Show that if C is a positively oriented simple closed contour, then the area of the region enclosed by $C$ can be written $$\frac{1}{2i}\int_C{\bar{z}} \ dz$$

Here's what I've done:

$$\int_C{f(z)} \ dz =\int_C{(u\ dx-v\ dy)} + i\ \int_C{(v\ dx+ u\ dy)} $$

$$=\int_C{P\ dx +Q\ dy} $$

$$= \iint_R{(Q_x-P_y)\ dA} $$

$$= \iint_R{(-v_x-u_y)\ dA}\ +i\iint_R{(u_x-v_y)\ dA} $$

Here: $f(z) = \bar{z} \ =x-iy \ = u+iv$

Cauchy-Riemann: $u_x = 1$, $v_y=-1$, and $u_y=0=v_x$

then,

$$\int_C{\bar{z}}\ dz = \iint_R{(0)\ dA}\ +i \iint_R{(1-(-1))\ dA} $$

$$=2i\ \iint_R{1\ dA} $$

— and I guess from here, the integration of has me a confused because the final answer is $$\frac{1}{2i}\int{\bar{z}\ dz}$$

but I'm not sure why the $2i$ flips?

## Best Answer

As $f(z):=\bar z$ is not analytic, CR has no place here. But Green's theorem indeed helps: $$\eqalign{\int_C\bar z\>dz&=\int_C(x-iy)\>(dx+idy) =\int_C\bigl(\,(x-iy)\>dx+(ix+y)dy\,\bigr)\cr&=\int_C(P\>dx+Q\>dy)\cr}$$ with $P=x-iy$ and $Q=ix+y$. Greens theorem then says that $$\int_C\bar z\>dz=\int_R(Q_x-P_y)\>{\rm d}(x,y)=\int_R 2i\>{\rm d}(x,y)=2i\>{\rm area}(R)\ .$$