How to prove this integral $\iint_{D} \frac{\mathrm{d}\bar{z}\mathrm{d}z}{z-\zeta} = - 2{\pi}i{\bar{\zeta}} $ I am reading this paper and there is an integral in it:

$$\iint_{D} \frac{\mathrm{d}\bar{z}\mathrm{d}z}{z-\zeta} = - 2{\pi}i{\bar{\zeta}},$$ where $D$ is a disc of radius $R$ and $\zeta$ is a point in $D$.

I write the left in definition. Let $\zeta = a+ i b$, then
\begin{align*}\iint_{D} \frac{\mathrm{d}\bar{z}\mathrm{d}z}{z-\zeta} &=2i \iint_{D}\frac{\mathrm{d}x\mathrm{d}y}{(x+iy)-(a+ib)}  
\\&=2 \iint_{D}\frac{(y-b)\,\mathrm{d}x\mathrm{d}y}{(x-a)^2+(y-b)^2} +2i\iint_{D}\frac{(x-a)\,\mathrm{d}x\mathrm{d}y}{(x-a)^2+(y-b)^2},
\end{align*}
and it should be
$$\iint_{D}\frac{(x-a)\,\mathrm{d}x\mathrm{d}y}{(x-a)^2+(y-b)^2}=-{\pi}a.$$
Using polar coordinates and change variable to $t = \tan\frac{\theta}{2}$,
\begin{align*}
&\mathrel{\phantom{=}}\iint_{D}\frac{(x-a)\,\mathrm{d}x\mathrm{d}y}{(x-a)^2+(y-b)^2}\\
&= \int^R_0\mathrm{d}r\int^{\pi}_{-\pi}\frac{r(r\cos\theta -a)}{(r\cos\theta-a)^2+(r\sin\theta - b)^2}\,\mathrm{d}\theta\\
&=\iint\frac{(r(\cos^2 \frac{\theta}{2}-\sin^2 \frac{\theta}{2})-a(\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}))\,\mathrm{d}\theta\mathrm{d}r}{(r^2+a^2+b^2)(\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2})-2ra(\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}) -4rb\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\\
&=\int^R_02r\,\mathrm{d}r\int^{+\infty}_{-\infty}\frac{r(1-t^2)-a(1+t^2)}{((r^2+a^2+b^2)(1+t^2)-2ra(1-t^2)-4rbt))(1+t^2)}\,\mathrm{d}t
\end{align*}
and I don't know how to continue. Did I do something wrong? And I think the author use complex language for convenience. I calculate in real is the wrong way but I don't know how to do it in complex.
Thank you! 

The integral you have isn't $-2\pi i \bar{\zeta}$ unless the disk $D$ is centered at origin. 
Assume $D$ is centered at origin and $r$ is its radius. Using Stoke's theorem for complex coordinates, we have
$$\int_D \frac{d\bar{z} \wedge dz}{z-\zeta} 
= \int_D d\left( \frac{\bar{z}-\bar{\zeta}}{z-\zeta} dz \right)
= \int_{\partial D} \frac{\bar{z}-\bar{\zeta}}{z - \zeta} dz
$$
On the circle $\partial D$, we have $\displaystyle\;\bar{z} = \frac{r^2}{z}\;$.
When $\zeta \ne 0 $ and $\in D \setminus \partial D$, we can evaluate the integral using an ordinary contour integral.
$$\begin{align}\int_{\partial D}\left(\frac{r^2}{z} - \bar{\zeta}\right)\frac{1}{z-\zeta}dz
&= \int_{\partial D}\left[\frac{r^2}{\zeta}\left(\frac{1}{z-\zeta}-\frac{1}{z}\right)- \frac{\bar{\zeta}}{z-\zeta}\right]dz\\
&= 2\pi i \left[ \frac{r^2}{\zeta} (1 - 1) - \bar{\zeta}\right]
= -2\pi i \bar{\zeta}
\end{align}
$$
When $\zeta = 0$, the result is similar.
Note
Please note that there are some subtle issues in how to handle the singularity at $\zeta$ correctly.
We have chosen to use the relation
$$\frac{d\bar{z} \wedge dz}{z-\zeta} 
= d\left( \frac{\bar{z}-\bar{\zeta}}{z-\zeta} dz \right)
\quad\text{ instead of }\quad
\frac{d\bar{z} \wedge dz}{z-\zeta} 
= d\left( \frac{\bar{z}}{z-\zeta} dz \right)
$$
to evaluate the integral. In this way, the intermediate $1$-form remains bounded and we can forget about the singularity.
