Question Number 148174 by tabata last updated on 25/Jul/21

find the residue of f(z)=((sin(z))/(cos(z^3 )−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))

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)).