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Question Number 148174 by tabata last updated on 25/Jul/21
find the residue of f(z)=((sin(z))/(cos(z^3 )−1))
findtheresidueoff(z)=sin(z)cos(z3)−1
Answered by mathmax by abdo last updated on 25/Jul/21
cosu∼1−(u^2 /2) ⇒cos(z^3 )∼1−(z^6 /2) ⇒cos(z^3 )−1∼−(z^6 /2) ⇒ f(z)∼−2((sinz)/z^6 ) we have sinz =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!))z^(2n+1) =z−(z^3 /(3!))+(z^5 /(5!))−(z^7 /(7!)) ⇒f(z)∼−(2/z^6 )(z−(z^3 /(3!))+(z^5 /(5!))−(z^7 /(7!))+...) ⇒f(z)∼−(2/z^5 )+(2/(3!z^3 ))−(2/(5!z))+... ⇒ Res(f)=−(2/(5.4.3.2))=−(1/(60))
cosu∼1−u22⇒cos(z3)∼1−z62⇒cos(z3)−1∼−z62⇒f(z)∼−2sinzz6wehavesinz=∑n=0∞(−1)n(2n+1)!z2n+1=z−z33!+z55!−z77!⇒f(z)∼−2z6(z−z33!+z55!−z77!+…)⇒f(z)∼−2z5+23!z3−25!z+…⇒Res(f)=−25.4.3.2=−160
Answered by Olaf_Thorendsen last updated on 25/Jul/21
f(z) = ((sinz)/(cos(z^3 )−1)) f(z) = ((sinz)/((1−(z^6 /2)+(z^(12) /(24))−(z^(18) /(720))...)−1)) f(z) = ((sinz)/(−(z^6 /2)+(z^(12) /(24))−(z^(18) /(720))+...)) f(z) = −(2/z^6 ).((sinz)/(1−((z^6 /(12))−(z^(12) /(360))+...))) f(z) = −((2sinz)/z^6 )[1+((z^6 /(12))−(z^(12) /(360))+...)+((z^6 /(12))−(z^(12) /(360))+...)^2 +...] f(z) = −((2(z−(z^3 /6)+(z^5 /(120))...))/z^6 )[1+((z^6 /(12))−(z^(12) /(360))+...)+((z^6 /(12))−(z^(12) /(360))+...)^2 +...] f(z) = −(2/z^5 )+(1/(3z^3 ))−(1/(60z))−((421)/(2520)) z+.. The term in z^(−1) is −(1/(60)) ⇒The residue is −(1/(60)).
f(z)=sinzcos(z3)−1f(z)=sinz(1−z62+z1224−z18720…)−1f(z)=sinz−z62+z1224−z18720+…f(z)=−2z6.sinz1−(z612−z12360+…)f(z)=−2sinzz6[1+(z612−z12360+…)+(z612−z12360+…)2+…]f(z)=−2(z−z36+z5120…)z6[1+(z612−z12360+…)+(z612−z12360+…)2+…]f(z)=−2z5+13z3−160z−4212520z+..Theterminz−1is−160⇒Theresidueis−160.

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