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

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.

• Not an answer, but $\int_C f\,dz$ could physically be interpreted as the circulation of the vector field $\bar f$ over $C$ plus $i$ times the flux of $\bar f$ across $C$. With $f(z)=\bar z$ we can interpret the field to be $F(x,y)=(x,y)$. Note that $\nabla\times F=0$ and $\text{div}\, F=2$. Thus, by the general formulas for circulation and flux (typically Stokes theorem), you get the area back. I accept that this does not give a deeper mathematical understanding, but it might give you some concrete idea of what is going on(?). Commented Mar 12, 2017 at 19:00

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 $f$?