Let $f$ and $g$ be differentiable on the strip $D = \{z \in \mathbb{C}: -2 < \operatorname{Im}z <2$. Suppose that $ f(z) =g(z)$ for all $z$ such that $|z|<0.01$. By considering Taylor expansions first about $0$, then $1$, and so on by induction, prove that $f(z)=g(z)$ on $D$.
Is it safe simply to state the Identity Theorem? If not, I am lost and need some assistance
Identity Theorem: If $f$ and $g$ are differentiable in a domain $D$ and $f(z)=g(z)$ for all $z \in S \subseteq D$ where $S$ has a limit point in $D$, then $f=g$ throughout $D$.
The fact that $f=g$ for $|z| <0.01$ implies that the power series expansions of $f$ and $g$ in the disk $D(0,2)$ have the same coefficients: the coefficients are simply $\frac {f^{(n)} (0)} {n!}$ and $\frac {g^{(n)} (0)} {n!}$. This proves that $f=g$ in $D(0,2)$. Now we already know that $f=g$ is some disk around $1$ so we can repeat the argument (using power series around the point $1$) to conlude that $f=g$ in $D(1,2)$, and so on. Thus, $f=g$ in $D(n,2)$ for each $n$ and hence $f=g$ on the part of the domain where $\Re z \geq 0$. Similar argument holds for the part $\Re z \leq 0$.