By combining several things I read in several places, it seems that the area inside a parameterized closed curve in the complex plane ($\;f : [0,1] \to \mathbb C\;$ with $\;f(0)=f(1)\;$ and 'piecewise smooth') is $$ {1\over{4i}}\int_{[0,1]} (\overline{f(t)}f'(t) - f(t)\overline{f'(t)}) \text{d}t $$ Note that this is the 'directed area', i.e., it is positive if the curve is counterclockwise for increasing $\;t\;$, and otherwise negative; let's assume $\;f\;$ is counterclockwise for now. And having a self-intersecting curve adds more specialness, so let's ignore those for now... (I've asked another question on that topic here.)
My question: How would one prove that the above integral gives the same result as doing a $\;\iint_D 1\:\text{d}A\;$ (Lebesgue) integral over the region $\;D \subset \mathbb C\;$ enclosed by $\;f\;$?
Apparently this is related to Green's theorem in some way, but I haven't come further than that yet.
For completeness, here is how I arrived at the above formula. First, Titu Andreescu and Dorin Andrica, "Complex Numbers from A to... Z", 2nd Ed., Birkhäuser 2014, explains (p. 105) how the triangle between complex numbers $\;0,a,b\;$ has a directed area of $\;{1 \over {2i}} (a \times b)\;$, where $\;a \times b \;=\; {1 \over 2} (\overline{a}b - a\overline{b})\;$ is the "complex product" (~cross product?) of $\;a\;$ and $\;b\;$ (p. 104).
Now adding up the areas of the little triangles $\;0,f(t),f(t+dt)\;$ over $\;t \in [0,1]\;$, and using $\;f'(t) = {f(t+dt) - f(t) \over \text{d}t}\;$, results in the above integral.