Solutions of $ \tan(z) = \frac{z}{z^{2} + 1}$ in the complexes In an exam I got this question:
Show that if the equation
$$ \tan(z) = \frac{z}{z^{2} + 1} $$
has $z_{0}$ as a solution, then $ \Re(z) = 0 $ or $ \Im(z) = 0 $
Writing $z$ as $x + i y$ seems too tedious. Is there a smart way of doing this?
 A: As $z \to \infty$, $\frac z {z^2+1} \to 0$, while on the other hand, $\tan(z)$ has large closed curves where it stays away from $0$ .
Since $\tan(z) \to \pm i$ as $\Im z \to \pm \infty$, it should be easy to show that for, say $\varepsilon = \frac 1 {100}$ and for integers $n$ large enough, if you consider the rectangle $R_n$ delimited by the lines $\Re z = \pm(n+\frac 12)\pi+\varepsilon$ and $\Im z = \pm n$, you will have $|\tan(z)| > |\frac z{z^2+1}|$ on $R_n$ ($\varepsilon$ is only there so that the rectangle doesn't go right through the poles of $\tan$).
You can then adapt Rouché's theorem to $f(z) = \tan(z) - \frac z{z^2+1}$ on those rectangles to show that inside each rectangle, $f$ and $\tan$ have the same difference "number of zeros" minus "number of poles".
Since $f$ keeps $\tan$'s poles at the same place with the same multiplicity, and has two more poles of order $1$ at $\pm i$, this implies that $f$ has exactly two more zeroes than $\tan$ inside each of those rectangles.
Because $\tan$ has $2n+1$ zeroes inside $R_n$, $f$ must have $2n+3$ zeroes. Then a study of $f$ on the real axis shows that $f$ has a triple zero at $0$, then one zero in each interval $((k-\frac 12)\pi ; (k+ \frac 12)\pi)$ when $k \neq 0$.
Since the number of real zeros of $f$ in $((-n-\frac 12)\pi ; (n+\frac 12)\pi$ matches the number of complex zeros of $f$ in $R_n$, this shows that $f$ can't have any additional complex zero.
