Rotation matrix exponential form derivation (only acts in 2 dimension) 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}
