# Complex Analysis Proof of Constant Function

• moo5003
In summary, the problem is to prove that a function f:c->c is constant if it is analytic and its limit at infinity f(z)/z is 0. The attempt at a solution involves using the Cauchy Integral Formula for the first derivative, but there is uncertainty about how to show the integral is 0. Other methods, such as taking the limit of the integral or showing that f(x)/(x-z)^2 is bounded, have been considered but it is unclear how to incorporate them into the proof. The issue of when limits and integrals commute is also a factor in finding a solution. More guidance is needed to move forward with the proof.

## Homework Statement

Prove that if a function f:c->c is analytic and lim as z to infinity of f(z)/z = 0 that f is constant.

## Homework Equations

Cauchy Integral Formula for the first derivative (want to show this is 0 ie: constant)

f prime (z) = 1/(2ipi)*Integral over alpha (circle radius r) f(x)/(x-z)^2 dx.

## The Attempt at a Solution

So, I think I'm right in incorporating the cauchy integral formula for the first derivative into my proof. I hope that I can show that the integral over the closed curve is 0 for all z, meaning that 0 is the derivative of f and therefore f is constant.

Problems: Not sure how to show this integral is 0, since using this equation the new integral has a critical point over the point z. I'm assuming that the introduction of the assumption that the limit of f(z)/z = 0 needs to be employed to take care of this.

In other words, I believe I have equiped myself with the right tools, though I'm not sure how to start implamenting them, I'll update the post as I continue working on this problem though any help would be appreciated.

Hey, I think I may have gotten another method of solving the problem. Though when I took the limit I was a little unsure if I was allowed to say its 0.

Steps:

Since f is analytic its derivative is analytic.

f prime (0) = Lim z to 0 (f(z) - f(0))/(z-0) = Lim z to 0 (f(z)-f(0))/z

Taking the limit on both sides we get: (Lim assumes z to infinity, Lim' is z to 0)

Lim f'(0) = Lim' Lim f(z)/z - f(0)/z = Lim' Lim -f(0)/z
Lim f'(0) = Lim' Lim -f(0)/z

-Lim f(0)/z = 0 since f(0) is a fixed number.

Thus Lim f'(0) = Lim' 0 = 0

Since f'(z) is 0 at the origin and its domain (C) has an accumilation point (anything on the plane) the entire connected domain must also be 0. Because the entire domain is connected we can state that f'(z) is 0 everywhere and therefore the function f(z) is constant.

Edit: Nvm, I can see that I can't commute the limits like I did. :*(

Though, I may be able to use the same logic using the cauchy integral formula and show that the limit of the integral is 0 as x goes to infinity.

Question: Integrals and limits commute right?

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Question: Integrals and limits commute right?

It does make sense but is not necessarily true! The whole question of when limits (and the integral itself involves a limit) commute is a major part of analysis.

HallsofIvy said:
It does make sense but is not necessarily true! The whole question of when limits (and the integral itself involves a limit) commute is a major part of analysis.

Our class hasnt covered (at all) cases in which limits and integrals commute (obviously by my question that is evident). Is the way I'm using *taking the limit of an integral* seem like the correct route to take or should I try finding another method to prove the claim. I'm a little lost on how to incorporate the limit of f(x)/x as x goes to infinity without taking the limit explicitley.

The only idea I have right now is to show that f(x)/(x-z)^2 is bounded for all x. Though I have no idea how I can do this without knowing f(x). I can try using the limit to prove that it must be bounded but again I'm unsure how to extrapolate this information.

Lim x to infinity of f(x)/x implies that for any epsilon > 0 I can find a S > 0 such that for all ABS(X) > S (Numbers outside the disk of radius S) it implies that ABS(f(x)/x) < Epsilon.

So... it doesn't really tell me anything is bounded, only that the complex numbers outside of a growing disk grow arbitrarly close to 0. It doesn't say that anything inside the disk namely f(0) is bounded by anything, nor does it state these facts about the actual function in question f(x)/(x-z)^2.

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## What is a complex analysis proof of a constant function?

A complex analysis proof of a constant function is a mathematical proof that shows a function is constant in the complex plane. This means that the function has the same output for all complex numbers as its input.

## How is a complex analysis proof of a constant function different from a regular proof?

Complex analysis proofs involve using techniques from complex analysis, such as Cauchy's integral theorem and the Cauchy-Riemann equations, to show that a function is constant. Regular proofs may use more general mathematical techniques.

## Why is the complex analysis proof of a constant function important?

The complex analysis proof of a constant function is important because it provides a rigorous mathematical foundation for understanding the behavior of complex functions. It also allows for the identification of constant functions, which can be useful in solving more complex problems in mathematics and physics.

## What are some common techniques used in a complex analysis proof of a constant function?

Some common techniques used in a complex analysis proof of a constant function include the Cauchy-Riemann equations, Cauchy's integral theorem, and the maximum modulus principle. These techniques help to establish the constancy of a function in the complex plane.

## Can a function be constant in the complex plane but not in the real plane?

Yes, a function can be constant in the complex plane but not in the real plane. This is because complex numbers have two components, a real part and an imaginary part, while real numbers only have one component. A function can be constant in the complex plane if the real and imaginary parts of the function are both constant, but it may not be constant in the real plane if only the real part is constant.