# Rouché's Theorem/Argument Principle Application

Preparing for some prelim. exams, I encountered this problem:

Show that $$p(z)=z^6+3z^4+1=0$$ has precisely two zeros in the upper half of the unit disk.

There are two good ways to solve this. The standard trick seems to be to notice that $$p(z)$$ sends $$\mathbb{R}\to\mathbb{R}$$, so that by unique continuation and the Schwartz Reflection Principle, we know that $$p(z)=\overline{p(\overline{z})}$$ for $$z$$ in the lower half of the unit disk. So, the number of zeros above is the same as the number of zeros below. We can then perform a straightforward application of Rouché's Principle to finish.

However, the less "clever" (but more general) way to solve this problem is to consider $$p(z)=z^6+3z^4+1$$ and compute $$\Delta\text{Arg}(p)$$ around the contour enclosing the upper half unit disk. We can see that $$\Delta Arg(p)=0$$ on the axis. Along $$e^{i\theta}$$ for $$\theta\in [0,\pi]$$, we know that the angle changes by $$\Delta \theta=\pi$$, so that considering the "dominant term" of $$3e^{4i\theta}$$, we calculate $$\Delta\text{Arg}(p)\approx 4\pi i$$. By the argument principle, we find that there are two zeros in the upper half disk.

My question: How do we make the second approach rigorous? I know how to use it, but I'm not really sold on the idea that the contributions of the terms of lesser modulus than $$3$$ are negligible. Can someone explain to me how to see this precisely?

We can construct a proof based on homotopy invariance, borrowing ideas from the proof of Rouche's theorem.

Let $$\gamma : [0, 1] \to\mathbb C$$ parametrise the semi-circular contour: $$\gamma(t) = \begin{cases} -1 + 4t & t \in [0, \tfrac 1 2] \\ e^{2\pi i (t - \tfrac 1 2 )} & t \in [\tfrac 1 2 , 1]\end{cases}$$

The number of zeroes of $$f$$ in the upper half disk is equal to the winding number around the origin for the curve $$f \circ \gamma : [0,1] \to \mathbb C^\star$$: $$f \circ \gamma(t) = \begin{cases} (-1 + 4t)^6 + 3(-1+4t)^4+1 & t \in [0, \tfrac 1 2] \\ e^{12\pi i (t - \tfrac 1 2 )} + 3e^{8\pi i (t - \tfrac 1 2 )} + 1 & t \in [\tfrac 1 2 , 1]\end{cases}$$

As you say, this winding number is hard to evaluate. However, since the $$3e^{8\pi i (t - \tfrac 1 2 )}$$ term is "dominant" for $$t \in [\tfrac 1 2 , 1]$$, we would expect the winding number of $$f \circ \gamma$$ around the origin to be the same as the winding number of the simpler-looking curve $$g : [0, 1] \to \mathbb C^\star$$, which is defined as: $$g(t) := \begin{cases} 3 & t\in [0, \tfrac 1 2] \\ 3e^{8\pi i (t - \tfrac 1 2 )} & t \in [\tfrac 1 2 , 1]\end{cases}.$$

To make this intuition rigorous, we exhibit a homotopy $$F: [0,1] \times [0,1] \to \mathbb C^\star$$ between $$g$$ and $$f \circ \gamma$$. A possible homotopy is $$F(s , t) = \begin{cases} 3(1-s) + \left((-1 + 4t)^6 + 3(-1+4t)^4+1\right)s & t \in [0, \tfrac 1 2] \\ se^{12\pi i (t - \tfrac 1 2 )} + 3e^{8\pi i (t - \tfrac 1 2 )} + s & t \in [\tfrac 1 2 , 1]\end{cases}$$ The key thing we need to check is that this homotopy avoids the origin, i.e. $$F(s, t) \neq 0$$ for all $$s$$ and $$t$$:

• If $$t \in [0, \tfrac 1 2 ]$$, then $$F(s, t) \neq 0$$ for all $$s \in [0,1]$$, because $$3$$ and $$(-1 + 4t)^6 + 3(-1 + 4t)^4 + 1$$ are both strictly positive real numbers.

• If $$t \in [\tfrac 1 2 , 1]$$, then $$F(s, t) \neq 0$$ for all $$s \in [0,1]$$, because $$\left|3e^{8\pi i (t - \tfrac 1 2 )}\right| > \left| e^{12\pi i(t - \tfrac 1 2 )} + 1\right|$$.

Thus, having shown that $$f \circ \gamma$$ and $$g$$ are homotopic within $$\mathbb C^\star$$, we deduce that they have the same winding number. As you say, the winding number of $$g$$ is obviously $$2$$, so this must be the winding number of $$f \circ \gamma$$ too.

• So the whole idea here is that $3$ is a sufficiently large coefficient precisely because we have this bound on the terms so that we avoid zero. In essence we just recycle the proof of Rouché's Theorem. Perfect, thanks. – Antonios-Alexandros Robotis Dec 30 '18 at 2:22