# Complex Polynomial Inequality Proof

I'm trying to solve this question but I am having trouble connecting the dots. The question reads:

Assume that we have a complex polynomial: $$P(z) = a_0+a_1z+...+a_nz^n$$ Satisfies $$|P(z)|\leq 1$$ whenever $$|z|=1$$. Show that $$|a_n|\leq 1 \>\>\forall n$$.

So, I have simplified $$|P(z)|\geq|a_n|\left|1-\frac{|a_{n-1}|}{|a_n|} -...-\frac{|a_0|}{|a_n|}\right|$$, using the fact that $$|z|=1$$.

Now, I'm confused by how to proceed. There's no indication that the sequence is decreasing? Am I on the right track?

• Maybe it pays to think about integrating $z^{-k}P(z)$ around the unit circle. Commented Nov 12, 2018 at 4:55

This is very easy with a simple trick. Let $$q(z)=a_n+a_{n-1}z+\cdots+ a_0z^{n}$$. Then $$q(z)=z^{n}p(\frac 1 z)$$. $$q$$ is a polynomial and $$|q(z)| \leq 1$$ if $$|z|=1$$. By MMP $$|q(0)| \leq 1$$ which is what we need.
• Ah, I misunderstood the question – I thought it was asking for $|a_k|\le1$ for $0\le k\le n$. Commented Nov 13, 2018 at 10:06
One way to show this is via Cauchy estimates. From Cauchy integral formula, for the $$n$$-th derivative of $$P$$ we have $$P^{(n)}(z) = \frac{n!}{2\pi i } \int_{|\xi| = 1} \frac{P(\xi)}{(z - \xi)^{n+1}}d\xi.$$ Hence, using the fact that $$|P|\leq 1$$ on $$|\xi| = 1$$, we obtain $$n! |a_n| \leq \frac{n!}{2\pi } \int_{|\xi| = 1} \frac{|d\xi| }{|z - \xi|^{n+1}} = \frac{n!}{2\pi} \int\limits_0^{2\pi} \left|\frac{ie^{i\theta}} {(e^{i\theta})^{n+1}}\right| d\theta = n!,$$ which gives $$|a_n|\leq 1$$.