suppose f is analytic on a domain g and the area of f(g) is 0. show f is constant on g
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.
If you have any doubt please post your comment Kindly rate the answer accordingly.
Get Answers For Free
Most questions answered within 1 hours.