# 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.

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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 '13 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 $n$ – Jack Dawkins Jan 5 '13 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 '13 at 23:48
@AntonioVargas Could you please explain your answer in detail? – Arpan Dutta Aug 11 '13 at 5:28
@ArpanDutta sure, I've just posted an answer. – Antonio Vargas Aug 12 '13 at 21:01

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 $-1$.

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