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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.

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  • $\begingroup$ I don't see why not. The derivative is again a rational function, whose numerator is the polynomial $p'q-q'p.$ By the fundamental theorem of algebra, if this is non-constant, then it has a zero. $\endgroup$ – Allawonder May 30 at 12:01
  • $\begingroup$ And it cannot be constant since if we differentiate it we get that the condition $$\frac{p'''}{q'''}=\frac pq,$$ which has no solution since $p$ is not less than quadratic. $\endgroup$ – Allawonder May 30 at 12:07
  • $\begingroup$ @Allawonder Thank you, that makes sense. However, can we eliminate the possibility that $p'q-q'p$ and $q^2$ share enough common factors for them to cancel out everything but a constant? $\endgroup$ – Stavroula Anna May 30 at 12:25
  • $\begingroup$ If $(p/q)'$ is a nonzero constant, that means $p/q$ is linear, which means $p=(az+b)q.$ This violates the assumption that $p,q$ should not share a root. $\endgroup$ – Allawonder May 30 at 12:34
  • $\begingroup$ @Allawonder Thanks again. But I do still have the following question: Consider the rational function $\frac{(z-1)(z+1)}{(z-i)(z+i)}$. Its derivative is $\frac{4z}{(z-i)^2(z+i)^2}$, so the degree of the numerator is reduced to $1$. Could it be possible to construct an example like this, where the numerator and denominator share a factor, so that it cancels to fraction, where the numerator is constant and the denominator isn't? The unsolvable equation $\frac{p''}{q''}=\frac{p}{q}$ only appears when we cannot clear any factors appearing in both numerator and denominator. $\endgroup$ – Stavroula Anna May 30 at 12:48
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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!

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