I used Green's Theorem to calculate the integral of this curve, I got $4i\sqrt{2}$ . However, I also want to calculate it through the definition of the integral $\int_{\gamma}{\bar{z}}dz$, by splitting the line segments and assigning bounds. Can someone please help me out?
For Green's Theorem, I did $\frac{1}{2} \int_{2e^{-i\pi/4}}^{2e^{i\pi/4}}{(2cos\theta*2cos\theta - {2sin\theta*(-2sin\theta))d\theta}}$ = $4i\sqrt{2}$.
Am I headed in the right direction, also am I applying Green Theorem properly? Of course, both answers should be equivalent.
$\newcommand{\i}{\mathrm{i}}$ $\newcommand{\C}{\mathbb{C}}$ $\newcommand{\e}{\mathrm{e}}$ $\newcommand{\d}{\mathrm{d}}$
Let $\zeta^s=\e^{\i \pi s/4}$, $\zeta = \zeta^1$.
Write $I$ for $[0,1]$. $$\begin{align} f_0 &: I \to \C & f_0(t) &= 2\zeta^{-1}t \\ f_1 &: I \to \C & f_1(t) &= 2\zeta^{-1-2t} \\ f_2 &: I \to \C & f_2(t) &= 2\zeta t\text{.} \end{align}$$
Write $C_i$ for the image of $f_i$. Then
$$\gamma = C_0 + C_1 + C_2\text{.}$$
Note that $\bar{z}=-\i z$ on $C_0$, $\bar{z}=4/z$ on $C_1$, and $\bar{z}=\i z$ on $C_2$. Then
$$\begin{split} \int_{\gamma}\bar{z}\d z &= \int_{C_0}\bar{z}\d z + \int_{C_1}\bar{z}\d z + \int_{C_2}\bar{z}\d z \\ &= -\i \int_{C_0}z\d z + 4 \int_{C_1}\frac{\d z}{z} + \i \int_{C_2}z \d z \\ &= -\i \int_0^{2\zeta^{-1}} z \d z + 4\i \int_{-\pi/4}^{-7\pi/4}\d\theta + \i \int_{2\zeta}^0 z \d z \\ &= \left.-\i\frac{z^2}{2}\right\rvert_{z=0}^{2\zeta^{-1}} + 4\i \left.\theta\right\rvert_{\theta=-\pi/4}^{-7\pi/4} + \left.\i\frac{z^2}{2}\right\rvert_{z=2\zeta}^{0} \\ &= -2\i\zeta^{-2} - 6\pi \i - 2\i\zeta^2 \\ &= -6\pi\i \end{split}$$
Using Green's theorem to convert to an area integral: write A for the region whose boundary is $-\gamma$. Then
$$\begin{split} \int_{\gamma}\bar{z}\d z &= -\int_A \d\bar{z}\wedge \d z\\ &= -2\i\int_A \d x\wedge \d y \\ &= -2\i\lvert A\rvert \\ &= -2\i(3\pi) \\ &= -6\pi \i\text{.} \end{split}$$