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
$$
S_k=\left\{z\in \mathbb C: 0< Arg(z) < \dfrac{2\pi k}{n} \right\}
$$
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. 
 A: Let
$$
C_n(a) = \left\{z\in\mathbb C\,\colon |z^n+n| = a\right\}.
$$
There are a few ways to see that $C_n(a)$ consists of exactly $n$ simple closed loops, one in each sector
$$
\frac{2\pi k}{n} < \arg z < \frac{2\pi (k+1)}{n}, \qquad k=1,2,\ldots,n,
$$
when $0 < a < n$.  The simplest (and least general) is to note that the circle $|w + n| = a$ in the $w$-plane does not intersect the non-negative real axis, so under the conformal mapping $w=z^n$ it has exactly $n$ smooth preimages, one strictly inside each such sector.
The point of $|w + n| = a$ with largest modulus lies at $w=-a-n$, so the point of $C_n(a)$ with largest modulus lies on the circle $|z| = (a+n)^{1/n}$.  Thus if $0 \leq a < n$ then
$$
z \in C_n(a) \quad \Longrightarrow \quad |z| < (2n)^{1/n}.
$$
Therefore if $n \geq 3$ and $z \in C_n(6^{1/3})$ we have
$$
|z| < (2n)^{1/n} \leq 6^{1/3} = |z^n + n|
$$
since $6^{1/3} < n$ and the function
$$
f(x) = (2x)^{1/x}
$$
is decreasing for $x \geq 3$.
By Rouché's theorem we may conclude that $z^n + z + n$ has precisely as many zeros inside each component of $C_n(6^{1/3})$ as does $z^n+n$.  Since $6^{1/3} < n$ each component surrounds exactly one zero of $z^n+n$ and lies strictly inside one of the sectors
$$
\frac{2\pi k}{n} < \arg z < \frac{2\pi (k+1)}{n}, \qquad k=1,2,\ldots,n.
$$
It follows that $z^n + z + n$ likewise has exactly one zero in each of these sectors.
We assumed above that $n \geq 3$.  We can address the case when $n=2$ by observing that the discriminant of $z^2 + z + 2$ is $-7$ and so its zeros are complex conjugates, one lying in the upper half plane and one in the lower.

Aside: This argument can be adapted to show that, for $n \geq 3$, the polynomial $z^n+z+n$ has a zero in each disk
  $$
\left|z-\zeta_n^k n^{1/n}\right| \leq n^{1/n} - \left(n-(2n)^{1/n}\right)^{1/n} = O\left(\frac{1}{n^2}\right),
$$
  $k = 1,2,\ldots,n$, where $\zeta_n$ is the principal $n^\text{th}$ root of $-1$.

