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For n ≥ 2, let z₁,..., Zn ben distinct points on C, with |zi| < 1. Let P(z) = (z-2₁)(z-Zn). Prove that 1 fix=1 P(2) dz=0. |z|=1 Hint: change variable w = 1/z, so that the integrand in the interior of disk |w| < 1 has no poles.

For n ≥ 2, let z₁,..., Zn ben distinct points on C, with |zi| < 1. Let P(z) = (z-2₁)(z-Zn). Prove that 1 fix=1 P(2) dz=0. |z|=1 Hint: change variable w = 1/z, so that the integrand in the interior of disk |w| < 1 has no poles.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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For n ≥ 2, let z₁,..., Zn ben distinct points on C, with |z₁| < 1. Let P(z) = (z − z₁)... (z — Zn). Prove that 1 fax-₁ P(2) |z|=1 Hint: change variable w = = 1/2, so that the integrand in the interior of disk |w| < 1 has no poles. -dz = 0. Remark: this also holds for more general numerators. For polynomial Q(z) with deg Q≤n - 2, we have Q(z) -dz = 0. |2|=1 P(2) You can prove this more general form (instead of the above one) if you want.
Transcribed Image Text:For n ≥ 2, let z₁,..., Zn ben distinct points on C, with |z₁| < 1. Let P(z) = (z − z₁)... (z — Zn). Prove that 1 fax-₁ P(2) |z|=1 Hint: change variable w = = 1/2, so that the integrand in the interior of disk |w| < 1 has no poles. -dz = 0. Remark: this also holds for more general numerators. For polynomial Q(z) with deg Q≤n - 2, we have Q(z) -dz = 0. |2|=1 P(2) You can prove this more general form (instead of the above one) if you want.
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