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|
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.