TITLE: two problems on complex analysis
QUESTION [5 upvotes]: 1.Pick out the true statements:
a. Let $f$ and $g$ be analytic in the disc $|z| < 2$ and let $f = g$ on the interval
[$−1, 1$]. Then$ f ≡g$.
b. If $f$ is a non-constant polynomial with complex coefficients, then it can be factorized into (not necessarily distinct) linear factors.
c. There exists a non-constant analytic function in the disc $|z| < 1$  which assumes only real values.   
2.Let  $\omega⊂\mathbb{C}$ be an open and connected set and let $f : \omega →\mathbb{C} $ be an analytic function. Pick out the true statements:
a. f is bounded if $\omega$ is bounded.
b. f is bounded only if $\omega$ is bounded.
c. f is bounded if, and only if, $\omega$ is bounded   
for the 1 st question
By fundamental theorem of algebra we can say that (b) is true.
For (a) I am little confused that am I able to apply identity theorem or not.
For (c) by applying Cauchy Riemann equation we get  it is false. Are my approaches correct?
for the 2nd question
here I am little confused. I guess none of them is correct but not sure.can anyone provide me some counter examples

REPLY [2 votes]: a. As $[-1,1]$ has accumulation points, we can apply identity theorem.
b. and c. Your approaches are good.
a. Consider $\omega:=\{z\in\Bbb C, |z|<1\}$ and $f(z):=(1-z)^{—1}$. 
b. Consider a constant function on $\Bbb C$. 
c. No by the previous answers.