Is the following statement true?
Let $p, q $ be complex polynomials in one variable, with $\deg (p)\geq 2$, not sharing a root. Let $\deg(p)\geq\deg(q)$.
Then $(\frac{p}{q})'$ has a zero within the complex plane.
The context where this question arises from is the following:
I want to show that a nonconstant holomorphic function $f\colon \hat{\mathbb{C}}\to\hat{\mathbb{C}}$ of degree $\geq 2$ has a ramification point. I do not have any methods like the Riemann-Hurwitz formula at hands. So my reasoning for this problem is:
- We know that $f$ is rational function. It is biholomorphic iff its degree is 1.
- I have been able to show that if $f'(z_0)=0$ for some point $z_0$ in $\mathbb{C}$ with $f(z_0)\in \mathbb{C}$, then $z_0$ is a ramification point.
If the above statement was true this would solve the problem, after taking the reciprocal of $f$, if necessary.
Thank you.
Best Answer
Note that $\frac{z^2-2z+2}{(z-1)^2}=1+\frac{1}{(z-1)^2}$ and it is obvious that its derivative doesn't have zeroes in the plane, so the arguments in the comments are incorrect and the OP claim is incorrect too.
If the degree of $p$ is strictly higher than the result is true as if $\deg p=n >m =\deg q$, then $p'q-q'p$ has degree $n+m-1$ as the leading coefficients cannot cancel and that is strictly bigger than $2m$ unless $n=m+1$ so we cannot have full simplification except if the latter holds.
In this last case we would need $p'q-q'p=cq^2$ and then $q/q'p$ so $q,p$ must have acommon factor as $\deg q' < \deg q$ so contradiction!