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.

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    $\begingroup$ 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. $\endgroup$ – Antonios-Alexandros Robotis Dec 30 '18 at 2:22

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