# Complex Analysis Qn

1. Aug 1, 2009

### LostEngKid

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

Suppose z0 = x0 + iy0 2 C, and r > 0. Further, suppose that f(z) is a real valued function that is analytic on the open box

B(z0; r) = { x + iy | x0 < x < x0 + r; y0 < y < y0 + r }.

Then show that f(z) must, in fact, be constant on the box B(z0; r).

3. The attempt at a solution
I really have no idea at all what this question is asking me to do. Can someone please point me in the right direction? Im not asking anyone to do it for me, im happy to figure it out by myself once i have some idea of what im supposed to do.

2. Aug 1, 2009

### winter85

hi LostEngKid,
the question is asking you to show that if f is an analytic function that takes on only real values in some square region, then this function must infact be constant in that region (it cant take different real values inside the square).

could you start by stating what basic facts you know about analytic functions?

3. Aug 1, 2009

### LostEngKid

Okay so i know that a function is analytic over a given open set if it is differentiable at every point in the set....thats about all i've picked up so far, the next thing that was covered in my lectures was the cauchy reimann equations which if i remember correctly are used to show if a function is differentiable at a given point.

4. Aug 1, 2009

### winter85

That's right, the Cauchy-Riemann equations are necessary conditions for a function to be analytic, could you state them explicitely?
they give you information about how an analytic function varies in the neighborhood of a point, so they should be useful in this situation :)

5. Aug 1, 2009

### LostEngKid

You split the function into real and imaginary parts, u(x,y) and y(x,y) and then take the partial derivate of each with respect to x and y.
If du/dx = dv/dy and dv/dx=-du/dy for (xo, yo) then the equation is differentiable at zo.
And then the derivative is du/dx + i dv/y.

6. Aug 1, 2009

### winter85

ok so for $$z = x + iy$$ , we have $$f(z) = u(x,y) + i v(x,y)$$. since f is real valued, v(x,y) = 0 for all z = x+ iy. substitute that in the CR equations, what do you get? then pick two points in the square, join them with a special line, and try to figure out from the CR equations you got how the function f changes when you move along this path from one of the points to the other.

7. Aug 1, 2009

### LostEngKid

Okay, so u(x,y) = x and v(x,y) = 0
That means du/dx = 1 and all the others = 0 which means that the function isnt differentiable, but the question says its analytic so it is? So obviously ive done something wrong...
If v(x,y) = y then dv/dy =1 and it works. So im assuming this is what im supposed to do?

That would mean that the derivative is 1 + i?

I guess for my two points id pick (xo,yo) and (xo + r, yo +r)

8. Aug 1, 2009

### winter85

why did you assume u(x,y) = x? that's not true in general. you only know v(x,y) = 0.

9. Aug 1, 2009

### LostEngKid

Oh i see, for some reason i was thinking f(z) = x + iy
So if we know that v(x,y)=0 and that the function is analytic then du/dx and du/dy have to equal 0 to satisfy CR.

So if the derivative is 0 then the function has to be a constant?

10. Aug 1, 2009

### winter85

well, du/dx = 0 and du/dy = 0.
now if you fix a value for y, say y = y0, consider the function g(x) = u(x,y0). dg/dx = 0, so g is a constant.
similarly, fix a value for x, say x = x0, consider the function h(y) = u(x0,y). dh/dy = 0, so h is a constant.
imagine you start at a point z0, first walk parallel to the real axis, stop, make a 90 degrees turn, an continue walking parallel to the imaginary axis, then you reach a point z1. how did the function change along this path?

11. Aug 1, 2009

### LostEngKid

I know that the answer is supposed to be that it hasnt changed but im not sure how to prove it. The derivative is the rate of change, and its 0 no matter where you take it, so the function isnt changing. I find it really hard to get my head around this topic, in the real plane the function would obviously have changed as you move along it but here it doesnt...seems so counter intuitive.

12. Aug 1, 2009

### winter85

ok consider this... fixing y = y0, the function u(x,y0) is a real function of one real variable which is x. so you can treat it as a normal function R -> R. its derivitaive with respect to x is 0, so as x varies, u(x,y0) doesnt change. but holding y0 constant and varying x means walking in parallel with the real axis... so walking in paralell with the real axis doesnt change the value of u.. then you do the same in parallel with the y axis... so the function doesnt change value, can you see it?

13. Aug 2, 2009

### LostEngKid

Yes i think i do. Thanks for being patient with me and thanks for all the help, i never would have been able to answer this question on my own.

14. Aug 2, 2009

### winter85

ur welcome :)