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?