# Cauchy-Riemann equations help

1. Feb 9, 2006

### Benny

Hi, can someone help me understand the following question?

Q. Show that if f is analytic on D, then g(z) defined by

$$g\left( z \right) = \mathop {f\left( {\mathop z\limits^\_ } \right)}\limits^{ - - - - - - }$$..the variable which f is applied to is the conjugate of z, it's a little difficult to see it.

is analytic on the reflected domain $D^ * = \left\{ {\mathop z\limits^\_ \left| {z \in D} \right.} \right\}$, and that $$g'\left( z \right) = \mathop {f'\left( {\mathop z\limits^\_ } \right)}\limits^{ - - - - - - }$$.

There is a solution (I'm not sure if it is a partial or complete solution) which shows that:

$$g'\left( z \right) = \mathop {f'\left( {\mathop z\limits^\_ } \right)}\limits^{ - - - - - - }$$

What else needs to be done to show that g is analytic on D*? I'm quite confused by this question and I don't really know where to start. The definition I have for a function being analytic at a point z_0 is that the function is differentiable at all points in a neighbourhood of z. I get the feeling that I might need to use some standard theorems but nothing comes to mind at the moment. Any help would be good thanks.

Note: I've been working on topics related to this question at a fairly relaxed pace and treating the material at a rather superficial level to fill in some time before the start of semester. So I'm not as proficient with the applications of standard procedures. Just thought I'd include an overview of my situation in case someone wants to know why I'm having trouble with what looks to be a fairly simple question.

Last edited: Feb 9, 2006
2. Feb 9, 2006

### matt grime

analytic means *complex* differentiable, not just differentiable, let's make that explicit, ie the partial derivatives satisfy the Cauchy-Riemann equations.

Why don't you just calculate the (ordinary) derivative of f(z*)* (star to denote conjugate) which is the composition of three functions, and show that this is the complex derivate, ie that it satisfies the Cachy-Riemann equations. (It will do exactly because -1 times -1 is 1)

after all if f(z)=f(x+iy)=f(x,y)=u(x,y)+iv(x,y) then what is f(x,-y)*? What are the partial derivatives?

3. Feb 9, 2006

### Benny

Thanks for the help matt grime.

The other problem I'm having with this question is not understanding what the domain D* is supposed to 'look' like or what it is. To draw an analogy, it's like me being unable to integrate 2xy with respect to y because I don't know what is meant by with respect to y.

Edit: After reading matt grime's reply I wrote f(z) = f(x+yi) = u(x,y) + iv(x,y).

f is analytic so the Cauchy-Riemann relations are satisfied. Anyway I used the definition given of g in terms of f and after using some properties of conjugates I got down to g(z) = u(x,y) + iv(x,y). But isn't that exactly what f is?

Last edited: Feb 10, 2006
4. Feb 10, 2006

### matt grime

That isn't g. Try again. Show your working, it's the onyl way I'll give more hints on that.

As for what it does to D: complex conjugation sends x+iy to x-iy, or in vector notation (x,y) to (x,-y). And you know what that transformation of the plane is.

5. Feb 10, 2006

### Benny

This is what I did.

f = u + iv

$$g\left( z \right) = \mathop {f\left( {\mathop z\limits^\_ } \right)}\limits^{ - - - - - - }$$

$$= \mathop {f\left( {x - iy} \right)}\limits^{ - - - - - - - - - }$$

$$= \mathop {u\left( {x,y} \right) + i\left( { - v\left( {x,y} \right)} \right)}\limits^{ - - - - - - - - - - - - - - - - - - - - - - - - }$$

$$= u\left( {x,y} \right) + iv\left( {x,y} \right)$$

I'm not really sure what you were referring to when you said to calculuate the ordinary derivative of f(z*)*. I would've thought that I needed to calculate the first partial derivatives of the real (u) and imaginary (v) parts of f somewhere in my answer.

Edit: Nevermind, having typed it out I saw the error in my expression for g. If f(z) = f(x+yi) = u(x,y) + iv(x,y) then f(z*)* = [u(x,-y) + iv(x,-y)]* = u(x,-y) - iv(x,-y) = u(x,-y) + i(-v(x,-y)). So now would I just differentiate that for example (-1)(-1)v_y = v_y to obtain g's version of v_y. Also; g's version of u_x is simply u_x. By definition of f, u_x = v_y. The way I've worded it is a little confusing but I think it's what I need to do.

Last edited: Feb 10, 2006
6. Feb 10, 2006

### matt grime

Right you went wrong exactly where i thought you'd gone wrong.

why is this line there?

f(x-iy) = u(x,y)-iv(x,y)

that is not correct.

f(x,y)=u(x,y)+iv(x,y)

if i change y to -y what happens?

Last edited: Feb 10, 2006
7. Feb 10, 2006

### Benny

Changing y to -y changes the variable z to it's conjugate. Ie. f(z) goes to f(z*) when y goes to -y. That's basically what I worked on in my edit.

Last edited: Feb 10, 2006