Let $\displaystyle \Gamma$ be polygon with its ordered vertices (the vertices you take in order as you travel around it) be $\displaystyle z_1,z_2,...,z_n$.

Also say they are taken in a counterclockwise fashion.

Let $\displaystyle \Gamma_j = [z_j,z_{j+1}]$ be the line segment from $\displaystyle z_j$ to $\displaystyle z_{j+1}$ for $\displaystyle j=1,2...,n-1$. But for $\displaystyle j=n$ define $\displaystyle \Gamma_n = [z_n,z_1]$.

Define $\displaystyle f: \mathbb{C}\to \mathbb{C}$ by $\displaystyle f(x+iy) = (-y+ix)$.

Let $\displaystyle R$ denote the interior of the polygon, then by Green's theorem:

$\displaystyle \frac{1}{2}\oint_{\Gamma} f = \frac{1}{2}\iint_R \frac{\partial (x)}{\partial x} - \frac{\partial (-y)}{\partial y} = \iint_R 1 = \text{area}(R)$.

Thus, the area is equal to: $\displaystyle \frac{1}{2}\oint_{\Gamma}f = \frac{1}{2}\sum_{j=1}^n \int_{\Gamma_j}f = \frac{1}{2}\sum_{j=1}^{n-1} \left( \int_{[z_j,z_{j+1}]} f \right) + \frac{1}{2}\int_{[z_n,z_1]} f$

For $\displaystyle j=1,2,...,n-1$ define $\displaystyle \gamma_j(t) = z_j + (z_{j+1}-z_j) t$ for $\displaystyle 0\leq t\leq 1$. Notice that $\displaystyle \gamma_j(t)$ parametrizes $\displaystyle [z_j,z_{j+1}]$.

Thus, $\displaystyle \int_{[z_j,z_{j+1}]} f = \int_0^1 f(\gamma_j(t))\dot {\gamma}_j(t) dt$.

Now, $\displaystyle f(x+iy) = -y + ix = i(x-iy)$, so $\displaystyle f(z) = i\bar z$, while $\displaystyle \dot{\gamma}_j(t) = z_{j+1}-z_j$.

Therefore, $\displaystyle \int_0^1 f(\gamma_j(t))\dot {\gamma}_j(t) dt = i(z_{j+1}-z_j)\int_0^1 \overline{\gamma}_j(t) dt = i(z_{j+1}-z_j)\bar z_j + \tfrac{1}{2}i|z_{j+1}-z_j|^2 $

So if you add these up including $\displaystyle [z_n,z_1]$ you get:

$\displaystyle \tfrac{1}{2}i( (z_2 - z_1)\bar z_1 + ... + (z_1 - z_n)\bar z_n) + \tfrac{1}{4}i(|z_1-z_2|+...+|z_n-z_1|) $

Work on simplying that expression now.