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### Messages - Ziqi Zhang

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##### Final Exam / Re: FE-P3
« on: December 18, 2018, 11:28:38 AM »
Residue at z=0: 0+cos2(0)=1

Residue at z=kπ, but k≠0: 0-2(kπ)cos(kπ)sin(kπ)+cos2(kπ)=(-1)n

Residue at z=0.5π+kπ: -1+0=-1

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##### Final Exam / Re: FE-P3
« on: December 18, 2018, 11:17:50 AM »
I think when z=0, it should have pole of order 1. Because at z=0, the numerator is not 0 when taking derivative once and denominator is not 0 when taking derivative twice.

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##### Final Exam / Re: FE-P5
« on: December 18, 2018, 11:06:00 AM »
(a) P(z)=z3+2z−3−i=(z-1)3+3(z-1)2+5(z-1)-i
at |z−1|=1, |5(z−1)|≥|(z-1)3+3(z-1)2-i|
so at |z−1|<1, |5(z−1)| and P(z) have the same number of roots which is 1 (z=1).

(b) At |z|=2, |z3|≥|2z-3-i|
so at |z|<2, |z3| and P(z) have the same number of roots which is 3 (z=0 has the multiplicity of 3).
but we know |z−1|<1 is in |z|<2, z≠1, so at |z−1|>1,|z|<2, P(z) has the number of roots of 3-1=2.

(c) Because P(z) has the highest power of 3, so it has in all 3 roots, but we already find 3 roots at |z|<2,
so at |z|>2, the number of roots is 0.

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##### End of Semester Bonus--sample problem for FE / Re: FE Sample--Problem 3
« on: November 27, 2018, 01:17:09 PM »
f(z)=(cos(z/6)+zsinz)/(sinz)^2

case 1: when z=3pi+6k*pi (where k is an integer), they have poles of order 1 because numerator has order 1 and denominator has order 2.

case 2: z=k*pi (where k is an integer and does not equal to 3+6n where n is an integer) have poles of order 2

let w=1/z

f(w)=(wcos(6/w)+sin(1/w))/w(sin(1/w))^2

when w approach 0,

f(w)=1/(wsin(1/w)) so f(w) approach infinity

so at infinity it is a pole

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