# Question #c0643

Aug 13, 2016

See below

#### Explanation:

The theorem (from Wiki):

$\setminus {\oint}_{\setminus \gamma} f \left(z\right) \setminus , \mathrm{dz} = 0$

The contour integral around the closed path should be zero.

I am going to use vector notation for the linear bits as its clearer and the algebra is the same

$\vec{A B} = \left(\begin{matrix}2 \\ 2\end{matrix}\right)$

So the param is $z = \left(\begin{matrix}1 \\ 1\end{matrix}\right) + t \left(\begin{matrix}1 \\ 1\end{matrix}\right) , t \in \left[0 , 2\right]$

$\mathrm{dz} = \left(\begin{matrix}1 \\ 1\end{matrix}\right) \mathrm{dt}$

$\implies {\int}_{t = 0}^{2} \left(\left(\begin{matrix}1 \\ 1\end{matrix}\right) + t \left(\begin{matrix}1 \\ 1\end{matrix}\right) + \left(\begin{matrix}- 1 \\ 1\end{matrix}\right)\right) \left(\begin{matrix}1 \\ 1\end{matrix}\right) \mathrm{dt}$

$= \left(\begin{matrix}1 \\ 1\end{matrix}\right) {\int}_{0}^{2} \left(\begin{matrix}0 \\ 2\end{matrix}\right) + t \left(\begin{matrix}1 \\ 1\end{matrix}\right) \mathrm{dt}$

$= \left(\begin{matrix}1 \\ 1\end{matrix}\right) {\left[\left(\begin{matrix}0 \\ 2\end{matrix}\right) t + {t}^{2} / 2 \left(\begin{matrix}1 \\ 1\end{matrix}\right)\right]}_{0}^{2}$

$= \left(\begin{matrix}1 \\ 1\end{matrix}\right) \left[\left(\begin{matrix}0 \\ 4\end{matrix}\right) + \left(\begin{matrix}2 \\ 2\end{matrix}\right)\right]$

$= \left(\begin{matrix}1 \\ 1\end{matrix}\right) \left(\begin{matrix}2 \\ 6\end{matrix}\right)$

$= \left(1 + i\right) \left(2 + 6 i\right)$

$= 2 + 6 i + 2 i - 6$

$= - 4 + 8 i$

$\vec{B C} = \left(\begin{matrix}- 4 \\ 0\end{matrix}\right)$

$z = \left(\begin{matrix}3 \\ 3\end{matrix}\right) + t \left(\begin{matrix}- 1 \\ 0\end{matrix}\right) , t \in \left[0 , 4\right]$

$\mathrm{dz} = \left(\begin{matrix}- 1 \\ 0\end{matrix}\right) \mathrm{dt}$

$\implies {\int}_{0}^{4} \left(\left(\begin{matrix}3 \\ 3\end{matrix}\right) + t \left(\begin{matrix}- 1 \\ 0\end{matrix}\right) + \left(\begin{matrix}- 1 \\ 1\end{matrix}\right)\right) \left(\begin{matrix}- 1 \\ 0\end{matrix}\right) \mathrm{dt}$

$= \left(\begin{matrix}- 1 \\ 0\end{matrix}\right) {\int}_{0}^{4} \left(\begin{matrix}2 \\ 4\end{matrix}\right) + t \left(\begin{matrix}- 1 \\ 0\end{matrix}\right) \mathrm{dt}$

$= \left(\begin{matrix}- 1 \\ 0\end{matrix}\right) {\left[\left(\begin{matrix}2 \\ 4\end{matrix}\right) t + {t}^{2} / 2 \left(\begin{matrix}- 1 \\ 0\end{matrix}\right)\right]}_{0}^{4}$

$= \left(\begin{matrix}- 1 \\ 0\end{matrix}\right) \left(\left(\begin{matrix}8 \\ 16\end{matrix}\right) + 8 \left(\begin{matrix}- 1 \\ 0\end{matrix}\right)\right)$

$= \left(\begin{matrix}- 1 \\ 0\end{matrix}\right) \left(\begin{matrix}0 \\ 16\end{matrix}\right)$

=$- 1 \cdot 16 i$

$= - 16 i$

$\vec{C A} = \left(\begin{matrix}2 \\ - 2\end{matrix}\right)$

$z = \left(\begin{matrix}- 1 \\ 3\end{matrix}\right) + t \left(\begin{matrix}1 \\ - 1\end{matrix}\right) , t \in \left[0 , 2\right]$

$\mathrm{dz} = \left(\begin{matrix}1 \\ - 1\end{matrix}\right) \mathrm{dt}$

$\implies {\int}_{0}^{2} \left(\left(\begin{matrix}- 1 \\ 3\end{matrix}\right) + t \left(\begin{matrix}1 \\ - 1\end{matrix}\right) + \left(\begin{matrix}- 1 \\ 1\end{matrix}\right)\right) \left(\begin{matrix}1 \\ - 1\end{matrix}\right) \mathrm{dt}$

$= \left(\begin{matrix}1 \\ - 1\end{matrix}\right) {\int}_{0}^{2} \left(\begin{matrix}- 2 \\ 4\end{matrix}\right) + t \left(\begin{matrix}1 \\ - 1\end{matrix}\right) \mathrm{dt}$

$= \left(\begin{matrix}1 \\ - 1\end{matrix}\right) {\left[\left(\begin{matrix}- 2 \\ 4\end{matrix}\right) t + {t}^{2} / 2 \left(\begin{matrix}1 \\ - 1\end{matrix}\right)\right]}_{0}^{2}$

$= \left(\begin{matrix}1 \\ - 1\end{matrix}\right) \left(\left(\begin{matrix}- 4 \\ 8\end{matrix}\right) + \left(\begin{matrix}2 \\ - 2\end{matrix}\right)\right)$

$= \left(\begin{matrix}1 \\ - 1\end{matrix}\right) \left(\begin{matrix}- 2 \\ 6\end{matrix}\right)$

$= \left(1 - i\right) \left(- 2 + 6 i\right)$

$= - 2 + 6 i + 2 i + 6$

$= 4 + 8 i$

$\implies {\int}_{A B} + {\int}_{B C} + {\int}_{C A}$

$\left(\begin{matrix}- 4 \\ 8\end{matrix}\right) + \left(\begin{matrix}0 \\ - 16\end{matrix}\right) + \left(\begin{matrix}4 \\ 8\end{matrix}\right) = \left(\begin{matrix}0 \\ 0\end{matrix}\right)$

$= 0$