Calculating the matrix exponential in this case? I am wondering how in general one can calculate the matrix exponential. For example, I am working in the Lie algebra $\mathfrak{su}(3)$. I have two matrices $$ m_1 = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$ and $$m_4 = \begin{pmatrix} 0 & i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. $$ Their linear combination $\alpha m_1 + \beta m_4$ can be written as $$ X := \alpha m_1 + \beta m_4 = \begin{pmatrix} 0 & - \overline{z} & 0 \\ z & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$ where $z = \alpha + i \beta$.
Now I wish to obtain a closed formula (if possible) for the matrix exponential $e^{tX}$ (where $t$ is some real number), perhaps in terms of sines and cosines? How does one in general calculate the matrix exponential like this?
 A: Notice that $X^3 = -|z|^2 X$. With that in mind, we can write
$$
\exp(tX) = \sum_{n=0}^{\infty}\frac{(tX)^n}{n!}
= I + \sum_{k=1}^\infty \frac{(tX)^{2k}}{(2k)!}
+ \sum_{k=0}^\infty \frac{(tX)^{2k+1}}{(2k+1)!}\\ 
= 
I + \left(\sum_{k=1}^\infty \frac{(-1)^k(|z|t)^{2k}}{(2k)!}\right) \frac{X^2}{|z|^2}
+ \left(\sum_{k=0}^\infty \frac{(-1)^k(|z|t)^{2k+1}}{(2k+1)!}\right) X
\\ 
= I  + \sin(t|z|) X + \frac{\cos(t|z|) - 1}{|z|^2} X^2.
$$
A: For an even more clear approach, lets work with $2$-by-$2$ matrices. Let $$Y =\begin{pmatrix} 0 & - \overline{z}  \\ z & 0  \end{pmatrix}$$
Notice that $Y^2 = -|z|^2I$ so $Y^{2k} = (Y^2)^k = (-|z|^2)^kI$ and we find
\begin{align*}
\exp(tY) &= \sum_{n=0}^{\infty}\frac{(tY)^n}{n!}\\
&= \sum_{k=0}^{\infty}\frac{(tY)^{2k}}{(2k)!} + \sum_{k=0}^{\infty}\frac{(tY)^{2k+1}}{(2k+1)!}\\
&= \sum_{k=0}^{\infty}\frac{t^{2k}(-|z|^2)^k}{(2k)!}I + \sum_{k=0}^{\infty}\frac{t^{2k+1}(-|z|^2)^k}{(2k+1)!}Y\\
&= \sum_{k=0}^{\infty}\frac{(-1)^k(t|z|)^{2k}}{(2k)!}I + \sum_{k=0}^{\infty}\frac{(-1)^k(t|z|)^{2k+1}}{(2k+1)!}\frac{Y}{|z|}\\
&= \cos(t|z|)I + \frac{\sin(t|z|)}{|z|}Y\\
&=\begin{pmatrix} \cos(t|z|) & -\frac{\overline{z}}{|z|}\sin(t|z|)  \\ \frac{z}{|z|}\sin(t|z|) & \cos(t|z|)  \end{pmatrix}
\end{align*}
We can easily extend this to the matrix $X$ that you gave:
$$\exp(tX) =\begin{pmatrix} \cos(t|z|) & -\frac{\overline{z}}{|z|}\sin(t|z|)& 0  \\ \frac{z}{|z|}\sin(t|z|) & \cos(t|z|)& 0 \\ 0 & 0 & 1 \end{pmatrix}$$
