Looking for some intuition behind why the area enclosed by a simple closed curve $C$ can be obtained by computing $\frac{1}{2i}\int_C {\bar{z}} \ dz$.

Question

By some manipulation and an application of Green's Theorem, I am able to show that $$Area = \frac{1}{2i}\int_C {\bar{z}} \ dz $$

To me, this seems to be an unexpected result. Is there some intuition I can use to understand why this must be true? The proof is solid, but I'd like to try to understand on a deeper level. This is my first time studying complex analysis.

Answer 1

While we wait for a better answer, I will answer with a question: do you understand, on a "deeper level", the fact that the area between the $x$-axis and the graph of the positive function $f$ and above the interval $[a,b]$ is given by $g(b)-g(a)$, where $g$ is an antiderivative of $f$?

After all, Green's theorem is an extension of the Fundamental Theorem of Calculus.