Finding complex eigenvalues and its corresponding eigenvectors 
$$A = \begin{bmatrix}3&6\\-15&-15\end{bmatrix}$$ has complex eigenvalues $\lambda_{1,2} = a \pm bi$ where $a =$____ and $b = $ ____. The corresponding eigenvectors are $v_{1,2} = c \pm di$ where $c =$ (____ , _____ ) and $d =$ (____ , ___)

So I got the char. poly. eqn $\lambda^2 + 12\lambda + 45 = 0$
Then using the quad. eqn I got $-6 \pm 3i$ which I know is in the form $a \pm bi$ so I was thinking $a = -6$, $b = 3$, but instead they have $b = -3$ , why? 
Also, I'm not sure how to obtain the corresponding eigenvectors because we are working with complex eigenvalues now.
 A: Note that $\pm (-3)=\mp3=\pm3$. Hence, whether you use $b=3$ or $b=-3$ is irrelevant. To find the corresponding eigenvectors, you need to solve the following:
$$
\begin{bmatrix}3&6\\-15&-15\end{bmatrix}\cdot\begin{bmatrix}x_{1}\\ x_{2}\end{bmatrix}=\lambda_{1,2}\cdot\begin{bmatrix}x_{1}\\ x_{2}\end{bmatrix}.
$$
A: I'll cover the how to find the eigenvectors part.
$$A = \begin{bmatrix}3&6\\-15&-15\end{bmatrix}$$
has eigenvalues $\lambda_{1,2} = -6\pm 3i$.
Now to find the associated eigenvectors, we find the nullspace of $$A-\lambda_{1,2}I = \begin{bmatrix}3-(-6\pm 3i)&6\\-15&-15-(-6\pm 3i)\end{bmatrix} = \begin{bmatrix}9\mp 3i & 6 \\ -15 & -9\mp 3i\end{bmatrix}$$
To find the nullspace I'll put this in REF:
$$\begin{align}\begin{bmatrix}9\mp 3i & 6 \\ -15 & -9\mp 3i\end{bmatrix} &\sim \begin{bmatrix} 5 & 3\pm i \\ 3\mp i & 2\end{bmatrix} \\ &\sim \begin{bmatrix} 5 & 3\pm i \\ 0 & 2-(3\pm i)\frac{-(3\mp i)}{5}\end{bmatrix} \\ &= \begin{bmatrix} 5 & 3\pm i \\ 0 & 0\end{bmatrix}\end{align}$$
Therefore all of the eigenvectors associated with $\lambda_{1,2}$ are of the form $w_{1,2} = \begin{bmatrix}\frac 15(-3\mp i)t \\ t\end{bmatrix}$.  Representative eigenvectors are then $$\bbox[5px,border:2px solid red]{v_{1,2} = \begin{bmatrix}-3\mp i \\ 5\end{bmatrix}}$$ which is obtained by setting $t=5$.
A: The if $b=-3$, $a=-6$, we have $a \pm bi=-6 \pm 3i$, if $b=3, a=-6$ we still have $a\pm bi=-6 \pm 3i$, both answers are correct but if you're submitting into an online homework service like cengage-brain or webassign your professor may have just forgotten to include both cases as acceptable. To find eigenvectors, you need to solve $Av=\lambda v$, or, equivalently $(A-\lambda I)v=0$ (where I denotes the identity matrix. Why are these two systems equivalent?). Nothing really changes with the complex case so long as you know how to do arithmetic in $\mathbb{C}$. To get you started, for an eigenvector corresponding to $\lambda=-6+3i$, we set up:
$$
\begin{bmatrix}
A-\lambda I 
\end{bmatrix}
v=
\begin{bmatrix}
9-3i & 6 \\
-15 & -9-3i
\end{bmatrix}
\begin{bmatrix}
v_1 \\
v_2
\end{bmatrix} =0
$$
Giving the system:
$$ 
\begin{array}
 ((9-3i)v_1+6v_2=0 \\
-15v_1+(-9-3i)v_2=0
\end{array}
$$
Can you take it from here?
