Find Fourier transform of $\delta(e^{i\pi t}-i)$

Question

Find Fourier transform of $\delta(e^{i\pi t}-i)$ I used distribution $\langle\delta(t),\phi(t)\rangle=\phi(0)$ but it didn't work.

Answer 1

Note: The OP was originally to find the inverse Fourier Transform of $\delta(e^{i\pi f}-i)$

Credits: Thanks @Skip for the comment below

Notice that the argument is zero only when $e^{i \pi f} = i$, or in other words \begin{equation} \pi f = \frac{\pi}{2} + 2k \pi \qquad k \in \mathbb{Z} \end{equation} hence for $f = \frac{1}{2 } + 2k$ where $k \in \mathbb{Z}$, the Dirac function peaks at $1$, else it is $0$. This results in a "Dirac comb" or "train of impulses" as \begin{equation} \delta(e^{i\pi f}-i) = \sum_{k = -\infty}^{\infty} \delta(f - f_k) \end{equation} where $f_k = \frac{1}{2 } + 2k$. Applying inverse Fourier transforms on both sides, we have \begin{equation} \mathcal{F}^{-1}\delta(e^{i\pi f}-i) = \sum_{k = -\infty}^{\infty} \mathcal{F}^{-1}\delta(f - f_k) \tag{1} \end{equation}

Using the Fourier transform table, we know that \begin{equation} \mathcal{F}(e^{j 2 \pi f_0 t}) = 2 \pi \delta(f - f_0) \end{equation} which means that \begin{equation} \mathcal{F}^{-1}(\delta(f - f_0)) = \frac{1}{2\pi}e^{j 2 \pi f_0 t}\tag{2} \end{equation} Using equation $(2)$ in $(1)$ we get \begin{equation} \mathcal{F}^{-1}\delta(e^{i\pi f}-i) = \frac{1}{2\pi}\sum_{k = -\infty}^{\infty} e^{j 2 \pi f_k t} \end{equation} Replacing $f_k =\frac{1}{2 } + 2k$, we get \begin{equation} \mathcal{F}^{-1}\delta(e^{i\pi f}-i) = \frac{1}{2\pi}\sum_{k = -\infty}^{\infty} e^{j 2 \pi (\frac{1}{2 } + 2k) t} = \frac{1}{2\pi}e^{j\pi t}\sum_{k = -\infty}^{\infty} e^{4 k \pi t } \end{equation}