Let $U$ be an open subset of $\mathbb C$. Let $f:U\rightarrow\mathbb C$ be an analytic function then which of the following are true?
(1) If $f$ is one one then $f(U)$ is open. (2) If $f$ is onto then $U=\mathbb C$. (3) If $f$ is onto then $f$ is one one. (4) If $f(U)$ is closed then $f(U)$ is connected.
Option (1) is correct using Open mapping theorem. Option (3) is not correct. For example take $f(z)=z^2$ then $f$ is onto but not one one. Option (4) is also correct. Because by open mapping theorem $f(U)$ is open and if $f(U)$ is also closed then it is a clopen subset of $\mathbb C$ which is either $\phi$ or $\mathbb C$. Now $f(U)\neq \phi$ so $f(U)=\mathbb C$ which is connected. Are the above reasons correct? Also I have no idea about (2).
Option (2) is false. Take, say, $f(z)=z^2-z$ and $U=\Bbb C\setminus\{0\}$.
Option (4) is also false. Take $f\colon\Bbb C\setminus\Bbb R\longrightarrow\Bbb C$ defined by$$f(x)=\begin{cases}1&\text{ if }\operatorname{Im}z>0\\0&\text{ if }\operatorname{Im}z<0.\end{cases}$$