Let $U$ be an open subset of $\mathbb{C}$ and $f:U \rightarrow \mathbb{C}$ be an analytic function.Then which of the following are true?
$(a)$ If $f$ is one-one, then $f(U)$ is open in $\mathbb{C}$
$(b)$ If $f$ is onto, then $U=\mathbb{C}$
$(c)$ If $f$ is onto, then $f$ is one-one.
$(d)$ If $f(U)$ is closed in $\mathbb{C}$ then $f(U)$ is connected.
My solution:
$(a)$ $f$ is one-one implies it is non constant and by open mapping theorem $f(U)$ is open in $\mathbb{C}$
For $(b)$ and $(c)$ , I have the following idea:-
Let $f(z)$ be a polynomial of degree $n$. Then it has $n$ roots in $\mathbb{C}$.
Now let $z_o$ be a particular zero of $f $ and$U=\mathbb{C}\setminus \{z_o\} $ is open. Then $f:U \rightarrow \mathbb{C}$ is a counter eg to $(b)$ and $(c)$
$(d)$ If $f$ is constant, then $(d)$ is obviously true. If non-constant, then $f(U)$ is clopen set in $\mathbb{C}$ which are only $\mathbb{C}$ or $\phi$ . The latter set is absurd, so image is $\mathbb{C}$ and hence connected. Thus $(d)$ is true.
All I want is verification or improvement of my answer. I would like to know any alternative ideas.Thanks for your time.
Your answer to (d) is incorrect; you mistakenly assumed that $f[U]$ must be open if $f$ is not constant. Suppose $U=V_1\cup V_2$ where $V_1, V_2$ are disjoint open non-empty subsets of $\Bbb C,$ with $f[V_1]=\{1\}$ and $f[V_2]=\{2\}.$