Let $f$ be analytic function of complex variable that assumes only purely imaginary values in a region $G$. Prove that $f$ is constant on $G$. I am stuck here. Will anybody please help?
f is constant on G
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$\begingroup$
complex-analysis
complex-numbers
complex-integration
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1How about using the CR equations? – 2017-01-20
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0Hint : open mapping theorem. – 2017-01-20
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0Will u please elaborate? – 2017-01-20
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0I got it thanks – 2017-01-20
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0But I did not get see-woo's point.pls help – 2017-01-20
1 Answers
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Choose any small disk $D\subseteq G$. If $f$ is constant on $D$, identity theorem implies $f$ is constant on $G$. Otherwise, by the open mapping theorem, $f(D)$ should be open subset of $\mathbb{C}$. However, it is impossible since $f(D)$ is contained in $i\mathbb{R}$. (Any open neighborhood of any point in $f(D)$ should contain other points in $\mathbb{C}\backslash i\mathbb{R}$.)