# Line integrals using Green's theorem

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 : $$\oint_C z^2dz = i\iint_R \left(\frac{\partial z^2}{\partial x}+i \frac{\partial z^2}{\partial y}\right)\; dA =0$$

• In (i), $\;C\;$ is not a "line", or path. You can't do line integrals on it. Did you mean, perhaps, $\;C=\{z\in\Bbb C\;/\;|z|=1\}\;$ ? – DonAntonio Apr 1 '17 at 9:57
• $C$ is a circle yes not an open disc sorry, will edit now – rapidracim Apr 1 '17 at 9:58
• I really don't understand your writing: $\;\overline z = x-iy\;$ , so applying Green to $\;\overline z\,dz=x\,dx+iy\,dy\;$ , one gets$$\int_C(x\,dx-iy\,dy)=\iint_R\left(0+0\right)dA=0$$ Am I missing something? Perhaps you're using other kind if differentials? For me, $\;\frac{\partial\overline z}{\partial x}\;$ doesn't mean much: what is your definition of this? – DonAntonio Apr 1 '17 at 10:04
• @DonAntonio $\frac{\partial \overline z}{{\partial x}} = \frac{\partial}{{\partial x}}(x-iy)$ – rapidracim Apr 1 '17 at 10:07
• @1 I see now. Then I think your work is completely fine. The results in the case of $\;z^2\;$ are logical since $\;z^2\;$ is a complex analytic function on the plane, so over any simple, nice closed path the integral will vanish...and that's exactly what you got. – DonAntonio Apr 1 '17 at 10:28