# Homework Help: Cauchy-Riemann Equations

1. Mar 30, 2009

### brendan

1. The problem statement, all variables and given/known data

Show that u(x,y) and v(x,y) satisfy the Cauchy-Riemann Equations.

x = r cos(Θ) y= r sin(Θ)

2. Relevant equations

Cauch-Riemann sum
∂u/∂x = ∂v/∂y

3. The attempt at a solution

My questions are:
1. Do I convert the x and y values to rectangular co-ordinates first ?

2. is u(x,y) equivalent to u(r cos(Θ)+r sin(Θ)) u(x^2+y^2) or u=R^2 ?

regards
Brendan

2. Mar 31, 2009

### CompuChip

I don't get the question at all. What are u and v? Why are you using polar coordinates?

3. Mar 31, 2009

### brendan

I have to take the partial derivatives of the functions u(x,y) and v(x,y).
I have been told that x = r cos(Θ) y= r sin(Θ)
So I suppose I have to find the partial derivatives of: u(x,y) and v(x,y).

And they should equal each other if they are Cauchy-Riemann Equations.

∂u/∂x = ∂v/∂y

My problem is I don't know how to represent both the functions u(x,y) and v(x,y) using

x = r cos(Θ) y= r sin(Θ)

To start and find the partial derivatives

4. Mar 31, 2009

### HallsofIvy

You can't do that if you don't know what u and v are! There is nothing in this that tells us what functions u and v are!

This is just x and y in polar coordinates. Nothing is said about u and v.

x and y are rectangular coordinates!

2. is u(x,y) equivalent to u(r cos(Θ)+r sin(Θ)) u(x^2+y^2) or u=R^2 ?[/quote]
u(x,y)= u(rcos(Θ), rsin(Θ)) of course. But even if you had "u(x^2+y^2)" that would not say that u was equal to x^2+ y^2.

Surely somewhere in that problem, or perhaps in an earlier problem, u and v are defined!

5. Mar 31, 2009

### brendan

Sorry Guys,
My mistake The functions for Cauchy-Riemann Equations are:

u(x,y)= excos(y)
v(x,y) = exsin(y)

Your were right they were in a previous question.

6. Apr 1, 2009

### CompuChip

Good. Now you can simply calculate
∂u/∂x = ∂v/∂y
right? I don't see why you would want to go to polar coordinates for this?
u directly depends on x, and v directly depends on y, so you can just take the partial derivatives...

OK, so I suppose the point is that once you show that u(x, y) and v(x, y) satisfy the Cauchy-Riemann equations, you have proven that
$$f(x, y) = e^x \left( \cos(y) + \mathrm i \sin(y) \right) = e^{x + i y}$$
is complex differentiable...