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suppose f is analytic on a domain g and the area of f(g) is 0. show...

suppose f is analytic on a domain g and the area of f(g) is 0. show f is constant on g

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Answer #1

Since g is a domain, it is open.

By the open mapping theorem, every non-constant analytic function takes open set to an open set. So f(g) must be an open set.
Since every open set in the complex plane have a non-zero area, so f(g) must have non-zero area unless f is constant.  
Since area of f(g) is 0, so f must be constant on the domain g.

Let x belongs to f(g) be any point. Since f(g) is open, there exist an open ball of radius r (say) B(x,r) such that B(x,r) is a subset of f(g) .
So area of B(x,r)f(g).
Now, area of B(x,r) = .
So area of f(g).

Thus if f is analytic on a domain g and the area of f(g) is 0, then we conclude that f must be constant on g.

Hence the proof. ​​​​

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