Let $f(z)$ be an analytic function on a open connected subset $\Bbb{G}$ of $\Bbb{C}$ with $|f(z)|= z_{0}$ for some fixed $z_{0}$ then prove that $f(z)$ is a constant function.
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2What theorems do you know? This is a very simple consequence of the open mapping theorem for instance. – JSchlather Mar 03 '13 at 02:49
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2Or maximum modulus principle... – N. S. Mar 03 '13 at 02:53
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1@N.S. And the maximum modulus principle is a very simple consequence of the open mapping theorem ;). – JSchlather Mar 03 '13 at 02:59
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If $z_0 = 0$, there is nothing to prove, as $f(z) = 0$ for $z \in G$. So we may presume that $z_0 > 0$.
Suppose $z_1 \in G$ such that $f'(z_1) \neq 0$, and consider $\phi(\lambda) = |f(z_1 + \lambda \frac{f(z_1)}{f'(z_1)})|^2$ for small $\lambda$. $\phi$ is differentiable at $\lambda = 0$, and $\phi'(0) = 2 |f(z_1)|^2$. However, by assumption, $\phi(\lambda) = z_0$, which implies that $\phi'(0) = 0$, a contradiction. Hence $f'(z_1) = 0$, and it follows that $f$ is locally constant. Since $f$ is connected , it follows that $f$ is constant on $G$.
copper.hat
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@CityOfGod: The assumption in the question was that $|f(z)| = z_0$ for all $x \in G$. I am splitting this into two cases, $z_0 =0$ and $z_0 >0$. – copper.hat Mar 03 '13 at 17:07