Criterion for local integrability of $1/f$ for $f$ a smooth function from $\mathbb R^2$ to $\mathbb C$ Suppose I have a function $f=u+iv$, $f:\mathbb{R}^2\to \mathbb{C}$ which is smooth, in the sense that $u$ and $v$ are smooth. I am wondering what some criteria are for
$$
\int_{B_r(x_0,y_0)}\frac1 f\mathrm dA(x,y)
$$
to converge in some suitable sense, where $\mathrm dA(x,y)$ is the area measure. For example, if $f(x_0,y_0)=0$ but $f$ has no other zeros in the ball, is it enough to have $\nabla f(x_0,y_0)\ne 0$?
Presumably, what I am asking is equivalent to asking if
$$
\int_{B_r(x_0,y_0)}\frac {u} {u^2+v^2}\mathrm dA(x,y)
$$
and
$$
\int_{B_r(x_0,y_0)}\frac {v} {u^2+v^2}\mathrm dA(x,y)
$$
converge separately.
 A: I set $z_0 = (x_0,y_0)$.
I strengthen a bit the assumptions.
If the differential $\mathrm{d}f(z_0)$ is invertible and if $f$ has no zero but $z_0$ on the closed ball $\overline{B}_{r}(z_0)$, the desired conclusion holds. Indeed, by local incversion theorem, $f$ restricted to some neighborhood of $z_0$ to its image is invertible, and $\mathrm{d}f^{-1}(0) = \mathrm{d}f(z_0)^{-1}$.
For all small enough $h$, we have $h = f^{-1}(f(z_0+h)) - f^{-1}(f(z_0))$. Since $f$ is differentiable at $z_0$ and $f^{-1}$ is differentiable at $f(z_0)$, we derive that when $h \to 0$,
$$h =  \mathrm{d}f^{-1}(f(z_0))(f(z_0+h)-f(z_0)) + o(||f(z_0+h)-f(z_0)||).$$
$$h = \mathrm{d}f^{-1}(f(z_0))(f(z_0+h)-f(z_0)) + o(||h||).$$
Since $f(z_0)=0$, we get
$$||h|| \le ||\mathrm{d}f^{-1}(0)|| \times ||f(z_0+h)|| + o(||h||).$$
The map $h \mapsto ||f(z_0+h)||/||h||$ does not vanish and is is continuous on $\overline{B}_{r}(0) \setminus \{(0,0)\}$, and it is bounded away from $0$ at the neighborhood of $(0,0)$.
Using the compactness of $\overline{B_r(0)} \setminus B_\delta(0)$
for $0<\delta<r$, one sees that there exists some $c>0$ such that $||f(z_0+h)|| \ge c||h||$ for every $h \in \overline{B}_r(0) \setminus \{(0,0)\}$.
Hence
$$\int_{\overline{B_r(0)}} \frac{1}{||f(z_0+h)||}dh \le \int_{\overline{B}_r(0)} \frac{c}{||h)||}dh = \int_0^r \frac{c}{\rho}\rho \mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta = 2\pi cr < +\infty.$$
Hence, the integral of $1/f$ on $\overline{B}_r(0)$ is well-defined.
A: "For example, if $f(x_0,y_0)=0$ but $f$ has no other zeros in the ball, is it enough to have $\nabla f(x_0,y_0)\ne 0$?"
Suppose $f$ is real valued (which implies $f$ is complex valued). Then by the implicit function theorem, $f$ cannot satisfy the above, because the zero set of $f$ near $(x_0,y_0)$ will be a smooth nonconstant curve through $(x_0,y_0).$
