I am trying to understand the Fourier transformation and the math behind it, so I was trying to use this formula:
$$ x_k = \sum_{n=0}^{N-1} x_n e^{-\frac{j2{\pi}kn}{N}} $$
to calculate all $ x_{0\ldots3} $ for $N = 4$ and
$$ x = \begin{pmatrix} x_0 \\ x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 - {j} \\ -{j} \\ -1 + 2{j} \end{pmatrix} $$
If I want to calculate $ x_1 $:
\begin{align} x_1 & = x_0\cdot e^{-\frac{{j}2{\pi}1 \cdot 0}{4}} + x_1\cdot e^{-\frac{{j}2{\pi}1{\cdot}1}{4}} + x_2\cdot e^{-\frac{j2\pi1\cdot2}{4}} + x_3\cdot e^{-\frac{j2\pi1\cdot 3}{4}} \\[8pt] x_1 & = x_0\cdot e^0 + x_1\cdot e^{-\frac{j\pi}{2}} + x_2\cdot e^{-j\pi} + x_3\cdot e^{-\frac{3j\pi}{2}} \\[8pt] x_1 & = 1 + (2-j)\cdot e^{-\frac{j\pi}{2}} - j\cdot e^{-j\pi} + (-1 + 2j)\cdot e^{-\frac{3j\pi}{2}} \end{align}
so according to Wikipedia's articles for DFT this should be equal to $ x_1 = -2 - 2j $ but I am missing the last step.
I got a hint that I should use the Euler's formula that $ e^{j\pi} + 1 = 0 $ but I am still unable to simplify it to $ x_1 = -2 - 2j $
[EDIT] I wanted to share my solution here, which might help some people in the future: According to Euler's formula: $ e^{jx} = \cos(x) + j\sin(x) $, $ j^2 + 1 = 0 $. We can then simplify the following expressions to: \begin{align} e^{-\frac{j\pi}{2}} & = \cos\Big(-\frac{\pi}{2}\Big) + j\sin\Big(-\frac{\pi}{2}\Big) = 0 - j = -j \\[6pt] e^{-j\pi} & = \cos(-\pi) + j\sin(-\pi) = -1 + j\cdot0 = -1 \\[6pt] e^{-\frac{3j\pi}{2}} & = \cos\Big(-\frac{3\pi}{2}\Big) + j\sin\Big(-\frac{3\pi}{2} \Big) = 0 + j\cdot1 = j \\[6pt] e^{-2j\pi} & = \cos(-2\pi) + j\sin(-2\pi) = 1 \end{align} Knowing that the cosinusoidal and sinusoidal functions have the same value every $ 2\pi $ and opposite sign every $ \pi $ intervals we can easily calculate the other functions if needed.
$$ x_1 = 1 + (-j)(2 - j) - j(-1) + j(-1 + 2j) = 1 + j(-2 + j + 1 - 1 + 2j) $$ $$ x_1 = 1 + j(-2 + 3j) = 1 - 2j + 3j^2 = 1 - 2j - 3(-1) = -2 - 2j $$
Hint: Using Euler's formula tells you that $$ e^{-j\pi/2} = -j, \quad e^{-j \pi} = -1, \quad e^{-3j\pi/2} = j. $$