- #1

- 265

- 7

- Summary:
- How does the Fourier transform handle functions of complex variables when only integrating over the reals?

Given the domain of the integral for the Fourier transform is over the real numbers, how does the Fourier transform transform functions whose independent variable is complex?

For example, given

\begin{equation}

\begin{split}

\hat{f}(k_{\mathbb{C}}) &= \int_{\mathbb{R}} f(z_{\mathbb{C}}) e^{2 \pi i k_{\mathbb{C}} z_{\mathbb{C}}} d z_{\mathbb{C}}

\end{split}

\end{equation}

where ##z_{\mathbb{C}} = z_1 + z_2 i## and ##k_{\mathbb{C}} = k_1 + k_2 i##.

It seems that evaluating ##f(z_{\mathbb{C}}) e^{2 \pi i k_{\mathbb{C}} z_{\mathbb{C}}}## over the reals includes only ##z_1## and ignores ##z_2 i## in ##f(z_{\mathbb{C}}) ##.

For example, given

\begin{equation}

\begin{split}

\hat{f}(k_{\mathbb{C}}) &= \int_{\mathbb{R}} f(z_{\mathbb{C}}) e^{2 \pi i k_{\mathbb{C}} z_{\mathbb{C}}} d z_{\mathbb{C}}

\end{split}

\end{equation}

where ##z_{\mathbb{C}} = z_1 + z_2 i## and ##k_{\mathbb{C}} = k_1 + k_2 i##.

It seems that evaluating ##f(z_{\mathbb{C}}) e^{2 \pi i k_{\mathbb{C}} z_{\mathbb{C}}}## over the reals includes only ##z_1## and ignores ##z_2 i## in ##f(z_{\mathbb{C}}) ##.