1. ## Analytic functions

The question
Let z = x + iy, with x and y real

If $\displaystyle f(z) = x^3 + iy^3$, find the value of the derivative of f at every point z where the derivative exists. Where is f analytic?

My attempt:
I used Cauchy Riemann sums as follows,

$\displaystyle u_x = 3x^2$
$\displaystyle u_y = 0$
$\displaystyle v_x = 0$
$\displaystyle v_y = 3y^2$

The equations hold when $\displaystyle 3x^2 = 3y^2$. This equals |x| = |y| which are intersecting lines through the origin of the x-y plane. So, the function has a derivative on these intersecting lines. If my understanding of analyticity is correct, this function is analytic nowhere since there's no epsilon neighbourhood where the function is differentiable.

My issue is, how do I find the value of the derivative? Do I use the definition of a derivative, or is there an easy way to find it using my result above?

Thanks.

2. Originally Posted by Glitch
The question
If my understanding of analyticity is correct, this function is analytic nowhere since there's no epsilon neighbourhood where the function is differentiable.

Right, according to the definition of analytic function, $\displaystyle f$ must be complex differentiable on a region $\displaystyle R\subset\mathbb{C}$ .

3. Just restrict h to real values when applying the limit definition of the derivative. Since you know that f is differentiable at |x|=|y|, then the real-restricted limit must be equal to the complex limit. We end up getting

$\displaystyle f'(x+iy)=u_x(x,y)+iv_x(x,y)$.

By the way, the reason we know f is differentiable on those lines is because all the first-order partial derivatives (1) satisfy the Cauchy-Riemann equations; and (2) are continuous. You mentioned (1) but don't forget that (2) is also required.

4. Thanks guys. So should I be using the definition of the derivative, or is there a faster way to compute it using the partial derivatives?

5. Well it's generally true that $\displaystyle f'(x+iy)=u_x(x,y)+iv_x(x,y)$. So if all you need to do is compute it, then your task is easy. But if you need to prove it, then the definition of the derivative will have to do, unless you have a theorem at your disposal.

6. Thank you again. That saves me a lot of time!