"DFT". Understanding the formula $e^{-i2\pi k}$ $k$ is a real number

Question

I am studying about fast Fourier transform.

Assuming that $x_0$, $x_1, \ldots, x_{n-1}$ are complex numbers, the DFT is defined as follows.

$$f_j = \sum\limits_{k=0}^{n-1} x_k e^{-\frac{2\pi i}{n}jk},\quad j= 0,\ldots, n-1$$

In the definition, $$W = e^{-i2 \pi/n}$$ and reorganize it,

$$f_j = \sum\limits_{k=0}^{n-1} x_k W^{jk},\quad j= 0,\ldots, n-1.$$

At this time, I looked at the equation below and solved $W$

$$ \begin{align} W(j) &= e^{-2 * pi * i * j / n}\\ &= e^{ (\pi \cdot i) \cdot (-2 \cdot j / n)}\\ &= \left(e^{\pi \cdot i}\right)^{-2 \cdot j / n} & \text {and } {e^{\pi \cdot i}= -1 }\\ &\text{so}\\ &= (-1)^{-2 \cdot j / n}\\ &= \left( -1^{-2} \right)^{j / n}\\ &= 1^{j / n}\\ &= 1 \end{align}$$

I couldn't figure out where was wrong by solving the equation. Help me out on what part I did wrong.

Answer 1

$$e^{ (\pi \cdot i) \cdot (-2 \cdot j / n)} \color{red}{\ne} \left(e^{\pi \cdot i}\right)^{-2 \cdot j / n}$$

You can't just $e^{a\cdot b}= \left(e^a\right)^b$ for arbitrary complex $a,b$, because $e^a$ is not necessarily a positive real number .

Answer 2

Please recognise $N$-th root of unity (on the complex-plane):

$$ 1^{1/N} = e^{j \frac{2\pi}{N} k} ~~,~~ k = 0,1,...,N-1 $$

where $j$ is the imaginary unit in DSP notation. Hence $1^{1/N} = 1$ is true but incomplete, for you ignore the complex roots.

So in your derivation last two lines should follow as:

$$ 1^{m / N} = (1^{1/N})^m $$ $$ e^{j\frac{2\pi}{N} k m } $$

where $j$ is the imaginary unit and $k$ is an integer. (I've replaced $n$ with $N$ and $j$ with $m$ for consistency of notation)