Complex differential 1-form question

kvt
Messages
2
Reaction score
0

Homework Statement


I am trying to solve Nakahara Ex. 1.5. I have already solved part (1), but I am stuck trying to generalize the equation of (1) to prove part (2). I think I will be able to complete the proof if I can establish the following equation:


Homework Equations



\int dz d\overline{z} \exp({-z\overline{z}}) = \int dx dy \exp({- x^2 - y^2})



The Attempt at a Solution


Using z = x + iy, it is obvious that both exponents are the same, but the Jacobian from the coordinate transformation does not seem to be equal to 1. Is it true that dz d\overline{z} = dx dy ? If so, why?
 
Physics news on Phys.org
What you have is dz\wedge d\bar{z}, compute dz and d\bar{z} and take their wedge product.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top