<s>[INST] Computing the Fourier transform of a complex-valued function I have the following function: $e^{-(a+bi)|x|^2}$. 
While trying to compute the fourier transform of the following function, I know that fourier transform of the real part remains the same and the result would be $\sqrt{\frac{\pi}{a}}e^{{-\xi^2}/4a}$. I'm not quite sure on how to proceed with the imaginary part. 
 [/INST] Completing the square in the exponent and the well known integral of $e^{-\pi x^2}$ are tools you can use:
$\begin{align}{F}\left\{e^{-(a+bi)|x|^2}\right\} &= {F}\left\{e^{-ax^2}e^{-ibx^2}\right\}\\
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&= \dfrac{1}{2\pi}{F} \left\{e^{-ax^2}\right\} * {F}\left\{e^{-ibx^2}\right\}\\
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&= \dfrac{1}{2\pi}\left( \sqrt{\frac{\pi}{a}}e^{{-\xi^2}/4a} \right)* \int_{-\infty}^{\infty}e^{-ibx^2}e^{-i\xi x} dx\\
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&= \dfrac{1}{2\pi}\left( \sqrt{\frac{\pi}{a}}e^{{-\xi^2}/4a} \right)* \int_{-\infty}^{\infty} e^{-ib\left(x^2+\frac{\xi}{b}x+\frac{\xi^2}{4b^2}\right)}e^{ib\frac{\xi^2}{4b^2}}dx\\
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&= \dfrac{1}{2\pi}\left( \sqrt{\frac{\pi}{a}}e^{{-\xi^2}/4a} \right)*\left(e^{{-\xi^2}/{4bi}}\int_{-\infty}^{\infty} e^{-ib\left(x+\frac{\xi}{2{b}}\right)^2}dx \right)\\
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&= \dfrac{1}{2\pi}\left( \sqrt{\frac{\pi}{a}}e^{{-\xi^2}/4a} \right)*\left(e^{{-\xi^2}/{4bi}}\int_{-\infty}^{\infty} e^{-\pi\left(\frac{1+i}{\sqrt{2}}\sqrt{\frac{b}{\pi}}x+\frac{1+i}{\sqrt{2}}\frac{\xi}{2\sqrt{\pi b}}\right)^2}dx \right)\\
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& \quad\quad\quad\quad u = \frac{1+i}{\sqrt{2}}\sqrt{\frac{b}{\pi}}x+\frac{1+i}{\sqrt{2}}\frac{\xi}{2\sqrt{\pi b}}\\
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& \quad\quad\quad\quad \frac{1-i}{\sqrt{2}}\sqrt{\frac{\pi}{b}}du = dx\\
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&= \dfrac{1}{2\pi}\left( \sqrt{\frac{\pi}{a}}e^{{-\xi^2}/4a} \right)*\left(\dfrac{1-i}{\sqrt{2}}\sqrt{\dfrac{\pi}{b}}e^{{-\xi^2}/{4bi}}\int_{-\infty+\frac{i}{\sqrt{2}}\frac{\xi}{2\sqrt{\pi b}}}^{\infty+\frac{i}{\sqrt{2}}\frac{\xi}{2\sqrt{\pi b}}} e^{-\pi u^2}du \right)\\
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&= \dfrac{1}{2\pi}\left( \sqrt{\frac{\pi}{a}}e^{{-\xi^2}/4a} \right)*\left(\dfrac{1-i}{\sqrt{2}}\sqrt{\dfrac{\pi}{b}}e^{{-\xi^2}/{4bi}}\right)\\
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&= \dfrac{1}{2\pi}\left( \frac{\sqrt{\pi}}{\sqrt{a}}e^{{-\xi^2}/4a} \right)*\left(\dfrac{\sqrt{\pi}}{\sqrt{b}e^{i\frac{\pi}{4}}}e^{{-\xi^2}/{4bi}}\right)\\
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&= \left[\dfrac{1}{2\sqrt{ab}e^{i\frac{\pi}{4}}}\right]\left[\left( e^{{-\xi^2}/4a} \right)*\left(e^{{-\xi^2}/{4bi}}\right)\right]\\
\end{align}$
where $*$ indicates convolution.  I'll also note that $\left(e^{i\frac{\pi}{4}}\right)^2 = i$ 
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