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let-f-z-sin-z-z-2-calculate-Res-f-0-




Question Number 63031 by mathmax by abdo last updated on 28/Jun/19
let f(z) =((sin(z))/z^2 )  calculate Res(f,0)
letf(z)=sin(z)z2calculateRes(f,0)
Commented by mathmax by abdo last updated on 28/Jun/19
0 is a double pole ⇒Res(f,0) =lim_(z→0)    (1/((2−1)!)){z^2 f(z)}^((1))   =lim_(z→0)  {sinz}^((1))  =lim_(z→0)  cosz =1
0isadoublepoleRes(f,0)=limz01(21)!{z2f(z)}(1)=limz0{sinz}(1)=limz0cosz=1
Commented by mathmax by abdo last updated on 28/Jun/19
another way if f(z) =Σ_(k=0) ^∞  a_k (z−z_0 )^k  +Σ_(k=1) ^∞   a_(−k) (z−z_0 )^(−k)   (tbe laurent serie of f(z))  → Res(f,z_0 ) =a_(−1)  =coefficient of (1/(z−z_o )) in this serie so we have  sin(z) =Σ_(n=0) ^∞  (((−1)^n  z^(2n+1) )/((2n+1)!)) =z−(z^3 /(3!)) +(z^5 /(5!)) −.... ⇒  ((sinz)/z^2 ) =(1/z) −(z/(3!)) +(z^3 /(5!)) −.... ⇒Re(f,0) =1   (coefficent of (1/z)).
anotherwayiff(z)=k=0ak(zz0)k+k=1ak(zz0)k(tbelaurentserieoff(z))Res(f,z0)=a1=coefficientof1zzointhisseriesowehavesin(z)=n=0(1)nz2n+1(2n+1)!=zz33!+z55!.sinzz2=1zz3!+z35!.Re(f,0)=1(coefficentof1z).

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