I want to calculate these line integrals using green's formula :$\textit{a)}$ $\oint_C \overline{z}dz$ $\textit{b)}$ $\oint_C z^2dz$
in the following cases : $\textit{i)}$ $C = \{z\in \mathbb{C}\;/\;|z|=1\}$ $\textit{ii)}$ $C$ is a square with vertices: $0$ , $1$ , $i$ , $1+i$
this is my first time computing line integrals and I want someone to confirm if I'm doing it right.
my try :
$\textit{a)}$ $\textit{i)}$ using Green's theorem :
$$\oint_C \overline{z}dz = i\iint_R \left(\frac{\partial \overline{z}}{\partial x}+i \frac{\partial \overline{z}}{\partial y}\right)\; dA =i\iint_R 2\; dA = 2i\iint_RdA =2\pi i$$
$\textit{ii)}$ again ,using Green's theorem : $$\oint_C \overline{z}dz = i\iint_R \left(\frac{\partial \overline{z}}{\partial x}+i \frac{\partial \overline{z}}{\partial y}\right)\; dA = i\iint_R 2\; dA=2i \int_0^1\int_0^1dydx = 2i $$
$\textit{b)}$ $\textit{i)}$ using G.T :
$$\oint_C z^2dz = i\iint_R \left(\frac{\partial z^2}{\partial x}+i \frac{\partial z^2}{\partial y}\right)\; dA =i\iint_R 2x+2iy +i(-2y+2ix)\; dA =i\iint_R0dA =\;\;i\iint_R(1-1)dA = i\iint_RdA - i\iint_RdA = 0$$ The zero makes me doubt it.
$\textit{ii)}$ this one is pretty similar to the previous one so : $$\oint_C z^2dz = i\iint_R \left(\frac{\partial z^2}{\partial x}+i \frac{\partial z^2}{\partial y}\right)\; dA =0 $$