Question #232174

Find the residue of the function f(z)=1/z(ez-1)


1
Expert's answer
2021-09-02T07:25:25-0400

 If f(z)f(z) has a pole of order kk at z=z0z=z_0 then


rez(f,z0)=limzz0(1(k1)!dk1dzk1((zz0)kf(z)))rez(f,z_0)=\lim\limits_{z\to z_0}\big(\dfrac{1}{(k-1)!}\dfrac{d^{k-1}}{dz^{k-1}}((z-z_0)^kf(z))\big)

ez=1+z1!+z22!+z33!+...e^z=1+\dfrac{z}{1!}+\dfrac{z^2}{2!}+\dfrac{z^3}{3!}+...

ez1=z1!+z22!+z33!+...e^z-1=\dfrac{z}{1!}+\dfrac{z^2}{2!}+\dfrac{z^3}{3!}+...

limz0(zf(z))=limz0(z(1z(ez1)))=limz0(1ez1)\lim\limits_{z\to0}(zf(z))=\lim\limits_{z\to0}(z(\dfrac{1}{z(e^z-1)}))=\lim\limits_{z\to0}(\dfrac{1}{e^z-1})

=does not exist=does\ not \ exist

1(21)!d21dz21((z0)2f(z))\dfrac{1}{(2-1)!}\dfrac{d^{2-1}}{dz^{2-1}}((z-0)^2f(z))

=ddz(z2(1z(ez1)))=\dfrac{d}{dz}(z^2(\dfrac{1}{z(e^z-1)}))

=ddz(zez1)=\dfrac{d}{dz}(\dfrac{z}{e^z-1})

=ez1zez(ez1)2=\dfrac{e^z-1-ze^z}{(e^z-1)^2}


rez(f,0)=limz0(ez1zez(ez1)2)rez(f,0)=\lim\limits_{z\to0}\big(\dfrac{e^z-1-ze^z}{(e^z-1)^2}\big)

=limz0(ezezzez2ez(ez1))=\lim\limits_{z\to0}\big(\dfrac{e^z-e^z-ze^z}{2e^z(e^z-1)}\big)

=limz0(z2(ez1))=\lim\limits_{z\to0}\big(\dfrac{-z}{2(e^z-1)}\big)

=limz0(12ez)=12=\lim\limits_{z\to0}\big(\dfrac{-1}{2e^z}\big)=-\dfrac{1}{2}


rez(f,0)=12rez(f,0)=-\dfrac{1}{2}


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Comments

Fozia Sayda
04.09.21, 09:19

Thank you so much for your answer.

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