Let $A$ be a unital Banach algebra and $R:\mathbb{C}\rightarrow A$ be a differentiable function.
If $\phi\circ R=0$ for all $\phi\in A^*$, then is $R$ necessarily zero?
Suppose $R$ is nonzero so that there exists $z\in \mathbb{C}$ such that $x:=R(z)\neq 0$. If there is a bounded linear functional $\phi$ such that $\phi(x)\neq 0$, then we are done. But how?
Assume there exist $z \in \Bbb C$ with $R(z) = v \ne 0$. Let $V = \langle v \rangle$. Define $\varphi : V \to \Bbb C$ by $\varphi (t v) = t \| v \|, \ \forall t \in \Bbb C$. Notice that $| \varphi (tv) | = |t| \| v \| = \| tv \|$. Using the Hahn-Banach theorem, there exist a linear extension $\tilde \varphi : A \to \Bbb C$ such that $\tilde \varphi \big| _V = \varphi$ and $| \tilde \varphi (a) | \le \| a \| \ \forall a \in A$ (which means that $\tilde \varphi$ is continuous). This means that $\tilde \varphi \in A^*$, and $\tilde \varphi (v) = \varphi (v) = \| v \| \ne 0$ (because $v \ne 0$ by assumption).
On the other hand, $\tilde \varphi (v) = (\tilde \varphi \circ R) (z) = 0 (z) = 0$, which contradicts the findings in the previous paragraph. Therefore, no such $z$ exists, so $R(z) = 0 \ \forall z \in \Bbb C$.