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Suppose a complex-valued function $f$ is analytic in the domain $D$ where $D$ is the disk $|z| < R$. If $f(0) = i$, and $|f(z)| \le 1$, what is $f$? I'm thinking that $f$ is just the constant function $f(z) = i$, but I'm not sure how to justify this.

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What were the last theorems in the course/book? – Berci Nov 29 '12 at 0:33
Do you know maximum principle? – Paul Nov 29 '12 at 1:18

Your answer is correct and follows from the maximum principle: You have $|f(z)|=1$ for all $z \in U:=\{w \in \mathbb{C}; |w|<R\}$. Moreover, $|f(0)|=|\imath|=1$ and clearly $0 \in U$, which means that $|f|$ attains its maximum value in the interior of $U$. Therefore you obtain from the maximum principle that $f$ is constant, hence $f=\imath$. (Observe that $f$ analytic implies $f$ holomorphic.)

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