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the aim is to calculate $$\int dz\,d\bar{z} e^{-z\bar{z}}$$

when interpreting $dz\,d\bar{z}$ term, I got confused:

$$dz\,d\bar{z}=(dx+i\,dy)(dx-i\,dy)=(dx)^2-(dy)^2 $$ $$dz\,d\bar{z}=\det\begin{bmatrix} 1 &i \\ 1 &-i \end{bmatrix}dx\,dy=-2i\,dx\,dy $$

why doesn't the first expression agree with the second one?

$dx$ and $dy$ are just numbers, $dx\,dy=dy\,dx$, right?

Then what is the geometry interpretation of this kind of integral: $$\int dz\,d\bar{z} f(z,\bar{z}) \text{ ?}$$

Michael Hardy
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Jian
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1 Answers1

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$dx$ and $dy$ are not numbers. Put in the $\wedge$ symbol so you remember what kind of product this is.

$\begin{eqnarray*}dz \wedge d\bar{z} &=& ( dx + i \,dy ) \wedge ( dx - i \,dy )\\ &=& dx \wedge dx + i\, dy \wedge dx + dx \wedge -i dy + i \,dy \wedge -i \,dy \\ &=& i dy \wedge dx - i \,dx \wedge dy \\ &=& - 2 i\, dx \wedge dy \end{eqnarray*}$

Michael Hardy
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AHusain
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  • Thank you for the answer. Then how to interpret $dzd\bar{z}$ as area? It looks like a rhombus , rather than a squre $dxdy$. – Jian Jan 11 '16 at 16:29