Consider $\mathbb{R}$ to be an additive topological group, and $\mathbb{C}^{\ast}$ to be a multiplicative topological group.


Is the following statement true?
If so, then how can one prove it?


Statement: any continuous group homomorphism $\mathbb{R} \to \mathbb{C}^{\ast}$ is always of the form $exp(\lambda x)$, $\lambda \in \mathbb{C}^{\ast}$.



Is any continuous group homomorphism from R to C* an exponential map?
Yes. First, decompose $\mathbf{C}^\times$ as $\mathbf{R}\_{>0}\times\mathbf{T}$, where $\mathbf{T}=\{z\in\mathbf{C}^\times:\vert z\vert=1\}$. We have $\mathbf{R}\_{>0}\cong\mathbf{R}$ via the logarithm, and any continuous homomorphism $\mathbf{R}\rightarrow\mathbf{R}$ is necessarily $\mathbf{R}$-linear, and hence given by $x\mapsto sx$ for some (unique) $s\in\mathbf{R}$. The set of continuous homomorphisms $\mathbf{R}\rightarrow\mathbf{T}$ is the Pontryagin dual of $\mathbf{R}$, which is isomorphic to $\mathbf{R}$ via $s\mapsto(x\mapsto\exp(ixs))$. Combining these facts, one finds that the continuous homomorphisms $\mathbf{R}\rightarrow\mathbf{C}^\times$ are of the form $x\mapsto \exp(zx)$ for $z\in\mathbf{C}$.


