# If $f$ is analytic and $|f(z_0)|\geq |f(z)|$ for every $z$ such that $|z-z_0|=R$ then there is no point $z_1$ such that $|f(z_1)|<|f(z_0)|$.

Suppose $f$ is analytic in a domain $D$. Let $z_0\in D$ and $\overline{D}_R(z_0)\subset D$. What I need to prove is that if $f$ is analytic and $|f(z_0)|\geq |f(z)|$ for every $z$ such that $|z-z_0|=R$ then there is no point $z_1$ such that $|f(z_1)|<|f(z_0)|$. I just have no idea where to even begin so any hint will be much appreciated. I apologize for not showing any effort. Any help will be appreciated. Thanks

• Do the inequalities above go in the right direction? Should it perhaps be $|f(z_0)|\leq|f(z)|$? – mickep Dec 23 '14 at 17:33
• Nope the given is correct – Heisenberg Dec 23 '14 at 17:33
• Hint: Use Maximum Modulus Principle for the Disc $\overline{D}_R(z_0).$ – Krish Dec 23 '14 at 17:41
• The Mean Value Property implies that $f$ is constant on $\partial D_R(z_0)$. Then Cauchy's Integral Formula implies that $f$ is constant on $\overline{D}_R(z_0)$. – robjohn Dec 23 '14 at 17:53

First Version: Let $G$ be a bounded domain and let $f : G \rightarrow \mathbb{C}$ be an analytic function. Suppose there exists an element $a \in G$ such that $|f(a)| \geq |f(z)|, \forall z \in G.$ Then $f$ is constant.
Second Version: Let $G$ be bounded open set in $\mathbb{C}$ and let $f : \overline{G} \rightarrow \mathbb{C}$ be a continuous function which is analytic on $G.$ Then max$\{|f(z)| : z \in \overline{G}\} =$ max$\{|f(z)| : z \in \partial G \}.$
(Ref: J. B. Conway: Functions of one complex variable, Chapter $VI.$)
Apply this to $G := \{z : |z - z_0| < R\}$ to conclude that $f$ is constant on $G.$ Now construct a function $g : D \rightarrow \mathbb{C}$ by $g(z) = f(z) - f(z_0).$ Then $g$ is an analytic function on $D$ and $D(z_0, R) \subseteq Z(g),$ the zero set of $g$. So $g = 0$ on $D.$