# Mathematical Olympiad question (complex variable)

complex numberscomplex-analysiscontest-math

I found this question in an old Mathematical Olympiad:

Let $$0 be a real number, and let $$f(z)$$ be a complex polynomial such that $$|f(z)|\leq \frac{1}{|z-a|}$$ on the unit disk $$|z|\leq 1$$. Prove that $$|f(a)| \leq \frac{1}{1-a^2}.$$

My attempt: Since $$f$$ is analytic in $$\{z\in\mathbb{C}:|z|\leq 1\}$$, we have that $$|f(a)| \leq \max_{|z|=1}|f(z)|.$$ Because of the triangle inequality, for all $$z\in \mathbb{C}$$ with $$|z|=1$$ we have $$|z-a|\geq |z|-a = 1-a.$$ By applying the hypothesis, we get $$|f(a)| \leq \max_{|z|=1}|f(z)| \leq \max_{|z|=1}\frac{1}{|z-a|}\leq \frac{1}{1-a}.$$ Nevertheless, we know that $$1-a^2>1-a$$ since $$a\in (0,1)$$. Therefore, we cannot get the desired result from the above inequality.

What can I apply to complete the proof?

Let $$F(z)=(z-a)f(z)$$ and $$G(z)=F(\frac {z+a} {1+az})$$. [Recall that $$z \to \frac {z+a} {1+az}$$ is an analytic bijection of the unit disk into itself whose inverse is $$z \to \frac {z-a} {1-az}$$]. Now $$|G(z)| \leq 1$$ and $$G(0)=F(a)=0$$. By Schwarz Lemma we get $$|G(z)| \leq |z|$$. This can be written as $$|F(z)|\leq |\frac {z-a}{1-az}|$$. Hence, $$|f(z)|\leq |\frac 1{1-az}|$$ and putting $$z=a$$ finishes the proof.