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