How Does the Imaginary Part of a Holomorphic Function Relate to the Real Part?

[cheeez]

does the behavior the imaginary part behave in anyway similar to the real part of a holomorphic function. say if the real part if bounded or positive, what can you conclude about the imaginary part.

 

[HallsofIvy]

The Cauchy-Riemann equations relate the real and imaginary parts of holomorphic functions.

Since the only bounded holomorphic functions are the constants, If the real part of a holomorphic function is bounded, then either it and the imaginary part are constants or the imaginary part cannot be bounded.

 

[cheeez]

so what about sinz or cos those aren't constant but clearly holomorphic right'

 

[HallsofIvy]

They also are not bounded!

$$cos(z)= \frac{e^{iz}+ e^{-iz}}{2}$$

If z is imaginary, say, z= ix, then
$$sin(ix)= \frac{e^{-x}+ e^{x}}{2}= -cosh(x)$$
which is not bounded.

 

[micromass]

An easy corollary of the Cauchy-Riemann equations is the following:

A complex differentiable function which takes on purely real (or purely imaginary) values is locally constant.

From this we obtain the following interesting theorem:

The real part of a function which is analytic in a connected open set is uniquely determined by it's imaginary part, up to an additive constant.

This is probably something which you are looking for?

 

[cheeez]

They also are not bounded!

$$cos(z)= \frac{e^{iz}+ e^{-iz}}{2}$$

If z is imaginary, say, z= ix, then
$$sin(ix)= \frac{e^{-x}+ e^{x}}{2}= -cosh(x)$$
which is not bounded.

doh I am still not used to thinking in complex

@micromass that's very helpful

 

[jakeunna]

I am looking at a past complex analysis paper and the following question comes up:

Let $$M\in\mathbb{R},f:\mathbb{C}\to\mathbb{C}$$ where $$f$$ is analytic on $$\mathbb{C}$$. Suppose for all $$z\in\mathbb{C}$$ we have that $$Re(f(z))\leq M$$. Show that $$f$$ is constant.

Can this just be solved by pointing out that the imaginary part is determined by the real part plus or minus some constant, and therefore the imaginary part is also bounded so by Liouville's theorem f is bounded?

 

[exponentiate]

The trick is to exponentiate. $$e^f$$ is analytic and bounded, hence is constant. Thus f is constant.

Tricky business.

 

[snipez90]

Hahaha, someone made a user account just to reply to a post made a month ago.

Tricky business