Given a complex-valued function $f(z)$ holomorphic in $|z|<2$. Prove that $$\max_{|z|=1} \left|f(z)-\frac{1}{z}\right|\geq1$$
I know the behaviour of holomorphic function but I can't do so much about that. Please help me.
Holomorphic Function in Disk and its Maximal
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complex-analysis
complex-numbers
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0Perhaps there's an 'easy' way to show there exists $|z_0|<1$ such that $| f(z_0)-1/z_0| \ge 1$, and then apply the maximum modulus principle. – 2017-01-20
1 Answers
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Let $$ c = \max_{|z|=1} \left| f(z)-\frac 1z \right|. $$ Then $$ \left|\int_{|z|=1} \left(f(z) - \frac1z\right)\,dz \right| \le 2\pi c $$ by the standard estimation lemma. On the other hand (by Cauchy's integral theorem and/or formula), $$ \int_{|z|=1} \left(f(z) - \frac1z\right)\,dz = -2\pi i $$ so $c \ge 1$.