# Real part of a analytic function

• Pouyan
In summary, the problem asks to show that xux + yuy is the real part of an analytic function if u(x,y) is and to find the analytic function to which the real part of u corresponds. The approach involves using Cauchy-Riemann condition and harmonic functions to determine an analytic function that satisfies the given condition.
Pouyan

## Homework Statement

Show that xux + yuy is the real part of an analytic function if u(x,y) is.
To which analytic function is the real part of u = Re (f(z))?

## Homework Equations

What I know about analytic functions is Cauchy-Riemann condition
(∂u/∂x) =(∂v/∂y) and (∂y/∂y)=-(∂v/∂x)

I know actually Harmonic functions and Laplace equation (2-dim) but I don't know if I need it here:
(∂2φ/∂x2) + (∂2φ/∂y2) =0

## The Attempt at a Solution

[/B]I say that there is a analytic function : f(z)=u(x,y)+iv(x,y)
(∂u/∂x)=ux+xuxx+yuxy =(∂v/∂y)
(∂u/∂y)=xuxy+uy+yuyy=-(∂v/∂x)
But further , should I integrate to find v(x,y) ?!

Am I in right path ?!

If I say there is a function F(z)=U(x,y)+iV(x,y)
and
U(x,y) = x (∂u/∂x) + y(∂u/∂y)
Can I use Cauchy-Riemann to find V(x,y)?

Is this right ?!

Please can somebody help me ?!

Pouyan said:

## Homework Statement

Show that xux + yuy is the real part of an analytic function if u(x,y) is.
To which analytic function is the real part of u = Re (f(z))?

## Homework Equations

What I know about analytic functions is Cauchy-Riemann condition
(∂u/∂x) =(∂v/∂y) and (∂y/∂y)=-(∂v/∂x)
In the second equation, it should be (∂u/∂y)=-(∂v/∂x)
Pouyan said:
I know actually Harmonic functions and Laplace equation (2-dim) but I don't know if I need it here:
(∂2φ/∂x2) + (∂2φ/∂y2) =0

## The Attempt at a Solution

[/B]I say that there is a analytic function : f(z)=u(x,y)+iv(x,y)
(∂u/∂x)=ux+xuxx+yuxy =(∂v/∂y)
(∂u/∂y)=xuxy+uy+yuyy=-(∂v/∂x)
But further , should I integrate to find v(x,y) ?!

Am I in right path ?!
For the first part, I don't think you are. The statement to prove is: If u(x, y) is the real part of an analytic function f(z), then Re(f(z)) = xux + yuy.

Pouyan
Mark44 said:
In the second equation, it should be (∂u/∂y)=-(∂v/∂x)

For the first part, I don't think you are. The statement to prove is: If u(x, y) is the real part of an analytic function f(z), then Re(f(z)) = xux + yuy.
OK if I say f(z)= U(x,y) + i V(x,y), then U(x,y) = x (∂u/∂x) + y(∂u/∂y) ?! (U is a new function and is not the same like u)
But how can I use it ?! It's not easy to use Cauchy-Riemann to find V

Pouyan said:
OK if I say f(z)= U(x,y) + i V(x,y), then U(x,y) = x (∂u/∂x) + y(∂u/∂y) ?! (U is a new function and is not the same like u)
What you're calling U above is, I believe, the same as u in ∂u/∂x. If they are the same, you should use just one letter and be consistent.
The situation as I see it is that f(z) = f(x + iy) = u(x, y) + iv(x, y).

If I'm wrong in this assumption, what's the meaning of u?
Pouyan said:
But how can I use it ?! It's not easy to use Cauchy-Riemann to find V

Pouyan
Pouyan said:
Show that xux + yuy is the real part of an analytic function if u(x,y) is.
Is this the exact wording of the problem?

It suggests to me that if f(z) = f(x + iy) = u(x, y) + iv(x, y), then u(x, y) = xux(x, y) + yuy(x, y). I don't see why that would be true, but I have to admit it's been a long time since I took a class in complex analysis.

Pouyan
OK somebody told me F(z) = U(x,y)+iV(x,y) and U(x,y) =xux + yuy
But if I use Cauchy -Riemann , it's going to be very hard to find V(x,y) ... I can come with more post and show why

Mark44 said:
Is this the exact wording of the problem?

It suggests to me that if f(z) = f(x + iy) = u(x, y) + iv(x, y), then u(x, y) = xux(x, y) + yuy(x, y). I don't see why that would be true, but I have to admit it's been a long time since I took a class in complex analysis.
OK thank you anyway

∂V/∂x = (-∂u/∂y) - x(∂2/∂x∂y) -y(∂2/∂y2)

∂V/∂y = y(∂u/∂x) +x(∂2u/∂x2)+y(∂2u/∂x∂y)

But I don't know how I can integrate to find V

Pouyan said:
∂V/∂x = (-∂u/∂y) - x(∂2/∂x∂y) -y(∂2/∂y2)
∂V/∂y = y(∂u/∂x) +x(∂2u/∂x2)+y(∂2u/∂x∂y)
These parts in the first line above don't make sense:

Pouyan said:
x(∂2/∂x∂y)
y(∂2/∂y2)
This -- ∂2/∂x∂y -- and the other one are operators. You have to indicate what you are taking the partial of, like you did in the second line of the first quote above.
Pouyan said:
But I don't know how I can integrate to find V

Last edited:
Pouyan said:

## Homework Statement

Show that xux + yuy is the real part of an analytic function if u(x,y) is.
To which analytic function is the real part of u = Re (f(z))?

## Homework Equations

What I know about analytic functions is Cauchy-Riemann condition
(∂u/∂x) =(∂v/∂y) and (∂y/∂y)=-(∂v/∂x)

I know actually Harmonic functions and Laplace equation (2-dim) but I don't know if I need it here:
(∂2φ/∂x2) + (∂2φ/∂y2) =0

## The Attempt at a Solution

[/B]I say that there is a analytic function : f(z)=u(x,y)+iv(x,y)
(∂u/∂x)=ux+xuxx+yuxy =(∂v/∂y)
(∂u/∂y)=xuxy+uy+yuyy=-(∂v/∂x)
But further , should I integrate to find v(x,y) ?!

Am I in right path ?!

Basically, you start with some analytic ##f(z) = f(x+iy) = u(x,y) + i v(x,y)## and are then asked to show that there is an analytic ##F(z)## such that ##F(x+iy) = [x u_x + y u_y]+ i V(x,y)## for some ##V##. Apply Cauchy-Riemann to the pair ##U(x,y) = x u_x(x,y) + y u_y(x,y)## and ##V(x,y)##, so see what can be said about ##V## in terms of ##u## and ##v##. In particular, can you show that an appropriate ##V## actually exists?

Pouyan
Ray Vickson said:
Basically, you start with some analytic ##f(z) = f(x+iy) = u(x,y) + i v(x,y)## and are then asked to show that there is an analytic ##F(z)## such that ##F(x+iy) = [x u_x + y u_y]+ i V(x,y)## for some ##V##. Apply Cauchy-Riemann to the pair ##U(x,y) = x u_x(x,y) + y u_y(x,y)## and ##V(x,y)##, so see what can be said about ##V## in terms of ##u## and ##v##. In particular, can you show that an appropriate ##V## actually exists?

Cauchy-Riemann : ∂U/∂x = ∂V/∂y and ∂U/∂y = - ∂V/∂x

∂V/∂x = (-∂u/∂y) - x(∂2/∂x∂y) -y(∂2/∂y2)

∂V/∂y = y(∂u/∂x) +x(∂2u/∂x2)+y(∂2u/∂x∂y)

But I don't know how I find V in this equation!

I know that the U(x,y) is harmonic (∂2U/∂x2) +(∂2U/∂y2) =0
But that gives me nothing to find V!

Pouyan said:
Cauchy-Riemann : ∂U/∂x = ∂V/∂y and ∂U/∂y = - ∂V/∂x

∂V/∂x = (-∂u/∂y) - x(∂2/∂x∂y) -y(∂2/∂y2)

∂V/∂y = y(∂u/∂x) +x(∂2u/∂x2)+y(∂2u/∂x∂y)

But I don't know how I find V in this equation!
You could integrate both sides of the first equation above with respect to x, and integrate the second equation with respect to y.Keep in mind that when you integrate Vx with respect to x, you don't end up with an arbitrary constant, you end up with an arbitrary function of y alone, say g(y). Similar idea when you integrate the second equation with respect to y -- you end up with an arbitrary function of x alone, say f(x).

This will give you two "views" of V, which you can set equal. I believe that's what Ray was suggesting.
Pouyan said:
I know that the U(x,y) is harmonic (∂2U/∂x2) +(∂2U/∂y2) =0
But that gives me nothing to find V!

## What is the "real part" of an analytic function?

The real part of an analytic function refers to the portion of the function that contains only real numbers. It is the part of the function that can be graphed on a traditional x-y coordinate plane.

## How is the real part of an analytic function different from the imaginary part?

The real part and the imaginary part of an analytic function are two separate components that make up the entire function. The real part contains only real numbers, while the imaginary part contains only imaginary numbers (numbers with a coefficient of "i"). Together, they make up the complex function.

## What is the importance of the real part in analyzing a function?

The real part of an analytic function is important because it allows us to visualize and understand the behavior of the function. It helps us determine the behavior of the function as the independent variable (usually represented by x) changes.

## Can the real part of an analytic function be negative?

Yes, the real part of an analytic function can be negative. This means that the function will have a negative value for certain values of the independent variable. The real part can also be positive or zero, depending on the function.

## How is the real part of an analytic function used in practical applications?

The real part of an analytic function is used in a variety of practical applications, such as signal processing, control systems, and image processing. It helps us analyze and manipulate data in these fields by providing information about the behavior of the function.

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