Answer
See below
Work Step by Step
Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
a-\lambda & b\\
-b & a-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
a-\lambda & b\\
-b & a-\lambda
\end{bmatrix}=0$
$(a-\lambda)^2+b^2=0$
$a-\lambda)^2-i^2b^2=0$
$\lambda_1=a+ib, \lambda_2=a-ib$
The eigenvalues of $A$ are $\lambda_1 = a+ib$ and $\lambda_2=a-ib$, and therefore A is nondefective.
A straightforward computation yields the following eigenvectors, which correspond respectively to $\lambda_1$ and $\lambda_2$:
$v_1=(1,i)\\
v_2=(1,-i)$
It follows from Theorem 7.4.3 that if we set $S =\begin{bmatrix}
1 & 1\\
i & -i
\end{bmatrix}$ and $D = diag(a+bi,a-bi)$
then $e^{At} = Se^{Dt}S^{−1}$
It is easily shown that
$S^{-1}=\begin{bmatrix}
1 & i\\
1 & -i
\end{bmatrix}$
Substituting:
$e^{At}=\frac{1}{2}\begin{bmatrix}
1 & 1\\
i & -i
\end{bmatrix}\begin{bmatrix}
e^{(a+bi)t} & 0\\
0 & e^{(a-bi)t}
\end{bmatrix}\begin{bmatrix}
1 & i\\
1 & -i
\end{bmatrix}=\frac{1}{2}\begin{bmatrix}
e^{(a+bi)t} & e^{(a-bi)t}\\
ie^{(a+bi)t} & -ie^{(a-bi)}
\end{bmatrix}\begin{bmatrix}
1 & i\\
1 & -i
\end{bmatrix}=e^{at}\begin{bmatrix}
\cos bt & \sin bt\\
-\sin bt & \cos bt
\end{bmatrix}$
