Let $a>0$. Let $f:\mathbb{R}\rightarrow\mathbb{C}$ be $f(x)= e^{(-a+bi)x^2}$. What is the Fourier transform of $f$?
Here is what I have tried:
The exponential decay of $e^{(-a+bi)x^2}$ means that $f$ is in the Schwartz space. So we can talk about its Fourier transform. And since $f$ is in the Schwartz space, we know that $$(f')^{\wedge}(\xi)=i\xi\hat{f}(\xi)$$ and that $$(-ixf(x))^{\wedge}(\xi)=(\hat{f})'(\xi).$$ Since $$f'(x)=2(-a+bi)xf(x),$$ using the above identities, after some simplification, we have that $$(\hat{f})'(\xi)+\frac{\xi}{2(a-bi)}\hat{f}(\xi).$$ Solving this differential equation using the method of integrating factors, I obtain that $$\hat{f}(\xi)=\hat{f}(0)e^{-\frac{\xi^2}{4(a-bi)}}$$
But how can I find $\hat{f}(0)$ i.e. what is $\int_\mathbb{R}e^{(-a+bi)x^2}dx$?
The function $$ F(z) = \int_{-\infty}^{\infty}e^{-zx^{2}}dx $$ is holomorphic in the right half plane $\Re z > 0$. For $z=a$ where $a$ is real and positive, one has $$ F(a) = \sqrt{\frac{\pi}{a}}. $$ Choose the branch of $\sqrt{\pi/z}$ which is holomorphic in the right half plane, and this chosen branch must equal $F(z)$ everywhere in the right half plane by the identity theorem.