Say I have a complex valued polynomial $f(z)= z^4 -6z +3$. I'm trying to use the quarter disk $0 \leq \theta \leq \pi/2$, but on the real axis, it seems $f$ is positive only when $z \leq 0$. Does that mean that the change of argument is $- \pi$ for this particular part of the quarter disk region? How would one apply Rouché's Theorem then to find how many roots of the equation have modulus between $1$ and $2$? The quarter disk I'm using to find how many roots lie in the first quadrant, just to clarify.
[Math] Argument principle and Rouché’s theorem for $f(z)$
complex-analysis
Related Solutions
We usually apply Rouché's theorem inside the disk and here is not the case. Instead, we use the argument principle.
It suffices to show that there is exactly one root in the first quadrant because it is a polynomial with real coefficients, and zeros of polynomials come in conjugate pairs.
Denote $P(z)=z^4+2z^2-z+1$.
$P(z)$ has no zeros on the real axis because $P(z)=z^4+z^2+(z-\frac12)^2+\frac34>0$.
$P(z)$ has no zeros on the imaginary axis because $P(iy)=y^4-y^2-iy+1\ne0$.
Let $L_1=[0,R]$, $L_2=[0,Ri]$, $\Gamma=Re^{i\theta}, 0\le \theta\le\frac\pi2$. Consider the quarter circle $C=L_1+ \Gamma- L_2$. From the argument principle, $$\underbrace{\text{number of zeros}}_{:=N} - \underbrace{\text{number of poles}}_{=0}=\frac1{2\pi i}\oint_C\frac{P'(z)}{P(z)}~\mathrm dz=\frac{\Delta_C\arg P(z)}{2\pi},$$ or \begin{align*} N&=\frac{\Delta_{L_1}\arg P(z)}{2\pi}+\frac{\Delta_\Gamma\arg P(z)}{2\pi}-\frac{\Delta_{L_2}\arg P(z)}{2\pi}. \end{align*}
We have $\Delta_{L_1}\arg P(z)=0$ because $P(x)>0~\forall~ x\in\Bbb R$, $$\Delta_\Gamma\arg P(z)=\Delta_\Gamma\arg z^4\left(1+\frac{2z^2-z+1}{z^4}\right)\to4\cdot\frac{\pi}{2}\quad\text{as}~ R\to\infty,$$ and \begin{align*} \Delta_{L_2}\arg P(z)&=\arg P(iR)=\arg{(R^4-R^2+1-iR)}\\ &=\arg\left(1-\frac{iR}{R^4-R^2+1}\right)\to 0 \quad\text{as}~ R\to\infty. \end{align*}
Therefore $N=1$, there is exactly one root in the first quadrant.
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.
Best Answer
Let $f(z)=z^4 -6z +3$. Then when $|z|=1+\epsilon$ where $\epsilon>0$ is sufficiently small, we have $$|f(z)-(-6z)|=|z^4+3|\leq |z|^4+3=(1+\epsilon)^4+3<6(1+\epsilon)=|-6z|.$$ By Rouche's theorem, $f(z)$ and $-6z$ has the same number of zeros in $|z|<1+\epsilon$. This implies that $f(z)$ has one zero inside $|z|<1+\epsilon$ for $\epsilon>0$ sufficiently small. On the other hand, when $|z|=2$, $$|f(z)-z^4|=|-6z+3|\leq 6|z|+3=15<16=|z^4|.$$ By Rouche's theorem again, $f(z)$ and $z^4$ has the same number of zeros in $|z|<2$. This implies that $f(z)$ has four zeros inside $|z|<2$. Combining all these, we can conclude that $f(z)$ has three zeros in $1 < |z| < 2$.