# Argument principle and Rouché's theorem for $f(z)$

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.

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Actually $f$ is also positive when $z \geq 2$. – Libertron Dec 9 '11 at 23:51

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 $1 < |z| < 2$.
Was my method for trying to find how many roots were in the first quadrant correct? I think it turns out that there is only one as I believe the total change in argument should be $2 \pi$. – Libertron Dec 10 '11 at 2:22