Riemann surface and order of zero Choose a local coordinate $t$ in a neighborhood of point $(1, 1)$ on Riemann surface $X = \{
(z, w) ∈ \mathbb{C}^2 : z^3 + w^3 = 2\}$    and evaluate the order of zero of the function $f(t) = z(t)w(t) − 1$ in the
point $(1, 1) ∈ X$. Hint: use power expansion with respect to $t$.
At this moment I just did it :
Let t↦(z(t),w(t)) of a neighborhood of the origin in C onto a neighborhood of p=(1,1) in X, $z(t) = \sum a_nt^n$ and $w(t) = \sum b_nt^n$.
SO now I compute with my expansion f(t), I obtain :
$f(t) = \sum a_nt^n \sum b_nt^n = (a_0 + a_1t + a_2t^2 +...)(b_0 + b_1t + b_2t^2+..)-1$
$f(t) = a_0b_0-1 + (a_0b_1 + a_1b_0)t + (a_1b_1 + a_0b_2 + a_2b_2)t^2 + ... $
To now the order we have to find which coefficient is different from $0$, by assumption we have that $(z(0),w(0)) = (1,1) \Rightarrow z(0) = w(0) = 1$
Order 0 : $z(t)^3 + w(t)^3 = 2$ at t = 0 we obtain 1 + 1 = 2  which is true and then $a0b_0 - 1 = 0$ so it is not order 0.
Order 1 : By differentiation we have:  $3z(t)^2z'(t)3 + 3w(t)^2w'(t) = 0$ so at t = 0 we know that $z'(0) = a_1$ and $w'(0) = b_1$ so we obtain that : $3z(0)^2z'(0)3 + 3w(0)^2w'(0) = 3a_1+3a_2 = 0 \Rightarrow a_1 = -b_1$ so that implies that $(a_0b_1 + a_1b_0) = 0$. So it's not order 1.
Order 2 : By a differentiation we have : $6z(t)z'(t)^2 + 3z(t)^2z'(t)z''(t) + 6w(t)w'(t)^2 + 3w(t)^2w'(t)w''(t)  = 0$ in a same way we know that : $z"(0) = 2a_2$  and $w"(0) = 2b_2$ so we obtain :$6z(0)z'(0)^2 + 3z(0)^2z'(0)z''(0) + 6w(0)w'(0)^2 + 3w(0)^2w'(0)w''(0)  = 6a_1^2 + 6a_1a_2 + 6b_1^2 + 6a_2b_1 = 12 a_1^2 +6a_1(a_2-b_2) =0 \Rightarrow a_2 - b_2 = -2a_1$
So we obtain :  $a_1b_1 + a_0b_2 + a_2b_2 = -a_1^2 + b_2 + a_2 = -a_1^2 + 2a_1 + 2a_2$
As you can see I don't know how to conlude for the order 2 any body can help me please ?
 A: You seem to have an error for the quadratic term, it should be
\begin{align}
f(t)&=\dots + (a_0b_2+a_1b_1+a_2b_0)t^2+\dots\\
&= (a_2+b_2-a_1^2)t^2 +\cdots\tag{$*$}
\end{align}
because $a_0=b_0=1$ and $b_1=-a_1$.
Also, your differentiation is wrong. When you differentiate once, you get (after dividing by 3... which you really should do to reduce the clutter)
\begin{align}
z(t)^2z'(t)+w(t)^2w'(t)&=0.
\end{align}
Now, differentiating again yields
\begin{align}
2z(t)z'(t)^2+z(t)^2z''(t)+2w(t)w'(t)^2+w(t)^2w''(t)&=0
\end{align}
So, plugging in $t=0$ yields
\begin{align}
0&=2(a_0)(a_1)^2+(a_0)\left(\frac{a_2}{2}\right)+2(b_0)(b_1)^2+(b_0)\left(\frac{b_2}{2}\right)\\
&=2a_1^2+\frac{a_2}{2}+2a_1^2+\frac{b_2}{2}
\end{align}
Hence, $a_2+b_2=-4a_1^2$. Now, plug this into $(*)$ to get that
\begin{align}
f(t)&=-5a_1^2\,t^2+\cdots
\end{align}
Now, recall that $(z'(0),w'(0))$ has full rank (by definition of a parametrization/inverse of a coordinate chart). This is equivalent to saying one of these quantities is non-zero, but in fact since they are negatives of each other, neither is $0$, so $a_1\neq 0$, and thus $f$ has a zero of order $2$.
