In order to test if a given nonzero vector $\mathbf{v}$ is a characteristic vector (aka eigenvector) of a given matrix $A$, you do not need to find the eigenvalues, or characteristic polynomial of the matrix! All you have to do is compute $A\mathbf{v}$ and see if what you get is a scalar multiple of $\mathbf{v}$ or not.
So you need to compute $\left(\begin{array}{cc}a&b\\-b&a\end{array}\right)\left(\begin{array}{c}1\\i\end{array}\right) = \left(\begin{array}{c}a+bi\\-b+ai\end{array}\right).$ Now, is $(a+bi,-b+ai)$ a scalar multiple of $(1,i)$? Yes: check that if $\lambda=a+bi$, then $\lambda(1,i) = (a+bi,-b+ai)$. Since $A(1,i)^t$ is a scalar multiple of $(1,i)$, then $(1,i)$ is a characteristic vector of $A$.
Similarly with $(1,-i)$.
(I'm reminded of how the vast majority of my Calculus students will, when faced with a problem like "Verify that $f(x)=2\sin x-3\cos x$ is a solution to the differential equation $y''+y=0$" will proceed to try to find the general solution to the equation and see if the given $f(x)$ is of that general form, instead of simply plugging in $f$ and checking the equality... even after I tell them to just plug in... then again, if they are asked to verify that $17$ is a solution to $x^2 -27x + 170=0$, they will proceed to solve the quadratic instead of just plugging in, too)