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[tex]\int_\mathbb{C} dz \: \exp \left( -\frac{1}{2}K|z|^2 + \bar{J}z \right)[/tex]

Where [itex]K \in \mathbb{R}[/itex] and [itex]J \in \mathbb{C}[/itex]

The book I am following (Kardar's Statistical Physics of Fields, Chapter 3 Problem 1) asserts that by completing the square this becomes [itex]Z \exp\left( \frac{- |J|^2}{2K} \right) [/itex] where [itex] Z = \int_\mathbb{C} dz \: \exp \left( -\frac{1}{2}K|z|^2 \right)[/itex]. I can't seem to reproduce this, and I think the trouble I'm running into arises from [itex]|z|^2[/itex] not being a square, but rather it involves conjugation as well. Therefore, I get the following

[tex]-\frac{1}{2}K|z|^2 + \bar{J}z = -\frac{1}{2}K\left( z\bar{z} -2 \frac{\bar{J}}{K}z \right)= -\frac{1}{2}K\left( z\bar{z} -2 \frac{\bar{J}}{K}z - 2 \frac{J}{K}\bar{z} +2 \frac{J}{K}\bar{z} + 4\frac{ |J|^2}{K^2} - 4 \frac{ |J|^2}{K^2} \right) = [/tex]

[tex] -\frac{1}{2}K\left( z - 2 \frac{J}{K} \right) \left( \bar{z} -2 \frac{\bar{J}}{K} \right) -J\bar{z} +2 \frac{ |J|^2}{K} [/tex]

Which means that I'm getting that

[tex]\int_\mathbb{C} dz \: \exp \left( -\frac{1}{2}K|z|^2 + \bar{J}z \right) = \left[ \int_\mathbb{C} dz \: \exp \left( -\frac{1}{2}K \left| z-2 \frac{J}{K} \right|^2 - J\bar{z} \right) \right] \exp\left( 2 \frac{ |J|^2}{K} \right) [/tex]

Which doesn't at all seem like

[tex] \left[ \int_\mathbb{C} dz \: \exp \left( -\frac{1}{2}K|z|^2 \right) \right] \exp\left( \frac{- |J|^2}{2K} \right) [/tex]

[itex][/itex]

[tex][/tex]