I am currently reading "Restrictions on harmonic maps of surfaces" by J. Eells and J. C. Wood and have trouble understanding the proof of the lemma on p. 265. The paper states:
Let $X$ and $Y$ be closed orientable smooth surfaces, and $\phi: X \to Y$ a smooth map. Relative to Riemannian metrics $g,h$ on $X,Y$ we say that $\phi$ is harmonic if the divergence of its differential vanishes identically. In isothermal charts, writing $h = \sigma^2(w)dwd\overline{w}$ and representing $\phi$ in the form $z \to w(z)$, that condition is expressed by $$ w_{z\overline{z}} + (2\sigma_w/\sigma)w_z w_\overline{z} = 0, $$ being the Euler-Lagrange equation associated to the energy functional $$ E(\phi) = \frac{1}{2}\int \vert d\phi(x) \vert^2 dx = \frac{1}{2} \int_X \sigma^2(w(z))\left[ \vert w_z \vert^2+ \vert w_\overline{z}\vert^2 \right] dx dy. $$ where subscripts denote partial derivatives and $w_z := \frac{1}{2}(w_x - iw_y), w_\overline{z} := \frac{1}{2}(w_x + iw_y)$, etc.
We have the complexified tangent bundle $T^c(X) = T(X) \otimes_\mathbb{R} \mathbb{C}$ and the decomposition $T^c(X) = T^{1,0}(X) \oplus T^{0,1}(X)$ with $$ \partial_{1,0} \phi: T^{1,0}(X) \to T^{1,0}(Y), \quad \partial_{0,1}\phi: T^{1,0}(X) \to T^{0,1}(Y). $$ Let us say that $\partial_{1,0}\phi$ has a zero of infinite order at a point if $$ \partial w/\partial z = o(\vert z \vert^m) \text{ as } z \to 0 \text{ for all } m \geq 0. $$
Corollary: If $\phi: X \to Y$ is harmonic but not $\pm$ holomorphic, then neither $\partial_{1,0}\phi$ nor $\partial_{0,1}\phi$ has a zero of infinite order.
Lemma: If $\phi: X \to Y$ is harmonic but not $\pm$ holomorphic, then in isothermal charts $\partial_{1,0}\phi$ and $\partial_{0,1}\phi$ have the forms \begin{align} &\partial w/\partial z = Az^m + o(\vert z \vert^m) \text{ for some } m \geq 0 \text{ and complex number } A \neq 0; \nonumber \\ &\partial \overline{w}/\partial z = Bz^n + o(\vert z \vert^n) \text{ for some } n \geq 0 \text{ and } B \neq 0. \end{align}
Proof: Taylor's expansion to $m^\mathrm{th}$ order for the first gives $$ \partial w/\partial z = Q_m(z,\overline{z}) + R(z,\overline{z}), $$ where $Q_m$ is a homogeneous polynomial of degree $m \geq 0$, and $R$ is $o(\vert z \vert^m)$ as $z \to 0$.
The preceding Corollary shows that $Q_m \neq 0$ for some $m$. Substituting the above equation in $$ w_{z\overline{z}} + (2\sigma_w/\sigma)w_z w_{\overline{z}} = 0 $$ gives $$ \frac{\partial}{\partial \overline{z}}(Q_m + R) + \frac{2\sigma_w}{\sigma}(Q_m + R)\frac{\partial w}{\partial \overline{z}} = 0. $$ Now $\partial Q_m/\partial \overline{z}$ is either a homogeneous polynomial of degree $m - 1$ or is identically $0$. Each of the other terms is $o(\vert z \vert^{m-1})$. It follows that $\partial Q_m/\partial \overline{z} = 0$. Therefore, $Q_m$ is a holomorphic homogeneous polynomial of degree $m$, and consequently has the form $Az^m$. The case $\partial_{0,1}\phi$ is handled similarly.
My main question is: Why is each of the other terms $o(\vert z \vert^{m-1})?$