How Does the Maximum Modulus Principle Apply to Polynomial Functions?

[podboy6]

So my professor threw in what he called an extra 'hard' question for a practice test. So naturally I have a question about it. It relates to the Maximum Modulus Principle:

a) Let $$p(z) = a_0 + a_1 z + a_2 z^2 + ...$$
and let M = max |p(z)| on |z|=1.
Show that $$|a_i|< M$$ for $$i = 0,1,2.$$

b) What is the order of the zero at infinity if f(z) is a rational function of the form

$$f(z) = \frac {p(z)}{q(z)}$$

where both p(z) and q(z) are both polynomials and $$deg(p) < deg(q).$$

 

[NateTG]

a) Let $$p(z) = a_0 + a_1 z + a_2 z^2 + ...$$
and let M = max |p(z)| on |z|=1.
Show that $$|a_i|< M$$ for $$i = 0,1,2.$$

As you have it, that's not necessarily true.
Consider:
$$p(z) \equiv 0$$
That is, the polynomial is constant zero.
Then $$M=0$$,but
$$a_i = M = 0$$ which contradicts $$|a_i|<M$$

 

[podboy6]

okay, it should be $$|a_i|$$ is less than or equal to M for $$i=0,1,2.$$

 

[NateTG]

So, have you tried anything?

 

[podboy6]

Well, for the first part, given that |p(z)| $$\leq$$|M| for |z|=1,
and with:
p(0)=$$a_0$$
p'(0)=$$a_1$$
p''(0)=$$2a_2$$,

then in general,

$$|p^k (0)| \geq \frac{k!}{2\pi i} \int_{|z|=1} \frac{f(z)}{z^(k+1)} dz \Rightarrow \frac{k!}{2\pi i} \int_{0}^{2\pi} f( e^(it) ) dt \leq k!M$$.

thats about as far for that one. the other one is

$$\frac{p(z)}{q(z)} = \frac{a_0 + a_1 z +...+a_k z^k}{b_0 + b_1 z +...+b_l z^l}$$ for some l>k. After that, I'm still working.

 

[shmoe]

a) check your inequalities a little more carefeully, specifically comparing the derivatives at 0 to the integral.

b) What is the order of the zero of f(1/z) at z=0?