How does this matrix math with complex numbers work/where is the mistake. I am a beginner to these types of math, so you will have to forgive me if this question has an obvious answer.
I just learned how to get the inverse of a matrix, so I thought it would be interesting to get the inverse of a matrix with complex numbers in it. This is the original matrix: 
$$ A =  \begin{pmatrix} i & i \\ 3 & i \end{pmatrix}$$
I then found the inverse of that matrix:
$$ A^{-1} =  \frac{1}{-1-3i} \begin{pmatrix} i & -i \\ -3 & i \end{pmatrix}$$
Therefore the top left corner of this matrix should be 1:
$$ A A^{-1} = \begin{pmatrix} i & i \\ 3 & i \end{pmatrix} \begin{pmatrix}\frac{1}{-1-3i} \begin{pmatrix} i & -i \\ -3 & i \end{pmatrix}\end{pmatrix}$$
Which means that :
$$ -1 \cdot \frac{1}{-1-3i} = 1  $$
Which simplifies into:
$$  \frac{1}{1+3i} = 1$$
And the only way for that to be true is if:
$$ 1+3i = 1$$
And finally that can only be true if: 
$$ 3i = 0$$

The error seems to be in how you did the matrix multiplication. Your setup is, after moving the scalar ($1/(-1-3i)$) to the left,
$$\frac{1}{-1-3i} \begin{bmatrix}
i & i \\ 
3 & i
\end{bmatrix} \begin{bmatrix}
i & -i\\ 
-3 & i
\end{bmatrix}$$
The top-left corner of the resulting matrix would be the dot product of the top row of the left matrix with the left column of the right matrix, all times that scalar on the left. Then you would get
$$\frac{i^2 - 3i}{-1-3i} = \frac{-1-3i}{-1-3i} = 1$$
as expected since a matrix times its inverse is the identity matrix.
