In the notes of my class, I've read the following application of Gershgorin theorems
If $A\in \mathbb{C}^{n\times n}$ is a strictly (row) diagonally dominant and its diagonal elements are in the semiplane $Re(z)>0$, then also the eigenvalues of $A$ are in the semiplane $Re(z)>0$.
Clearly the centers of the Gershgorin circles are in the semiplane $Re(z)>0$. Then my notes state that the strict diagonal dominance imposes that the Gershgorin circles cannot "escape" the semiplane $Re(z)>0$.
But I don't understand this implication. The strict diagonal dominance just implies that the radii of the Gershgorin circles are less than the distance of their center from the origin. Basically the Gershgorin circles cannot contain the origin.
Am I missing something here or the statement is simply false?
Counterexample: the eigenvalues of $\pmatrix{1+2i&-2i\\ 2i&1+2i}$ are $3+2i$ and $-1+2i$.