Question related to conformal mappings

[sari]

1. P(z) = a_0 + a_1*z + a_2*z^2 + ... + a_n*z^n is 1-1 on the open unit disk D = {z: |z|<1}. Prove that |a_n| < |a_1|\n.



2.



3. P(z) is analytic and 1-1, and therefore conformal, so |P'(z)|>0 for every z in D.
So 0 < |P'(z)|=|a_1 + 2a_2*z +...+ na_n*z^n-1| < |a_1| + ...+|na_n|.

Also the inverse function P^-1(z) is conformal in P(D) and |P^-1(z)|<1.
We also know that the derivative d\dz P^-1(z) = 1\P'(P^-1(z)) and that |d\dz P^-1(z)|>0.

Additionally, we know that a_k = (P^(k) (0)\ k!) .

The only other things I thought of that could be useful are the Cauchy inequality |a_k| < M(r)\ r^k, where M(r) = max{f(z): |z| = r }, and maybe Shwarz's Lemma, though since P(z) doesn't map 0 to 0 and doesn't map the unit disc to itself, I don't see how it would be relevant.

Would appreciate any help very much ... I don't really know where to go from here.

 

[mighty2000]

To prove that |a_n| < |a_1|, we can use the Cauchy inequality and the fact that P(z) is 1-1 on the open unit disk D.

From the Cauchy inequality, we know that |a_n| < M(1)\ 1^n = M(1) for all n, where M(1) is the maximum value of P(z) on the unit circle.

Since P(z) is 1-1 on D, it must map each point on the unit circle to a unique point inside the unit disk. Therefore, the maximum value of P(z) on the unit circle must be less than the maximum value of P(z) on the unit disk.

So, we have |a_n| < M(1) < M(r) for all n and for all r<1.

Using the Cauchy inequality again, we have |a_1| < M(r)\ r for all r<1.

Since |a_1| < M(r)\ r for all r<1 and |a_n| < M(r) for all n and for all r<1, we can conclude that |a_n| < |a_1| for all n.

Therefore, we have proven that |a_n| < |a_1| for all n, as required.