Q:

Intuitive understanding of some important exponential functions

After watching these 2 brilliant YouTube Videos--
1) Euler's formula with introductory group theory
2) But what is the Fourier Transform? A visual introduction.

I am facing some difficulty in understanding some of these exponential functions intuitively---

A:

Disclaimer: I did not watch the videos.
The ideas you are describing are from complex numbers and their geometry. For an introduction I recommend working through the early chapters in one of the many free complex analysis books available online.
One of things you learn early on about values in the complex plane is that they can be represented in rectangular form by an expression like $z = x + i y$ where $x$ and $y$ are simply the familiar rectangular coordinates. It turns out these numbers can also be represented by the expression $$z = r e^{i\theta}$$ where $\theta$ is an angle (in radians) and $r$ is a magnitude. When $r = 1$ the expression $z = e^{i\theta}$ gives points on the unit circle corresponding to the angle $theta$, and so the function $$f(t) = e^{it}$$ traces out a unit circle as $t$ varies from, say, $0$ to $2\pi.$
So now your question. Consider the equation $g(t) = 2\pi t$ as $t$ goes from $0$ to $1.$ You'll see that it gives you, in a linear fashion (you graphed it, right?), all values from $0$ to $2\pi.$ Therefore if we look at the function $$f(t) = e^{g(t)} = e^{2 \pi t}$$ we see that it traces out the unit circle as $t$ varies from $0$ to $1.$
Now suppose we look at yet another function $h(t) = g(t) * k$ (I'm using $k$ rather than $f$ to avoid confusion) where $k$ is a real number greater than $0$ but less than $1.$ What happens to the values as we increase $t$ from $0$? You should see that now in order that $h(t) = 2 \pi,$ we must have $2\pi = g(t) * k = 2 \pi t k$ so that, solving for $t$ we get $$t = \frac{1}{k}.$$ Since $0 < k < 1$ we know that $t > 1.$ Thus we now have a function that varies from $0$ to $2\pi$ as $t$ goes from $0$ to $1/k$ (rather than $1$). OK, now putting this again an exponential function, we obtain $$F(t) = e^{h(t)} = e^{2\pi kt}$$ which will trace out a unit circle as we put in values from $0$ to $1/k$. Since $1/k > 1$ we have that $F$ traces the unit circle more slowly than $f$ did!
Now what if the value $k$ were greater than $1$? Try it out!

