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Let and define for all n € N. p(z) = 25 + z¹ + 6z² +3z +1 Pn(z) = p(z) — 1/n (a) Show pn has 2 roots in B₁ (0) for all n. (b) How many roots does p have in B₁(0)? You may assume |p(z)| ≥ 4 for z € ƏB₁(0). (c) Why do these kinds of problems about locating roots of polynomials avoid directly using the Argument Principle?

Let and define for all n € N. p(z) = 25 + z¹ + 6z² +3z +1 Pn(z) = p(z) — 1/n (a) Show pn has 2 roots in B₁ (0) for all n. (b) How many roots does p have in B₁(0)? You may assume |p(z)| ≥ 4 for z € ƏB₁(0). (c) Why do these kinds of problems about locating roots of polynomials avoid directly using the Argument Principle?

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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plz solve questio (a) it with explanation within 30-40 mins I'll give you multiple upvote
Let p(2) = 25 + zª + 6z² + 3z + 1 and define Pu(2) = p(z) – 1/n for all n e N. (a) Show p, has 2 roots in B1(0) for all n. (b) How many roots does p have in B1(0)? You may assume [p(z)| > 4 for z E ƏB1(0). (c) Why do these kinds of problems about locating roots of polynomials avoid directly using the Argument Principle?
Transcribed Image Text:Let p(2) = 25 + zª + 6z² + 3z + 1 and define Pu(2) = p(z) – 1/n for all n e N. (a) Show p, has 2 roots in B1(0) for all n. (b) How many roots does p have in B1(0)? You may assume [p(z)| > 4 for z E ƏB1(0). (c) Why do these kinds of problems about locating roots of polynomials avoid directly using the Argument Principle?
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