# 'constant' functions on complex analysis

• Redsummers
I came across the following:If f is a complex function and z is in the domain of f, then there exists a point z0 such that f(z0) = const. This is stating that the function is a constant at z0. So, if we have a function that is approaching a maximum in some domain, then the function is constant at that point.In summary, the maximum principle states that if a holomorphic function reaches a maximum, then the function is constant at that point.f

#### 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

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"

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

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...

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.