1. ## Complex Differentiation

Hello,

I'm trying to find the set of all points z such that the function f is differentiable. I am having trouble comprehending the steps one needs to take. I have been playing around with the Cauchy-Riemann equations but to little success. The two different functions I am having trouble with are:

1) f(x + iy) = 2x + ixy^2

2) f(z) = cos(z*), where z* is z conjugate

2. Well, what trouble did you run into after using C-R equations?

Also, are you searching for domains where the function is differentiable or analytic?

To be honest, I have never fully understood the C-R equations so I keep ending up go in circles. I just attempted to use that to little success because I have read that that is the proper method.

And the domains I am looking for are where the function is differentiable. Is analytic when the function is differentiable everywhere?

Thanks again.

4. 1) C-R equations: let $\displaystyle f(x,y) = u(x,y) + iv(x,y)$. Then:
a) f is differentiable at $\displaystyle (x_0, y_0)$ if $\displaystyle u_x(x_0, y_0) = v_y(x_0, y_0)$ and $\displaystyle u_y(x_0, y_0) = -v_x(x_0, y_0)$
b) f is analytic at $\displaystyle (x_0, y_0)$ if it is differentiable there, and it is also differentiable at a small neighborhood around $\displaystyle (x_0, y_0)$

I'll do the first one:

$\displaystyle f(x,y) = 2x + ixy^2$, so $\displaystyle u = 2x, \ v = xy^2$
And so $\displaystyle u_x = 2, \ v_y = 2y$
$\displaystyle u_y = 0, -v_x = -1$

So you can see that $\displaystyle u_y$ is never equal to $\displaystyle -v_x$, and so the function is nowhere differentiable.

For the second one, there is a theorem which can be used to solve but I'm not sure you've learned it - it says that f is analytic in $\displaystyle z_0$ if and only if $\displaystyle \displaystyle \frac{ \partial f}{ \partial \bar{z}} =0$.

2) To further illustrate the point about the difference between a differentiable function and an analytic one --
Let $\displaystyle f(x,y) = (x-iy)(2 - x^2 - y^2)$.
Applying the C-R condition gives you the result that f is differentiable only on the unit circle (verify it!).
But since it is differentiable only on the unit circle, you don't have any points which have a neighborhood on which the function is differentiable, and so it is nowhere analytic.