Describe the complex function $f(z) = z + 1/z$. A question from my homework asks me to find the region where $f(z) = z+ \frac{1}{z}$ is real, and the region where $Im f(z) > 0$, the region where $Im f(z) < 0$.
But I don't know where to start, how do you tackle this kind of problems in general?
I am always very bad at this kind of visualizing problems on complex functions, any help would be appreciated.
What I would usually do is solve for
\begin{equation}
0 = Im ( (a + bi) + \frac{a-bi}{a^2 + b^2})
\end{equation}
But the algebra is usually  hideous. Is there any slick approach?
 A: Im$\left(a+b\color{blue}i+\dfrac{a-b\color{blue}i}{a^2+b^2}\right)=b-\dfrac b{a^2+b^2}=b\left(1-\dfrac1{a^2+b^2}\right)=0$
$\implies b=0$ or $a^2+b^2=1$, so $z+\dfrac1z$ is real when $z\in\mathbb R$ or $z=e^{i\theta}$ for $\theta\in\mathbb R$.
A: Without loss of generality (WLOG), $z \neq 0.$
Further, WLOG Im$(z) \neq 0$, else the entire problem (both questions) becomes trivial.
$$z + \frac{1}{z} = z + \frac{\overline{z}}{z\overline{z}}.\tag1$$
Further, $z + \overline{z} \in \mathbb{R}.$
Therefore, it is immediate that the RHS of equation (1) above will be real if and only if the denominator in the 2nd term of the RHS of equation (1) happens to equal $1$.
For the other question (WLOG) $|z| \neq 1$ :
Case 1
Im$(z) > 0.$ 
Im$(\overline{z})$ which happens to equal $(-1) \times ~$Im$(z)$, is being scaled by $\frac{1}{z\overline{z}} = \frac{1}{|z|^2}.$ 
Further $|z|^2 > 1 \iff |z| > 1.$
Therefore Im$\left(z + \frac{1}{z}\right) > 0 \iff |z| > 1.$
Case 2
Here, with Im$(z) < 0$, the situation is symmetrically opposite to Case 1. 
Therefore Im$\left(z + \frac{1}{z}\right) < 0 \iff |z| > 1.$
