Recent content by Likemath2014

How can we show that the following equation has infinitely many solutions e^z-z^2=0. Thanks

many thanks, my question is the second part

Hi there, I am not sure if it's the right place to ask the question. My question is how the GPS chooses the best way. I mean where I can find something about its idea Thx

Yes, the question is how is that?

of course that is a grade school arithmetic, but it was not my question. My question is: how a^2+b^2 +2a( \lambda -1)+( \lambda-1)^2\leq\lambda^2 implies 2a(λ-1) < 2(λ-1)?

How could that help :redface:.

How I can show the following \int _{\mathbb{T}} \frac{1}{|1-e^{-i\theta}z|^2}dm(e^{i\theta})= \frac{1}{1-|z|^2} , where z is in the unit disc dm is the normalized Lebesgue measure and T is the unite circle.

I thought I got it, but it seems not yet :confused:. We will start like that, let |z+\lambda-1|^2 <|\lambda|^2, fro all z in the disc. Let z=a+ib, hence a^2+b^2+2a(\lambda-1)+(\lambda-1)^2<\lambda^2. How could that mean 2a(λ-1) < 2(λ-1)? Thx.

:redface:

Maybe it was not clear in the question that the inequality is for all z in the disc, that was my fault, I am sorry. Thank you very much mathman for the solution.

thank, but how could that imply that lambda is bigger than one? On the other hand how did you get the last inequality, why you consider a is positive?

In fact z is inside the disc that means |z|<1.

hi there, I am trying to prove the following inequality: let z\in \mathbb{D} then \left| \frac{z}{\lambda} +1-\frac{1}{\lambda}\right|<1 if and only if \lambda\geq1. The direction if \lambda>1 is pretty easy, but I am wondering about the other direction. Thanks in advance

Now it is clear, and the last one is true because (1-|z|^2)(1-|a|^2)>0. Thanks

I want to show that the modulus of the automorphism \frac{a-z}{1-\overline{a}z} is strictly bounded by 1 in the unit disc. Applying Schwarz lemma gives the result immediately. But I am looking for a straight forward proof for that. Thanks in advance