Take $\mathbb C^2$ with coordinates $(z,w)$. Suppose that $J$ is a $C^{\infty}$ almost complex structure on $\mathbb C^2$ such that the line $w=0$ is $J$-holomorphic and $J(0,0)$ is given by $(z,w)\to (iz,iw)$.


Suppose that there is a smooth function $f(z):\mathbb D\to \mathbb C$ defined on a unit disk $\mathbb D$ such that the graph $(z,f(z))\subset \mathbb C^2$ is a $J$-holomorphic subvariety of $(\mathbb C^2, J)$ (we don't assume that $z\to (z,f(z))$ is a $J$-holomorphic map).


**Question.** Is it true that there is $n$ and a constant $c\in(0,1)$ such that $c|z^n|<|f(z)|<\frac{1}{c} |z^n|$ for sufficiently small $z$? 


*Remark.* The statement is trivial in the case when $J$ coincides with the standard complex structure on $\mathbb C^2$. This question is a follow up to [Two smooth tangent almost complex curves in a $4$-manifold](https://mathoverflow.net/questions/296460/two-smooth-tangent-almost-complex-curves-in-a-4-manifold)



An estimate on deviation of two smooth tangent $J$-holomorphic curves
Yes, this is true. In fact, a more precise statement holds: Unless $f$ vanishes identically on $\mathbb{D}$, there is an integer $n$ and a nonzero complex number $a$ such that $f(z) = a\,z^n + f\_{n+1}$, where $f\_{n+1}$ is a smooth function on the disk that vanishes to order $n{+}1$ at $z=0$.


The proof follows immediately from a Taylor series expansion argument, plus a fact about solutions of elliptic equations with the same symbol as the Cauchy-Riemann equations.


Here is the argument: Let $\Omega$ be a nonzero complex-valued $2$-form on $\mathbb{C}^2$ such that $\Omega$ is of type $(2,0)$ with respect to $J$. Thus, a $2$-dimensional submanifold $\Sigma$ of $\mathbb{C}^2$ is $J$-holomorphic if and only if $\Sigma^\*(\Omega)=0$. (I use the notation $\Sigma^\*$ to denote the operation of pulling back forms to $\Sigma$.) This $\Omega$ is uniquely defined up to multiplication by a nonzero complex function. Because you have assumed that $J(0,0)$ is the standard complex multiplication, it follows that, up to a nonzero multiple, $\Omega$ is equal to $\mathrm{d}z\wedge\mathrm{d}w$ plus a term that vanishes at $(z,w)=(0,0)$. In particular, we can take $\Omega$ to have the form
$$
\Omega = (\mathrm{d}z -a\_1\,\mathrm{d}\bar z - a\_2\,\mathrm{d}\bar w)\wedge
(\mathrm{d}w -b\_1\,\mathrm{d}\bar z - b\_2\,\mathrm{d}\bar w)
$$
where $a\_1,a\_2,b\_1,b\_2$ all vanish at $(z,w)=(0,0)$. Letting $L\subset\mathbb{C}^2$ be the line defined by $w=0$, we see that,
because $w=0$ is a $J$-holomorphic curve, we must have 
$$
0 = L^\*\Omega = (\mathrm{d}z -L^\*a\_1\,\mathrm{d}\bar z )\wedge
( -L^\*b\_1\,\mathrm{d}\bar z ) = -(L^\*b\_1)\,\mathrm{d}z\wedge\mathrm{d}\bar z.
$$
Consequently, $b\_1$ vanishes on $L$, so it must be of the form $b\_1 = c\_1\,w + c\_2\,\bar w$ for some smooth functions $c\_1$ and $c\_2$. 


Now, suppose that a $J$-holomorphic curve $C\subset\mathbb{C}^2$ is defined by an equation $w = f = f(z,\bar z)$ in a neighborhood of $(z,w) = (0,0)$, and that $f(0,0)=0$. If $f$ vanishes to infinite order at $z=0$ (i.e., its Taylor series at $z=0$ vanishes identically), then elliptic theory tells us that $f$ must vanish identically in a neighborhood of $z=0$. Let's assume that this is not the case, so that there is an $n\ge 1$ such that $f$ vanishes to order $n$ at $z=0$ but not to order $n{+}1$. Write $f = f\_n + f\_{n+1}$ where $f\_n$ is the order $n$ Taylor polynomial of $f$ at $z=0$ and $f\_{n+1}$ vanishes to order $n{+}1$. Now $C^\*(\Omega)=0$, and we can compute its $(n{-}1)$-th order Taylor expansion as 
$$
0 \equiv C^\*(\Omega) \equiv (\mathrm{d}z)\wedge(\mathrm{d}f\_n ) 
\equiv \frac{\partial f\_{n}}{\partial\bar z}\,\mathrm{d}z\wedge\mathrm{d}\bar z,
$$
where the equivalences mean that we ignore terms that vanish to order at least $n$ at the origin. (This uses the fact that $a\_1$, $a\_2$, and $b\_2$ vanish to order at least $1$ at the origin while $C^\*b\_1 = C^\*\!c\_1\,f + C^\*\!c\_2\,\bar f$ vanishes to order at least $n$, while $\mathrm{d}f$ and $\mathrm{d}\bar f$ vanish to order at least $n{-}1$.) It follows that $\partial f\_{n}/{\partial\bar z} = 0$ and, hence, that $f\_n = a\,z^n$
for some constant $a$, which is nonzero, since $f\_n$ is nonzero.


