Let $f : U \longrightarrow V $ be a holomorphic map germ defined by $f = (f_{1}, f_{2}) = (z_{2} + z_{1}^{2}, z_{1}^{2} + z_{2}^{2})$, where $U$ and $V$ are open in $\mathbb{C}^{2}$ both containing $0 \in \mathbb{C}^{2}$.
Note that $f^{-1}(0) = \lbrace (i, 1), (-i, 1), (0,0)\rbrace $. Let $\overline{U_{0}}$ the closure of a neighborhood $U_{0} \subset U$ of origin such that $\overline{U_{0}} \cap f^{-1}(0) = \lbrace 0 \rbrace$ (*)
According with: Multidimensional Residues and Their Applications (A.K.Tsikh) (in our example) the local residue of meromorphic form $\omega = \dfrac{hdz_{1} \wedge dz_{2}}{f_{1}.f_{2}}$ at the point $(0,0)$ is the integral : $$\text{res}_{(0,0)}\omega = \dfrac{1}{(2 \pi i)^{2}} \int_{\Gamma_{(0,0)}} \dfrac{hdz_{1} \wedge dz_{2}}{f_{1}.f_{2}}$$ where $\Gamma_{(0,0)} = \lbrace z \in U_{0} ; |f_{j}(z)| = \varepsilon_{j} \,, j = 1, 2 \rbrace$.
My doubts are:
1) Is it always possible to get a neighborhood like $U_{0}$ satisfying (*)?
2) If so, how to calculate the residue at $(0, 0) $ according to the definition above? Can the function $h$ be $\text{det}J(f_{1}, f_{2})$?
I tried to use power series in the denominator, but was unsuccessful in calculations.
References, suggestions and help will be welcome.
Thanks a lot.
1) It must be a finite map. Consider $f(x,y) = (x,xy)$, then $f^{-1}(0) = \{x=0\}$
2) The function $h$ may be any holomorphic thing you want... In the example we have $$ \text{res}_{(0,0)}\omega = \dfrac{1}{(2 \pi i)^{2}} \int_{\Gamma_{(0,0)}} \dfrac{hdz_{1} \wedge dz_{2}}{f_{1}.f_{2}} = \dfrac{1}{(2 \pi i)^{2}} \int_{\Gamma_{(0,0)}} \frac{h}{\det J(f_1,f_2)}\dfrac{df_{1} \wedge df_{2}}{f_{1}.f_{2}} $$ We have that $f_2 = (z_1-iz_2)(z_1+iz_2)$, hence $$ \text{res}_{(0,0)}\omega = \dfrac{1}{(2 \pi i)^{2}} \int_{\Gamma_{(0,0)}} \frac{h}{\det J(f_1,f_2)}\dfrac{df_{1} \wedge d(z_1-iz_2)}{f_{1}.(z_1-iz_2)}+\frac{h}{\det J(f_1,f_2)}\dfrac{df_{1} \wedge d(z_1+iz_2)}{f_{1}.(z_1+iz_2)} $$
For $h=\det J(f_1,f_2)$ we have that $$ \text{res}_{(0,0)}\omega = \dfrac{1}{(2 \pi i)^{2}} \int_{\Gamma_{(0,0)}} \dfrac{df_{1} \wedge d(z_1-iz_2)}{f_{1}.(z_1-iz_2)}+\dfrac{df_{1} \wedge d(z_1+iz_2)}{f_{1}.(z_1+iz_2)} = \dfrac{1}{(2 \pi i)^{2}} \int_{\Gamma_{(0,0)}^1} \dfrac{df_{1} \wedge d(z_1-iz_2)}{f_{1}.(z_1-iz_2)}+\dfrac{1}{(2 \pi i)^{2}} \int_{\Gamma_{(0,0)}^2}\dfrac{df_{1} \wedge d(z_1+iz_2)}{f_{1}.(z_1+iz_2)} = 1+1 = 2 $$ where $\Gamma_{(0,0)}^1 = \{|f_1| = \epsilon_1, |z_1-iz_2|=\epsilon_2\}$ and $\Gamma_{(0,0)}^2 = \{|f_1| = \epsilon_1, |z_1+iz_2|=\epsilon_2\}$.
For the other points note that one of the integrals vanish and the other is equal to one.