Prove: $\int_\gamma \bar z \, dz=2i\operatorname{Area}(G)$ Let $G$ be a bounded, open and connected set and $\gamma$ be its boundary. Prove that if $\gamma$ is a closed, smooth curve, then:  $$\int_\gamma \bar z \, dz = 2i \operatorname{Area}(G)$$
What I have done so far: I proved that the statment is correct when $G$ is a triangle, and then when the boundary of $G$ is a polygonal chain. Now for the general case, I thought about approximating $\gamma$ with polygonal chains somehow, but I'm pretty stuck. Is my thinking correct? Any ideas how to move forward?
 A: Area in polar coordinates is given by
$$A=\frac{1}{2}\int\limits_{\alpha}^{\beta}R(t)^2dt \tag{1}$$
Now, if we assume $\gamma(t)=R(t)e^{it}+z_0$, where $z_0$ is a constant and $t\in [0, 2\pi]$, using contour integration formula
$$\int\limits_{\gamma }f(z)dz=\int\limits_{\alpha}^{\beta}f(\gamma(t))\gamma'(t)dt$$ 
and considering $f(z)=\overline{z}$
$$\int\limits_{\gamma}\overline{z}dz=
\int\limits_{0}^{2\pi}\overline{\gamma(t)}\gamma'(t)dt=
\int\limits_{0}^{2\pi}\left(R(t)e^{-it} + \overline{z_0}\right)\gamma'(t)dt=\\
\int\limits_{0}^{2\pi} R(t)e^{-it} \cdot \left(R'(t)e^{it}+iR(t)e^{it}\right)dt+  \overline{z_0}\int\limits_{0}^{2\pi}d(\gamma(t))=...$$
$\gamma$ is closed, thus $\gamma(0)=\gamma(2\pi)$, as a result
$$...=\int\limits_{0}^{2\pi}R(t) \cdot \left(R'(t)+iR(t)\right)dt=\int\limits_{0}^{2\pi}R(t)  d\left(R(t)\right)+i\int\limits_{0}^{2\pi}R(t)^2dt=...$$
obviously $R(0)=R(2\pi)$, thus
$$...=i\int\limits_{0}^{2\pi}R(t)^2dt=...$$
comparing it to $(1)$
$$...=2i \cdot \frac{1}{2}\int\limits_{0}^{2\pi}R(t)^2dt=2i\cdot Area(G)$$
A: 
$$z\bar{w}=r_1r_2\cos(\phi_1-\phi_2)+ir_1r_2\sin(\phi_1-\phi_2)$$
then
$$\dfrac12{\bf Im}\, z\bar{w}=r_1r_2\sin(\phi_1-\phi_2)=r_1r_2\sin\theta=\text{Area of triangle}$$
this shows that the area of a shape is purely imaginary.

Now consider area $S$ bounded with $\gamma$ then with any partition $P=\{p_0,p_1,\cdots,p_n=p_0\}$ for $\gamma$, the area of every triangle $Op_ip_{i+1}$ we have
$$\text{Area of triangle} Op_ip_{i+1}=\dfrac12{\bf Im}\, \bar{z}\Delta z$$
Then adding these triangles together and obtain the Riemann sum of these area which concludes that 
$$\text{Area}=\lim_{||P||\to0}\sum_{i=1}^n\dfrac12{\bf Im}\, \bar{z_i}\Delta z_i=\dfrac12{\bf Im}\,\int_\gamma \bar{z}dz$$
but the area of a shape is purely imaginary, so $$\dfrac12{\bf Re}\, \bar{z}\Delta z=0$$ and thus
$$\dfrac{1}{2i}\int_\gamma \bar{z}dz=\dfrac{-i}{2}{\bf Re}\,\int_\gamma \bar{z}dz+\dfrac12{\bf Im}\,\int_\gamma \bar{z}dz=\text{Area of S}$$
