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IMC / 2021 / Problems / Day 2, P7

IMC 2021 · Day 2 · P7

very hard

Let DCD \subseteq \mathbb{C} be an open set containing the closed unit disk {z:z1}\{ z : |z| \le 1 \}. Let f:DCf : D \to \mathbb{C} be a holomorphic function, and let p(z)p(z) be a monic polynomial. Prove that f(0)maxz=1f(z)p(z).\bigl| f(0) \bigr| \le \max_{|z|=1} \bigl| f(z) p(z) \bigr|. (proposed by Lars Hörmander)

Solution (official)

Hint: Apply the maximum principle or the Cauchy formula to a suitable function f(z)q(z)f(z) q(z).

Let q(z)=znp(1/z)q(z) = z^n \cdot p(1/z), or more explicitly, if p(z)=zn+an1zn1++a0,p(z) = z^n + a_{n-1} z^{n-1} + \dots + a_0, let q(z)=1+an1z++a0zn.q(z) = 1 + a_{n-1} z + \dots + a_0 z^n. Note that for z=1|z| = 1 we have 1/z=zˉ1/z = \bar{z} and hence q(z)=p(z)|q(z)| = |p(z)|. Hence by the maximum principle or the Cauchy formula for the product of ff and qq, it follows that f(0)=f(0)q(0)maxz=1f(z)q(z)=maxz=1f(z)p(z).\bigl| f(0) \bigr| = \bigl| f(0) q(0) \bigr| \le \max_{|z|=1} \bigl| f(z) q(z) \bigr| = \max_{|z|=1} \bigl| f(z) p(z) \bigr|.

How the field did

contestants scored
514
average (of 10)
0.91
solved (≥ 80%)
5.6%
near-0 (≤ 10%)
91.6%
discrimination
0.42

Score distribution (field cohort)

Computed on contestants with a meaningful total (field cohort); discrimination is the corrected item–total correlation.

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