Let $A$ be the two-dimensional area integral, over the complex plane, of a gaussian function: $$A = \int_\text{plane} \frac{\mathrm{d}\bar{z} \wedge \mathrm{d}z}{2i} \ \exp[-z\bar{z}].$$
Of course one could evaluate this integral by converting to Cartesian (or alternatively to polar) coordinates, where the answer is evident, $$A = \int_\text{plane} \mathrm{d}x \wedge \mathrm{d}y \ \exp[-(x^2+y^2)] = \pi.$$
But, is there a way to do the complex area integration directly in the complex coordinates $z,\bar{z}$?
In particular, let us define the (non-holomorphic) function $f$, $$f=-\frac{1}{z}\exp[-z\bar{z}], \\ \frac{\partial f}{\partial \bar{z}} = \exp[-z\bar{z}].$$
Then I wonder whether one can proceed with the following manipulations: first "integrate" over $\bar{z}$,
$$A = \int_\text{plane} \frac{\mathrm{d}\bar{z} \wedge \mathrm{d}z}{2i} \ \exp[-z\bar{z}]
= \oint_\text{contour} \frac{\mathrm{d}z}{2i}\ f(\bar{z},z) .$$
and then evaluate the positive definite Gaussian part at the $z=0$ pole, to use the Residue theorem,
$$
= \oint_\text{contour} \frac{\mathrm{d}z}{2i} \ \frac{1}{z} = \pi ,$$
which produces the correct value of the integral. Is there a sense in which these manipulations are correct?
Can one perform the complex plane area integral by first integrating over $\bar{z}$, and then over $z$?