I'm working on an electrical engineering problem set. My confusion is not with the concepts of the topics I'm covering, rather with a bit of mathematics.
Does:
$$e^{j2\pi kt} = (e^{j2\pi k})^t, k \in \mathbb{Z}, |k| \leq 3, t \in \mathbb{R}, j = \sqrt{-1}$$
For context (off topic, but you might see the motivation for my question), the problem is asking me to find a function $f(t)$ with some period $T$ that has been filtered such that only the Fourier Coefficients (of the function) $c_k = 0, |k| > 3 $. If the above statement is true, I'd say $e^{j2\pi kt} = (e^{j2\pi k})^t = 1^t = 1$, which simplifies my calculations.
Wolfram tells me it's only true for $t=0$ but I'm clearly missing something fundamental here.
Thanks in advance!
My answer is NO.
For example, let $k=1$ and $t=1/2$, then $$ e^{j 2 \pi k t} = e^{j \pi} = -1 \tag{1} $$ and $$ (e^{j 2 \pi k})^t = 1^{1/2} = 1. \tag{2}$$
Clearly, (1) $\neq$ (2).