Complex Analysis 2nd Ed. by Stephen D. Fisher: Q&A

[becu]

Hi,
I'm studying complex analysis right now, I would like to use this thread to ask questions when I read books. Many questions will be very stupid, so please bear with me.
Also, English is my second language.

text: Complex Analysis (2nd edition)
author: Stephen D. Fisher

[question deleted] this first question is very stupid. i figured it out. thank you. i will come back for other questions.

Thanks.

 

[becu]

Variation of Maximum principle: "If u(x,y) is harmonic and nonconstant on a domain D, then |u(x,y)| has no local maximum in D". the proof of this is left as an exercise. i want to prove it.

case 1: u(x,y) is complex-valued function, then since u is harmonic, it is analytic on D. By the Maximum Modulus Principle, |u| has no local max in D
case 2: u(x,y) is real-valued function, then u has no local max and no local min in D. How do you go from this to |u|?

Thanks.

 

[HallsofIvy]

If u is harmonic on D, then there exist another harmonic function, v(x,y), such that f(z)= u(z)+ iv(z) where z= x+ iy, is analytic on D.

 

[jostpuur]

If u is harmonic on D, then there exist another harmonic function, v(x,y), such that f(z)= u(z)+ iv(z) where z= x+ iy, is analytic on D.

Not true 100%. For example D=\mathbb{C}\backslash\{0\} and u(z)=\log |z|.

 

[HallsofIvy]

True! I was assuming a simply connected domain.

 

[becu]

If u is harmonic on D, then there exist another harmonic function, v(x,y), such that f(z)= u(z)+ iv(z) where z= x+ iy, is analytic on D.

then? we're using this fact to prove when the case u is complex-valued function right? I got that part down, put I'm stuck when u is real-valued function.

actually when I read further down, the book suggest to use mean-value property to prove above statement.