Find all of the eigenvalues and eigenvectors of the 2x2 matrix

Question

$$A = \begin{pmatrix}1&2i\\ -3i&-4\end{pmatrix}$$ I got $λ^2 - 3λ - 4 + 6i^2 = 0$

-I isoloated down to $i^2 = -1/6 (λ -4)(λ+1)$ and wrote how λ must be -1 < λ < 4. I don't know what to do next

Answer 1

Just use the fact that $i^2=-1$ within your determinate of the matrix. Furthermore it should be $+3\lambda$ instead of $-3\lambda$. Therefore you got $\lambda^2+3\lambda-4+6(-1)=0$. Can you proceed from here?

Answer 2

$$A = \begin{pmatrix}1&2i\\ -3i&-4\end{pmatrix}$$ Now $|A-\lambda I|=0$ $$\begin{vmatrix} 1-\lambda & 2i \\ -3i & -4-\lambda \end{vmatrix}=0$$ $$(1-\lambda)(-4-\lambda)+6i^2=0$$ $$-4-\lambda+4\lambda+\lambda^2-6=0$$ $$\lambda^2+3\lambda-10=0$$ $$(\lambda+5)(\lambda-2)$$ $$\lambda=-5,2$$

To find the eigenvector you need to find the null space using the above eigen values