I know how I compute the exponential $e^{tA}$ where $A$ is this matrix
$$A=\begin{pmatrix}a&-b&0\\ b &a & 0\\ 0&0&c\end{pmatrix},\quad b\ne0,$$ using the expression $\exp(tA)=\sum\limits_{k=0}^\infty\frac{t^kA^k}{k!}$. How can find the result using a geometric method i.e. the propriety of matrix of rotation or other geometric transformation?
$A^2 = \begin{bmatrix} a^2-b^2 &-2ab\\2ab&a^2-b^2\\0&0&c^2\end{bmatrix}$
Now I am going to suggest that you should be thinking about complex numbers here where $z = a+bi$ and
$A^2 = \begin{bmatrix} Re[z^2] &-Im[z^2]\\Im[z^2]&Re[z^2]\\0&0&c^2\end{bmatrix}$
Where $Re[]$ is the real part and $Im[]$ is the imaginary part.
$A^n = \begin{bmatrix} Re[z^n] &-Im[z^n]\\Im[z^n]&Re[z^n]\\0&0&c^n\end{bmatrix}$
$e^{At} = \begin{bmatrix} Re[e^{zt}] &-Im[e^{zt}]\\Im[e^{zt}]&Re[e^{zt}]\\ 0&0&e^{ct}\end{bmatrix}$
$e^{zt} = e^{at}\cos bt + i e^{at}\sin bt$
$e^{At} = \begin{bmatrix} e^{at}\cos bt &-e^{at}\sin bt\\e^{at}\sin bt &e^{at}\cos bt\\ 0&0&e^{ct}\end{bmatrix}$
From a geometric point of view the block in the top right of $A$ suggests a scaling factor combined with a rotation factor, while the $c$ by itself on the bottom right suggest purely a scaling factor in that direction.