Mathematics Asked by Dimitri on December 23, 2021
As mentionned in the title, I’d like to get the function’s Laurent series and after its residue, I have tried to separate the two denominators to get a partial fraction but I still have a z at numerator that’s bothering me …
So far : $$
frac{z}{(z-1)(z-3)} = frac{-z}{2(z-1)} + frac{3z}{(z-3)}
$$
I can obviously try $omega = z-3$ and change my variable but I don’t feel it’ll will work … Because I will still have this z on top of my fraction …
Could you help me ?
Thank you very much 😉
$$frac1{z-1}=frac1{z-3+2}=frac12frac1{1+frac{z-3}2}=frac12left(1-frac{z-3}2+frac{(z-3)^2}4-ldotsright)=$$
$$=frac12-frac{z-3}4+ldots$$
So
$$frac z{(z-1)(z-3)}=frac12left(frac3{z-3}-frac1{z-1}right)=frac32frac1{z-3}-frac14+frac{z-3}8+ldots$$
so the residue is $,3/2,$ .
Answered by DonAntonio on December 23, 2021
Hint: $$ frac{3z}{z - 3} = frac{3(z - 3) + 9}{z - 3} $$ begin{align*} frac{-z}{2(z - 1)} &= frac{1}{2}frac{-(z - 3) - 3}{(z - 3) + 2}\ &= -frac{1}{2}frac{z - 3}{(z - 3) + 2} - frac{1}{2}frac{3}{(z - 3) + 2}\ &= -frac{z - 3}{4}cdotfrac{1}{(z - 3)/2 + 1} - frac{3}{4}cdotfrac{1}{(z - 3)/2 + 1} end{align*} I think you should be able to expand these with the help of well-known power series...
Answered by Stahl on December 23, 2021
$z=3$ is a simple pole so $Res(f)_{z=3}=lim_{zto 3} f(z)={3over 2}$
Answered by Marso on December 23, 2021
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