# Complex Analysis - Proving an analytic function f(z) is constant

• Daized
In summary, using the Cauchy-Riemann equation for the partial derivatives, it can be shown that f is constant and that u and v are constant as well.
Daized

## Homework Statement

Let f(z) be an analytic function in the complex plane ℂ, and let $\phi$ be amonotonic function of a real variable.

Assume that U(x,y) = $\phi$(V(x,y)) where U(x,y) is the real part of f(z) and V(x,y) is the imaginary part of f(z). Prove that f is constant.

## Homework Equations

The analytic function f(z) is constant if f'(z)= 0 everywhere.

The Cauchy Riemann equation...

∂u/∂x=∂v/∂y,∂v/∂x=−∂u/∂y

## The Attempt at a Solution

I'm honestly a bit lost on where to start. I know that if f(z) is analytic then it is differentiable, so I thought that using the Cauchy Riemann equations for the partial derivatives might be helpful where U$_{x}$=V$_{y}$ and U$_{y}$=-V$_{x}$, but I don't know how to work with these when the function $\phi$ is there.

Just any information on what theory is best to look at would be helpful. Thank you.

Last edited:
simply substitute into the C-R equation and make use of the chain rule of taking derivatives. Also note that phi is monotonic (amonotonic=a monotonic, as I assume), so that phi'≠0

Actually managed to figure it out just about

Assume that f(z) = u(x,y) + i v(x,y).

We know that u(x, y) = ϕ(v(x, y)).

By Cauchy-Riemann, we have
∂u/∂x = ∂v/∂y ==> ϕ'(v(x,y)) * ∂v/∂x = ∂v/∂y, and
∂u/∂y = -∂v/∂x ==> ϕ'(v(x,y)) * ∂v/∂y = -∂v/∂x.

Substituting the first equation into the second equation yields
ϕ'(v(x,y)) * [ϕ'(v(x,y)) * ∂v/∂x] = -∂v/∂x
==> [(ϕ'(v(x,y)))^2 + 1] ∂v/∂x = 0.
==> ∂v/∂x = 0, since (ϕ'(v(x,y)))^2 + 1 is nonzero.

Substituting this into the first equation yields ∂v/∂y = 0.

Since ∂v/∂x = ∂v/∂y = 0, we conclude by Cauchy-Riemann that ∂u/∂x = ∂u/∂y = 0 as well.
Hence, both u and v are constant.
Therefore, f = u + iv is also constant.

Thanks though.

good work, better than what I did.

## 1. What is an analytic function in complex analysis?

An analytic function in complex analysis is a function that is differentiable at every point within its domain. This means that it can be locally approximated by a linear function at each point.

## 2. How do you prove that a function is analytic?

To prove that a function is analytic, you need to show that it is differentiable at every point within its domain. This can be done by using the Cauchy-Riemann equations, which state that the partial derivatives of the real and imaginary parts of a complex function must exist and satisfy certain conditions.

## 3. What does it mean for an analytic function to be constant?

If a function is constant, it means that it has the same value at every point within its domain. In the context of complex analysis, this means that the function does not change in either its real or imaginary parts and remains constant throughout its domain.

## 4. How do you prove that an analytic function is constant?

To prove that an analytic function is constant, you can use the Cauchy-Riemann equations to show that the partial derivatives of both the real and imaginary parts of the function are equal to zero. This means that the function does not change in either direction, and therefore, must be constant.

## 5. What is the significance of proving that an analytic function is constant?

Proving that an analytic function is constant has significant implications in complex analysis. It means that the function is not only differentiable at every point within its domain, but it also does not vary in either its real or imaginary parts. This can be useful in many applications, such as in physics and engineering, where constant functions play a crucial role in modeling and solving problems.

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