Rational representation of the additive group $\mathbb{C}$ as an algebraic group The additive group $\mathbb{C}$ is a linear algebraic group via the embedding into $SL_2(\mathbb{C})$, $$z \mapsto \begin{bmatrix} 1 & z \\ 0 & 1 \end{bmatrix}.$$
In this way, the Lie algebra of $\mathbb{C}$ is the $\mathbb{C}$-span of the matrix $e_{12}$.
I know that the exponential map gives a bijection between nilpotent elements of $M_2(\mathbb{C})$ and unipotent elements of $GL_2(\mathbb{C})$, via $$ \begin{bmatrix} 0 & z \\ 0 & 0 \end{bmatrix} \leftrightarrow \begin{bmatrix} 1 & z \\ 0 & 1 \end{bmatrix}.$$
I want to show the following:

if $(\pi, V)$ is any regular representation of $\mathbb{C}$, there exists a unique nilpotent matrix $A \in \text{End} V$ for which $\pi(z) = \exp(zA)$ for all $z \in \mathbb{C}$.

I know that for $A \in \mathfrak{g}$, $\exp A = I + z e_{12}$ for some $z \in \mathbb{C}$, and so for any $z \in \mathbb{C}$, the bijection given by $\exp$ says that $\pi(z)$ is effectively equivalent to $\pi(\exp A)$, where $A = I + z e_{12}$.
I also know that in this setup, $\pi(\exp A) = \exp (d\pi(A))$, so I believe I need to show that $d\pi(A) = z A'$ for some $A' \in \text{End} V$.
By what I said earlier, we know the form that the matrix $A$ takes, and in particular, $$d\pi(A) = \pi \left( \begin{bmatrix} 1 & z \\ 0 & 1 \end{bmatrix} \right) d\pi \left( \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \right) = z A',$$ where $A' = d\pi \left( \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \right)$.
I believe this concludes the proof - is this correct?
 A: Given any rational representation $\pi:G\to GL(V)$ (that is, $\pi$ may be written down explicitly using rational functions and matrices), the derivative $\mathrm{d}\pi:\mathfrak{g}\to\mathfrak{gl}(V)$ is also a rational map and
$$ \pi(\exp X)=\exp\mathrm{d}\pi(X). $$
In the case of $G=(\mathbb{C},+)$, we have $\mathfrak{g}=\mathbb{C}$ too with $\exp z=z$. (I assume $G$ is being treated as a complex group, so $\mathfrak{g}$ is a complex vector space.) Then $\mathrm{d}\pi:\mathbb{C}\to\mathfrak{gl}(V)$ is a linear map, it must be $\mathrm{d}\pi(z)=z\mathrm{d}\pi(1)$, so if we define $X=\mathrm{d}\pi(1)\in\mathfrak{gl}(V)$ we have $\mathrm{d}\pi(z)=zX$ and hence
$$ \pi(z)=\pi(\exp z)=\exp\mathrm{d}\pi(z)=\exp(zX). $$
There is a choice of coordinates in which $X$ is block-diagonal with Jordan blocks (i.e. canonical form). So it suffices to examine the exponential of a Jordan block and see when it's rational.
Let $J=\lambda I+N$ be a Jordan block with $N$ strictly upper triangular. By looking at the power series it is straightforward to verify $\exp(zN)$ is rational, however $\exp(zJ)=e^{\lambda z}\exp(zN)$ is then not rational unless $\lambda=0$, which forces all of $X$'s Jordan blocks and hence $X$ itself to be nilpotent.
