These two questions are driving me mad as I need to help my daughter but I can't remember all this stuff.

$\,(1)\,\,$ Let $\,p(z)\,,\,q(z)\,$ be two non-constant complex polynomials of the same degree s.t. $$\text{whenever |z|=1}\,\,,\,|p(z)|=|q(z)|$$

If all the zeros of both $\,p(z)\,,\,q(z)\,$ are within the open unit disk $\,|z|<1\,$ , prove that $$\forall z\in\mathbb{C}\,\,,\,q(z)=\lambda\, p(z)\,,\,\lambda\in\mathbb{C}\,\,\text{a constant}$$

What've I thought: since the polynomials are of the same degree, I know that $$\lim_{|z|\to\infty}\frac{q(z)}{p(z)}$$ exists finitely, so we can bound $\,\displaystyle{\frac{q(z)}{p(z)}}\,$ say in $\,|z|>1\,$ . Unfortunately, I can't use Liouville's Theorem to get an overall bound as the rational function is not entire within the unit disk...

$\,(2)\,\,$ Let $\,f(z)\,$ be analytic in the punctured disk $\,\{z\in\mathbb{C}\;|\;0<|z-a|<r\,,\,\text{for some}\,\,0<r\in\mathbb R\}\,$.

Prove that if $\,\displaystyle{\lim_{z\to a}f'(z)}\,$ exists and finite, then $\,a\,$ is a removable singularity of $\,f(z)\,$ .

My thoughts: We have a Laurent expansion in the above disk$$f(z)=\ldots +\frac{a_{-n}}{(x-a)^n}+\ldots +\frac{a_{-1}}{z-a}+a_0+a_1(z-a)+\ldots$$ so taking the derivative term-term (is there any special condition that must be fulfilled in this particular case to do so?) we get$$f'(z)=\ldots -\frac{na_{-n}}{(z-a)^{n+1}}-\ldots -\frac{a_{-1}}{(z-a)^2}+a_1+2a_2(z-a)+\ldots$$Now, as the limit of the above when $\,z\to a\,$ exists finitely, it must be that all the terms with negative power of $\,z-a\,$ vanish, thus $$\ldots =\,a_{-n}=a_{-n+1}=\ldots =a_{-1}=0$$ and we get that the above Laurent series for $\,f(z)\,$ is, in fact, a Taylor one and, thus, the function's limit (not the derivative's!) exists and finite when $\,z\to a\,$ and $\,a\,$ is then a removable singularity.

Any help in (1) if I got right (2), or in both if there's some problem with the latter will be much appreciated.

  • 3
    $\begingroup$ +1 for your carefully formulated question, your solution of (2) and above all for being such a great daddy! $\endgroup$ – Georges Elencwajg Jun 18 '12 at 11:16
  • $\begingroup$ Hehe...thanks, @George! But this girl's exercises in complex functions are making me realize I should have taught that course or, at least, follow it more closely. Most of the time it was linear, abstract algebra, calculus...I need now to get into shape in order to be a good dad... $\endgroup$ – DonAntonio Jun 18 '12 at 12:33

The function $g(z)= \frac{p(\frac{1}{z})}{q(\frac{1}{z})}$ has a removable singularity at $0$ since the $ \lim_{z \rightarrow \infty }\frac{p(z)}{q(z)}$ exists. Furthermore since $p(z),q(z)$ have all their zeros inside $|z|=1\,$ , the functions $g(z)$ and $\frac{1}{g(z)}$ are analytic in $|z|<1+\epsilon\,$ , for some $\,\epsilon >0\,$ , so by the maximum principle both functions $\left(|g(z)|, \left|\frac{1}{g(z)}\right|\right)$ are bounded by $1$ for $ |z| \leq 1$ since the condition $|p(z)|=|q(z)|$ for $|z|=1$ that means $|g(z)|=1$ for $|z| \leq 1$ and you are done.

  • $\begingroup$ you needed that $\epsilon >0$ in order to apply the Maxixmum principle, also you can find it because the zeros inside the circe are finite so they have a positive distance from the circe. $\endgroup$ – clark Jun 18 '12 at 0:13
  • $\begingroup$ thanks for the answer but I still don't follow. At the end of your answer, did you mean $\,\displaystyle{|f(z)|:=\left|\frac{p(z)}{q(z)}\right|=1\,\,\,for\,\,\,|z|\leq 1}\,$? Or what is that function $\,f(z)\,$? And how exactly does any of the above end problem (1)? $\endgroup$ – DonAntonio Jun 18 '12 at 2:05
  • $\begingroup$ I fixed it, it was a typo mistake I meant $g(z)$ instead of $f(z)$ and you conclude $|g(z)|=1 $ for $ |z| \leq 1$ because it is analytic you can coclude that $g(z)$ is constant you can go there using the cauchy riemann equations for instance. And then $ \frac{p(z)}{q(z)} = e^{i \theta} = g(\frac{1}{z}) \forall z \geq 1$ and since they are polynomials and they take the same value for infinite $z$ they must coincide $\endgroup$ – clark Jun 18 '12 at 10:42

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