I am trying to derive the rotation matrix in expoonential form. I start by considering a rotation in three dimensions about an arbritrary axis, $\hat{\textbf{n}}$ by an infinitesimally small angle $\delta\theta$ as depicted above. The coordinate transforms as \begin{align*} \textbf{r}\rightarrow\textbf{r}'=\textbf{r}+\delta\textbf{r}\approx\textbf{r}+\delta\theta\hat{\textbf{n}}\times\textbf{r} \end{align*} which gives an infinitesimal rotation matrix of \begin{align*} R_{\hat{\textbf{n}}}(\delta\theta)=\begin{pmatrix} 1 & -\delta\theta n_{z} & \delta\theta n_{y} \\ \delta\theta n_{z} & 1 & -\delta\theta n_{x} \\ -\delta\theta n_{y} & \delta\theta n_{x} & 1 \end{pmatrix},\;\;\;\;\textbf{r}'=R_{\hat{\textbf{n}}}\textbf{r} \end{align*} I then see that this can be expressed as \begin{align*} R_{\hat{\textbf{n}}}(\delta\theta)=\mathbb{1}-i\delta\theta\hat{\textbf{n}}\cdot\textbf{J} \end{align*} Where $\textbf{J}$ is a vector of matrices, \begin{align*} J_{x}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0 \end{pmatrix},\;\; J_{y}=\begin{pmatrix} 0 & 0 & i \\ 0 & 0 & 0 \\ -i & 0 & 0 \end{pmatrix},\;\; J_{z}=\begin{pmatrix} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \end{align*} Using the composition property of rotations I then find \begin{align} \frac{d}{d\theta}R_{\hat{\textbf{n}}}(\theta)=-i\hat{\textbf{n}}\cdot\textbf{J} R_{\hat{\textbf{n}}}(\theta) \end{align} Which can be solved to give \begin{align*} R_{\hat{\textbf{n}}}(\theta)=e^{-i\hat{\textbf{n}}\cdot\textbf{J}\theta} \end{align*} So at this point I'm thinking I've obtained my solution, however when i try to consider a rotation about the $z$-axis I find that
\begin{align*} R_{\hat{\textbf{z}}}(\theta)&=e^{-iJ_{z}\theta}\\ &=\sum^{\infty}_{n=0}\left(\frac{(-i\theta)^{n}}{n!}\right)\\ &=\sum^{\infty}_{n=0}\left(\frac{(-1)^{n}\theta^{2n}}{2n!}\right)\mathbb{1}_{2}-i\sum^{\infty}_{n=0}\left(\frac{(-1)^{n}(\theta)^{2n+1}}{(2n+1)!}\right)J_{z}\\ &=\mathrm{Cos}(\theta)\mathbb{1}_{2}-i\mathrm{Sin}(\theta)J_{z} \end{align*}
which is \begin{align*} R_{\hat{\textbf{z}}}(\theta)=\begin{pmatrix} Cos(\theta) & -Sin(\theta) & 0 \\ Sin(\theta) & Cos(\theta) & 0 \\ 0 & 0 & 0 \end{pmatrix} \end{align*}
This is incorrect because it shouldn't map the z value to zero, but I can't for the life of me see where the 1 in the bottom right of the matrix would come from in the exponential.
Does the following help ?
\begin{align} &\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}+ \sum_{n=1}^\infty \frac{(-1)^n\theta^{2n}}{(2n)!}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix}\\ &=\begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}+\sum_{n=0}^\infty \frac{(-1)^n\theta^{2n}}{(2n)!}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix} =\begin{pmatrix}\cos \theta & 0 & 0 \\ 0 & \cos \theta & 0 \\ 0 & 0 & 1\end{pmatrix}\,, \end{align} and
\begin{align} \sum_{n=0}^\infty \frac{(-1)^n\theta^{2n+1}}{(2n+1)!}\begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}= \begin{pmatrix} 0 & -\sin \theta & 0 \\ \sin \theta & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}\,. \end{align} Therefore, with $$ V=\begin{pmatrix}0 & -\theta & 0\\\theta & 0 & 0\\ 0& 0& 0\end{pmatrix}\,, $$
\begin{align} e^V=\sum_{n=0}^\infty \frac{V^n}{n!} =\begin{pmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{pmatrix}\,. \end{align}