The exponention map of the matrix Let $sl(n)$ denote the set of all $n\times n$ real matrices with trace equal to zero and let $SL(n)$ be the set of all $n\times n$ matrices with determinant equal to one. Let $\varphi(z)$ be a real analytic function defined in a neighborhood of $z=0$ of the complex plane $\mathbb{C}$ satisfying the conditions $\varphi(0)=1$ and $\varphi'(0)=1$.
(1) If $\varphi$ maps any near zero matrix in $sl(n)$ into $SL(n)$ for some $n\geq 3$, show that $\varphi(z)=\exp(z)$.
(2) Is the conclusion of (1) still true in the case $n=2$? If it is true, prove it. If not, give a counterexample.
On (1), for near zero $A$ with $tr(A)=0$, then $det(\varphi(A))=1$. How can we deduce that $\varphi(z)=\exp(z)$.
 A: Because $\varphi$ is is analytic, it is enough to show that it agrees with the exponential in some small neighbourhood of zero. Let $\varepsilon>0$ such that $\varphi$ maps any trace-zero matrix with all entries less that $\varepsilon$ is absolute value to $SL(n)$. 
In particular, for any $s$ with $|s|<\varepsilon$,
$$
1=\det\varphi\left(\begin{bmatrix}s&0&0\\0&-s&0\\0&0&0\end{bmatrix}\right)=\varphi(s)\varphi(-s)\varphi(0)=\varphi(s)\varphi(-s),
$$
so $\varphi(-s)=1/\varphi(s)$. Now, for any small enough $s,t$
$$
1=\det\varphi\left(\begin{bmatrix}t&0&0\\0&s&0\\0&0&-s-t\end{bmatrix}\right)=\varphi(t)\varphi(s)\varphi(-s-t).
$$
We conclude that $\varphi(t)\varphi(s)=\varphi(t+s)$. Now
$$
\varphi'(s)=\lim_{h\to0}\frac{\varphi(s+h)-\varphi(s)}h=\lim_{h\to0}\frac{\varphi(s)\varphi(h)-\varphi(s)}h=\varphi(s)\,\lim_{h\to0}\frac{\varphi(h)-\varphi(0)}h=\varphi(s)\,\varphi'(0)=\varphi(s).
$$
So $\varphi'=\varphi$, and the initial conditions guarantee that $\varphi(s)=e^s$ in the neighbourhood of zero where it is analytic. 
For $n>3$, we can always work on the upper left $3\times 3$ corner. 
For $n=2$, the assertion is false. For example, let $\varphi(z)=e^{f(z)}$, where $f(z)$ is analytic and $f(-z)=-f(z)$ (for instance, $f(z)=z^3$). Then, if $A$ is $2\times 2$ and has trace zero, then its eigenvalues are $\lambda,-\lambda$ for some $\lambda\geq0$. Then $$\det\varphi(A)=\varphi(\lambda)\varphi(-\lambda)=e^{f(\lambda)}e^{f(-\lambda)}=e^{f(\lambda)+f(-\lambda)}=e^0=1.$$ 
