This is a question from a previous complex analysis qualifying exam that I'm working through to study for my own upcoming qual. I'm really struggling to know where to go with it and any help would be appreciated!
Question (exact wording from the exam):
Consider the function $f(z) = u(x,y) + iv(x,y)$, where
$$ u(x,y) = \frac{x^3 - y^3}{x^2 + y^2}, \quad v(x,y) = \frac{x^3 + y^3}{x^2+y^2} $$
for $z=x+iy\neq 0$ and $u(0,0) = v(0,0) = 0$.
(a) Verify that $u, \; v$ are continuous in a neighborhood of $z=0$ and satisfy the Cauchy-Riemann Equations at $z=0$.
(b) Show that $f'(0)$ does not exist.
Why I'm struggling from the very start:
(1) I know the epsilon-delta definition of continuity at a point. How do I apply this to a neighborhood of a point instead?
(2) When I try to apply the Cauchy-Riemann Equations, $$ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} $$
these would be undefined when evaluated at $(x,y) = 0$. Am I supposed to take a limit perhaps?
(3) If I successfully do part (a), it seems that's supposed to ensure that a derivative of $f$ DOES exist at $z=0$, but the question asks to show the opposite.
Can someone please walk me through this problem?
In fact \begin{eqnarray} \frac{\partial u}{\partial x}(0,0)&=&\lim_{\Delta x\to0}\frac{u(\Delta x,0)-u(0,0)}{\Delta x}=\lim_{\Delta x\to0}\frac{\Delta x}{\Delta x}=1\\ \frac{\partial v}{\partial y}(0,0)&=&\lim_{\Delta y\to0}\frac{v(\Delta y,0)-v(0,0)}{\Delta y}=\lim_{\Delta y\to0}\frac{\Delta y}{\Delta y}=1. \end{eqnarray} Similarly, $$ \frac{\partial u}{\partial y}(0,0)=-\frac{\partial v}{\partial x}(0,0)=-1. $$ So the Cauchy-Riemann equations hold at $z=0$. Now for $\Delta z=\Delta x+i\Delta y$, \begin{eqnarray} f'(0)&=&\lim_{\Delta z\to0}\frac{f(\Delta z)-f(0)}{\Delta z}\\ &=&\lim_{\Delta x\to0, \Delta y\to 0}\frac{u(\Delta x,\Delta y)+iv(\Delta x,\Delta y)}{\Delta x+i\Delta y}\\ &=&\lim_{\Delta x\to0, \Delta y\to 0}\frac{\frac{\Delta x^3 - \Delta y^3}{\Delta x^2 + \Delta y^2}+i\frac{\Delta x^3 + \Delta y^3}{\Delta x^2+\Delta y^2}}{\Delta x+i\Delta y}. \end{eqnarray} For $\Delta y=k\Delta x$, \begin{eqnarray} f'(0)&=&\frac{(1-k^3)+i(1+k^3)}{(1+k^2)(1+ki)} \end{eqnarray} which depends on the value of $k$. So $f'(0)$ does not exists.