Periodicity of the exponential function in a field

Question

Let $\mathbb{K}$ an exponential field, i.e. a field of characteristic $0$ with a (non constant) function $E:\mathbb{K}\to \mathbb{K}$ such that $E(x+y)=E(x)E(y)$ and $E(0)=1$.

It is easy to prove that:

if there exists an element $\alpha \in \mathbb{K}$ such that $E(\alpha)=-1$ than:

$1)$ in $\mathbb{K}$ there is also an element $\beta= E(\frac{\alpha}{2})$ that is a ''square root'' of $-1$, because $E(\alpha)=E(\frac{\alpha}{2} + \frac{\alpha}{2})=\beta^2=-1$

and

$2)$ the exponential function $E$ is periodic with period $2\alpha$, because $E(2\alpha)=(-1)\cdot(-1)=1$.

Is it true also the converse? I.e. : If in an exponential field there is an element $\beta$ such that $\beta^2=-1$ than there is also an element $\alpha$ such that $E(\alpha)=-1$ and the function $E$ is periodic?

This is obviously true if $\mathbb{K}=\mathbb{C}$, but I can't find a general result for a generic field.

Answer 1

Without assuming more (e.g. surjectivity) of the 'exponential' $E$, it is not true:

Let $\Bbb K:=\Bbb C$ and $E(z):=e^{\mathrm{Re}(z)}$ to give a counterexample to the existence of such an $\alpha$.