I'm having problems with the Gershgorin circle theorem with complex numbers...
Let $T:\mathbb{C} \to \mathbb{C}$ be a linear transformation $T(x, y, z, w) = (x+2z, 2αy+2z−w,12x + 3y − 5z, x + y − iz + 12iw)$ $α\in \mathbb{K}$
There's a bunch of options, but the correct one is "If $|α| > 10$ then T is diagonalizable."
I found all the circles:
$C_1=\{z \in \mathbb{C} : |z-1|\le 2\}$
$C_2=\{z \in \mathbb{C} : |z-2α|\le 3\}$
$C_3=\{z \in \mathbb{C} : |z+5|\le 7/2\}$
$C_4=\{z \in \mathbb{C} : |z-12i|\le 2-i\}$
But I don't know how to find out if they're disjoint. (I'm kind of dumb with complex numbers lol)
EDIT: My problem is just with $C_3$... I don't know what to do with a complex radius
EDIT: This is how I found all the circles:
$E$ being the standard basis, the associated matrix for T is:
$_{E}(T)_{E}=\begin{bmatrix} 1 & 0 & 2 & 0 \\ 0 & 2 \alpha & 2 & -1 \\ 1/2 & 3 & -5 & 0 \\ 1 & 1 & -i & 12i \end{bmatrix} $
So $C_1$'s center is $1$ and radius $0+2+0=2$, $(1, 2)$
$C_2$'s center is $2\alpha$ and radius $0+2+1=3$, $(2\alpha, 3)$
$C_3$'s center is $-5$ and radius $1/2+3+0=7/2$, $(-5, 7/2)$
$C_4$'s center is $12i$ and radius $1+1-i=2-i$, $(12i, 2-i)$