Thread: 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

Thank you for your time.

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?

Thanks for the reply!

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.

1) C-R equations: let . Then:
a) f is differentiable at if and
b) f is analytic at if it is differentiable there, and it is also differentiable at a small neighborhood around

I'll do the first one:

, so
And so


So you can see that is never equal to , 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 if and only if .

2) To further illustrate the point about the difference between a differentiable function and an analytic one --
Let .
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.