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let-f-z-1-z-2-e-2z-z-3-calculate-Res-f-0-




Question Number 37300 by math khazana by abdo last updated on 11/Jun/18
let f(z)=(((1−z^2 )e^(2z) )/z^3 )  calculate Res(f, 0)
letf(z)=(1z2)e2zz3calculateRes(f,0)
Commented by prof Abdo imad last updated on 15/Jun/18
0 is a triple pole of f  so  Res(f,0) =lim_(z→0)   (1/((3−1)!)){z^3  f(z)}^((2))   =lim_(z→0)  (1/2){ (1−z^2 )e^(2z) }^((2) ) but  {(1−z^2 )^  e^(2z) }^((1)) = −2z e^(2z)   +2(1−z^2 )e^(2z)   =(−2z +2−2z^2 )e^(2z)   ⇒  {(1−z^2 )e^(2z) }^((2)) =(−4z −2)e^(2z)   +2(−2z+2−2z^2 )e^(2z)   =(−4z −2 −4z +4 −4z^2 )e^(2z)   =(−4z^2  −8z +2)e^(2z)  ⇒  Res(ϕ,0) = (1/2) 2 =1 .
0isatriplepoleoffsoRes(f,0)=limz01(31)!{z3f(z)}(2)=limz012{(1z2)e2z}(2)but{(1z2)e2z}(1)=2ze2z+2(1z2)e2z=(2z+22z2)e2z{(1z2)e2z}(2)=(4z2)e2z+2(2z+22z2)e2z=(4z24z+44z2)e2z=(4z28z+2)e2zRes(φ,0)=122=1.

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