Suppose . Then if and only if , where is short for the function .
Proof By definition, means that the has at least one zero on the domain . That is for some , thus . The converse is also true by the same arguments.
Open Mapping Theorem
Let be a domain and let be a non-constant function. Then is open.
Proof Fix some , we must show that some open disk around is contained in . Since is non-constant, we can pick some such that for all . If such doesn’t exist, then there would exist a sequence such that for all , then by the uniqueness theorem, would be constant, which yields a contradiction. Now let We have since on . Notice that the minimum can be achieved as the boundary is compact and is continuous.
We will now show that . Let , considerFor , we have Hence we can apply the Rouché’s theorem to to obtainBy lemma, this implies that , therefore , and so is open.
Theorem
Let and suppose for some . Then is not injective.
Proof Since , we fix such similar to the above arguments, we can still obtain But since , . So there are two cases:
there exist such that but ;
if and , then .
In fact, the second option would imply that is zero on , and so is identically zero on by uniqueness theorem. That is saying that is constant, yielding a contradiction. Thus only the first option holds, which implies is not injective.
Corollary
If is injective, than for all .
Proof Simply contrapositive of the above theorem, yet might be more useful.
The above isn't true in
Attention that the above is not generally true in real analysi. Consider real function , it satisfies but is injective.
Theorem
Let be injective. Then is a domain and is biholomorphic.
Proof By the open mapping theorem, is open. is connected since is continuous. So is a domain. By the above corollary, as is injective, for all . Hence the Jacobian of is non-zero: with not both zeroes. And thus , which indicates that the Jacobian is invertible at every . Thus by the inverse function theorem, we know that is -differentiable, and Thus also satisfies the Cauchy-Riemann equations, therefore holomorphic.
Maximum Modulus Principle
Strong Maximum Modulus Principle
Let be a domain and . If attains its maximum at some , then is constant.
Proof Suppose for all , but is not constant. Then by open mapping theorem, is open. Thus there exists some such that . So has a greater modulus than , yielding a contradiction.
Weak Maximum Modulus Principle
If is a bounded domain and , then attains its maximum on . i.e.
Proof If is constant, then the result is trivial. Otherwise, by the open mapping theorem, is open. Similar to the weak maximum principle, we can show that the maximum cannot be achieved in the interior of .