a. $e^{iz}$ satisfies $|e^{iz}|=e^{-\text{Im}(y)}<1$
b. $iz$ as Tim Raczkowski told you.
c. $e^{z}$ is bounded on the imaginary axis. If $iz\in\mathbb{R}$ then $|e^{z}|=1$.
Did you already learn the identity theorem in you complex analysis course? If not, I can explain the proof if you ask. If you did learn the theorem already, this question is a direct application of it.
EDIT: Sorry, I misread your question. The theorem you are given at the beginning: "A function that is analytic in a domain $D$ is uniquely determined over $D$ by its values in a domain, or along a line segment, contained in $D$" is just a re-wording of the identity theorem. I use the wording of the theorem given on Wikipedia below because I find it to be clearer /EDIT
Specifically, using the hint given we start with,
Assume by means of contradiction, that $f(z)=w_0$ is constant on some neighborhood in $D$.
Then, by the identity theorem, which states
Given two holomorphic functions $f$ and $g$ on a domain (open and connected set), if $f=g$ on some neighborhood in $D$ (non-empty open subset), then $f=g$ on all of $D$.
and using the functions $f$ and $g \equiv w_0$ (the constant function equal to $w_0$ which is trivially holomorphic), we must conclude that $f=g=w_0$ everywhere in $D$, and that $f$ is constant everywhere in $D$. This contradicts our assumption that $f$ was to be non-constant in $D$.
Therefore it cannot be true that $f$ is constant on any neighborhood in $D$.
Best Answer
There is not. Any such function $f$ would be bounded near $0$. So, by Riemann's extension theorem, $f$ can be extended to an analytical function $\hat{f}$ in $\Bbb{C}$. But $\hat{f}$ is bounded and entire, so it is constant. Since $f=\hat{f}$ in $\Bbb{C} \setminus \{0\}$, $f$ is constant too.