# two problems on complex analysis

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

1. a. As $[-1,1]$ has accumulation points, we can apply identity theorem.

b. and c. Your approaches are good.

2. 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.

• ok [[][][]][][][][ – Marso May 4 '13 at 1:23

For the $1$st question:

a. Why not apply the identity theorem?

b,c. You are correct.

For the $2$nd question:

Hint: Aren't $\mathbb{C}$ and the open unit disc $D$ such two sets $\omega$?