I am reading An elementary proof of the existence of isothermal parameters on a surface
There is an integral in it
$$\int_{D} \frac{\mathrm{d}\bar{z}\mathrm{d}z}{z-\zeta} = - 2{\pi}i{\bar{\zeta}} $$ where $D$ is a disc of radius $R$ centered at origin and $\zeta$ is a point in $D$
I take $\dfrac{\partial}{\partial \zeta}$ on it
$$0=\frac{\partial}{\partial \zeta}(-2\pi i\bar{\zeta})=\frac{\partial}{\partial \zeta}(\int_{D} \frac{\mathrm{d}\bar{z}\mathrm{d}z}{z-\zeta}) =\int_D\frac{\partial}{\partial \zeta}\frac{\mathrm{d}\bar{z}\mathrm{d}z}{z-\zeta}=\int_D\frac{\mathrm{d}\bar{z}\mathrm{d}z} {(z-\zeta)^2}$$
but i don't know whether $\dfrac{\partial}{\partial \zeta}$ commute with $\int$
I try to compute it directly .i can write $\dfrac{\mathrm{d}\bar{z}\mathrm{d}z} {(z-\zeta)^2} $ as $\mathrm{d}(\dfrac{\mathrm{d}\bar{z}}{z-\zeta})$ but it has singular point .i can't use stoke's theorem
Is $\iint_D\dfrac{\mathrm{d}\bar{z}\mathrm{d}z} {(z-\zeta)^2} = 0$ true??
thank you!!