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Assume that $z_1, z_2, \ldots, z_n \in\mathbb{C}=\mathbb{R}^2$ are the corners of a convex polygon ($n\in\mathbb{N}$, $n\geq 3$). Assume that the corners are named in order around the polygon.

Now prove that the area of the polygon is given by: $$ \frac{1}{2}\cdot |\mathrm{Im}(\overline{z_1}z_2+\overline{z_2}z_3+\ldots +\overline{z_n}z_1)|. $$

I have already done the proof for $n=3$ (triangle), but when doing the proof for $n\geq 4$, I get problems with the absolute value. My result looks like that:

$$ \frac{1}{2}\cdot \left(|\mathrm{Im}(\overline{z_1}z_2)|+|\mathrm{Im}(\overline{z_2}z_3)|+\ldots +|\mathrm{Im}(\overline{z_n}z_1)|\right). $$

At this point I don't know how to join the separate absolute-value-terms.

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    $\begingroup$ Check out Shoelace formula for the area of a simple polygon. $\endgroup$ – g.kov Nov 11 '14 at 18:56
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If we can do it for a triangle, can't we do it inductively with any polygon? Given convex polygon with vertices $z_1, z_2, \dotsc, z_n$, note that the area of this polygon is equivalent to the sum of the areas of the triangles given by $\{z_1, z_2, z_3\}, \{z_1, z_3, z_4\}, \dotsc, \{z_1, z_{n-1}, z_n\}$.

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  • $\begingroup$ ... assuming the vertices are named in order aroung the polygon. $\endgroup$ – Arthur Nov 11 '14 at 18:53
  • $\begingroup$ I have tried to do it that way, but I end up in the situation described above. $\endgroup$ – B.math Nov 11 '14 at 18:54
  • $\begingroup$ Thank you Arthur, I have forgotten to write that in my post. $\endgroup$ – B.math Nov 11 '14 at 18:55
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Let $z_j = x_j+iy_j$. Then $\mathrm{Im}(\overline{z_j}z_{j+1}) = x_jy_{j+1}-x_{j+1}y_j$. Now apply the shoelace formula.

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