I have a homework question, and I'm having a hard time interpreting it.
Question: Where is the function $f(x+iy)=x^4y^5+ixy^3$ complex differentiable? Determine the derivative in such points.
My first plan was to find a region for which the following theorem applied:
Suppose $f=u+iv$ is a complex-valued function defined on an open set $\Omega$. If $u$ and $v$ are continuously differentiable and satisfy the Cauch-Riemann equations on $\Omega$, then $f$ is holomorphic on $\Omega$ and $f'(z)=\frac{\partial f}{\partial z}$.
Of course, $u$ and $v$ are going to be continously differentiable; the only question is, on what region are the Cauchy-Riemann equations satisfied?
So, I found that the Cauchy-Riemann equations in this case are the following:
$4x^3y^5=3xy^2$ and $5x^4y^4=-y^3$
and the only point at which these equations are satisfied is $(0,0)$.
There's a converse to this theorem, but a lone point is not a region, so I'm not sure if either of these theorems are relevant.
Is the only way to go back and find at which points
$lim_{h\to0} \frac{f(z_{0}+h)-f(z_{0})}{h}$
exists?
Thanks.
Best Answer
No, you don't need to check the limit (although you can if you want to). The existence of the complex limit is equivalent to the CR equations (when the function is continuously differentiable, as in this case).
Assuming your equations are correct, notice that there are more points you've missed - you divided by $y$ not taking into account the fact that $y$ could be 0. So for starters the entire line $\{y=0\}$ satisfies those equations. Finally, it's easy to show that those are the only points that satisfy the equations.