# Number of roots of a complex equation/ Rouche's theorem

For $n\geq2$ consider the equation $z^n+z+n=0$ for $z\in \mathbb C$. Show that if $k$ is an integer with $1\leq k \leq n$ then inside the sector

$$S_k=\{z\in \mathbb C: 0< Arg(z) < \dfrac{2\pi k}{n} \}$$

There are exactly $k$ roots of the above equation. $Arg (z)$ is the principal argument of $z$. (Hint: Prove that $x^n+n>x$ for real $x$)

The only thing I can think of is Rouche's theorem but then the region needs to be bounded to be able to use that. Can anybody give some pointers as to how I should proceed here. Thanks.

-
If you make the substitution $z = n^{1/n} \zeta$, you get the equation $\zeta^n + \frac{n^{1/n}}{n}\zeta + 1 = 0$. Does this give you an idea of where to look for your roots? – Antonio Vargas Jan 5 at 21:41
@AntonioVargas: I still don't see it. when $n$ is large the cofficient of $\zeta$ goes to zero. But that is not relevant, I think. We need to show this for all $n$ – user54755 Jan 5 at 23:28
Exactly, when $n$ is large the equation is very similar to $\zeta^n+1=0$. So you should be looking for the roots of $\zeta^n + \frac{n^{1/n}}{n}\zeta + 1 = 0$ near the roots of $\zeta^n+1=0$. This should give you an idea of what region to use in Rouché's theorem. – Antonio Vargas Jan 5 at 23:48