# Determining where a function is complex differentiable

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.

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The function is complex differentiable exactly on the on the real axis (where $y = 0$), not just at the origin. –  user18063 Jan 17 '12 at 19:52

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.

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They are not completely equivalent -- it is possible for the CR equations to be satisfied even when the complex derivative does not exist, such as for $f(x+iy)=\min(|x|,|y|)$ at $0$, where all terms in Cauchy-Riemann are zero. –  Henning Makholm Jan 17 '12 at 19:59
@HenningMakholm, indeed, although (and I should have stated that more clearly, I'll update momentarily) in this case where everything is smooth, as the OP already noted, they are equivalent. –  davin Jan 17 '12 at 20:06
Thanks. I don't know how I missed that the equations are always satisfied when y=0. It's quite obvious now. –  user18297 Jan 17 '12 at 20:11

It's easy to prove that if at any particular point the function does not satisfy the Cauchy-Riemann equations, then its complex derivative cannot exist at that point -- or contrapositively: If the complex derivative exists at some point, then the Cauchy-Riemann equations are satisfied there.

Consider $h$ going towards $0$ along either the real or the imaginary axis. If the complex difference quotient has the same limit in both these cases (as it must have if the general limit for $h\to 0$ exists), then by a very simple calculation that means that the Cauchy-Riemann equation are satisfied.

The theorem you quote is (more or less) the converse of this easy case, and is the harder direction.

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