z}$ is real. Is my solution good? I must find all z s.t. $z+\frac{1}{z}$ is real. I know that $z = a + bi$ is real when the Imaginary part is 0. So, there we go:
$$z+\frac{1}{z}= \frac{z^2+1}{z}=\frac{(a+bi)^2+1}{a+bi}=\frac{[(a+bi)^2+1](a-bi)}{(a+bi)(a-bi)}$$
$$\frac{[(a+bi)^2+1](a-bi)}{(a+bi)(a-bi)} = \frac{(a^2 +2abi-b^2+1)(a-bi)}{(a^2+b^2)}$$
$$\frac{(a^2 +2abi-b^2+1)(a-bi)}{(a^2+b^2)}=\frac{a^3 +2a^2bi-b^2a+a-ba^2i+2ab^2+b^3i-bi}{(a^2+b^2)}$$
$$2a^2b-ba^2+b^3-b=0$$
$$a^2b+b^3-b=0$$
$$b_1=0$$
$$a^2+b^2-1=0$$
$$b^2=1-a^2$$
$$b_2=\sqrt{1-a^2}$$
$$b_3=-\sqrt{1-a^2}$$
so, the solutions are:
$$z_1=a+0i$$
$$z_1=a+\sqrt{1-a^2}i$$
$$z_1=a-\sqrt{1-a^2}i$$
I am not sure if that should be done like I did it.
 [/INST] If $z$ is real, you know what happens.
Now if $z$ is not real, the vectors $z$, $1/z$, and $z+1/z$ form a triangle. You would like the complex angle of $z+1/z$ to be $0$.
The vectors $z$ and $1/z$ have complex angles that are negative of each other. So now this triangle has two equal angles with $z+1/z$ representing the edge between them. It's an isosceles triangle, and it follows that the magnitude of $z$ equals that of $1/z$.
So the only $z$ for which this can hold are the $z$ on the unit circle. And you can then verify that for all such complex numbers, the sum is real: $e^{it}+e^{-it}=2\cos(t)$.
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