'constant' functions on complex analysis

[Redsummers]

Okay, so, I don't understand this concept of 'maximum principle'.

A few weeks ago we did Liouville's theorem, which states that any bounded complex function is continuous. Okay... (I can't really imagine the picture of a function which is bounded to be constant, e.g. sin(z) is bounded, at least it should, because the Real part is bounded by [-1,1] but, if I picture the sine function it is clearly not constant. So I am confused).

And today we did the maximum principle, which is more general than Liouville's (but still a corollary). It states that if the function reaches a maximum (e.g. local), then the function is constant. The proof makes sense and all, it's pure logic.
But if I think about it graphically, it makes none.

So if anyone of you guys could help me understand this concept I would appreciate. Also, how would you apply the maximum principle to a complex function? Say, e^(z^2), which in the real setting reaches a maximum?

Thank you

 

[HallsofIvy]

Okay, so, I don't understand this concept of 'maximum principle'.

A few weeks ago we did Liouville's theorem, which states that any bounded complex function is continuous.

Nonsense! f(z)= 0 if the real part is rational, 1+ i if the real part is irrational, is a "bounded complex function" that is NOT continuous.

What Liouville's theorem says is that any bounded function that is holomorphic (analytic in the entire complex plane) is a constant- which is what you appear to be saying below.

Okay... (I can't really imagine the picture of a function which is bounded to be constant, e.g. sin(z) is bounded, at least it should, because the Real part is bounded by [-1,1] but, if I picture the sine function it is clearly not constant. So I am confused).0

Yes, you are. We are talking about functions of a complex variable here. Whether or not the real part of a function is bounded or not says nothing about whether the function itself is bounded. $sin(z)= (e^{iz}- e^{-iz})/2$. In particular if z is imaginary, if z= yi, then $sin(z)= sin(iy)= (e^{-y}+ e^{y})/2$ which is NOT bounded.

And today we did the maximum principle, which is more general than Liouville's (but still a corollary). It states that if the function

a holomorphic function- don't leave out important parts of theorems!

reaches a maximum (e.g. local), then the function is constant. The proof makes sense and all, it's pure logic.
But if I think about it graphically, it makes none.

So if anyone of you guys could help me understand this concept I would appreciate. Also, how would you apply the maximum principle to a complex function? Say, e^(z^2), which in the real setting reaches a maximum?

Thank you

First, of all the complex numbers are NOT an "ordered field"- there is no way to say that one complex number is "larger" or "smaller" than another so it makes no sense to say that $e^{z^2}$ reaches a maximum or minimum. You must talk about the absolute value of a complex number or complex valued function, $\left|e^{z^2}\right|$ to have a "maximum" or "minimum"

 

[l'Hôpital]

EDIT: Eh. Hall Did a better job. Haha.

 

[micromass]

Also, the function $$e^{z^2}$$ does not have a maximum when z is real. I think you mean $$e^{-z^2}$$. But the function $$|e^{-z^2}|$$ does not have a maximum when z is complex. For instance, if z=iy, then $$|e^{-z^2}|=|e^{y^2}|$$ which does not have a maximum.

Your complex analysis course is no doubt very good. But I feel that most courses lack some geometric intuitions. A great book that remedies this is "Visual complex analysis". It gives a great intuition to complex analysis! Maybe you can check it out...

 

[Redsummers]

Oops, sorry, at Liouville's theorem I meant constant, not continuous...

Oh, I see the truth with the sin(z) unboundess, thank you.
Still, I was thinking about the whole concept of functions in the complex sense, and I don't know if I am completely right. Let me explain myself:
What I understand is that if we have a complex domain D, and a function f(z), if f(z) is bounded, then for all $$\zeta$$ in D, we will have $$f(\zeta) = const.$$

That is, the function will plot the whole domain D into a single point (const.). Is that what Liouville states?

Then for the maximum principle, it is kind of different I see. (As Hall stated, it is nonsense to talk about bigger and smaller values in the complex plane.) It is saying that if the absolute value of a function reaches a maximum inside a compact domain K, then the function is constant inside that domain. Hmm, so now I am confused about the difference between being bounded and the radius of convergence (because, e.g. tan(z) attains a radius of convergence of pi/2... so is it bounded?).

And for instance, whilst looking through the internet I found that log f(z), attains a maximum at say, z_0. And that implies that the absolute value of f(z) is constant, as well as f(z). (the sketch of the proof used here: http://en.wikipedia.org/wiki/Maximum_modulus_principle ). And I don't really see this graphically, because one could plug any z into this function and always get different images.

Thanks for your answers so far!

 

[Redsummers]

Oh! I see how I was mistaken! I see now the theorems clear and reasonable.

Also, thank you for the suggestion, micromas. I basically read about the visualisation of complex analysis somewhere in the internet and things got clear. You're right that one needs some intution as well, in order to understand the theorems and all.
I sometimes find that our courses are just too theoretical and give no room for intution. With Real analysis was okay, because you can somehow imagine it; but when it comes to complex analysis, it is not as trivial to visualise maps of C->C, i.e. 4 dimensions.

Thank you all again. And sorry for the double post.