Answer to Question #232175 in Complex Analysis for Fozia Sayda

Question #232175

Determine the poles and the residue at each pole of the function f(z)=1/(z-1)2(z+2)


1
Expert's answer
2021-09-02T07:25:28-0400

 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)

z=2,z=-2, simple pole

rez(f,2)=limz2((z+2)f(z))rez(f,-2)=\lim\limits_{z\to-2}((z+2)f(z))

=limz2((z+2)(1(z+2)(z1)2))=\lim\limits_{z\to-2}((z+2)(\dfrac{1}{(z+2)(z-1)^2}))

=limz2(1(z1)2)=1(21)2=19=\lim\limits_{z\to-2}(\dfrac{1}{(z-1)^2})=\dfrac{1}{(-2-1)^2}=\dfrac{1}{9}

z=1,z=1, order 22 pole

rez(f,1)=limz1(1(21)!d21dz21((z1)2f(z)))rez(f,1)=\lim\limits_{z\to 1}\big(\dfrac{1}{(2-1)!}\dfrac{d^{2-1}}{dz^{2-1}}((z-1)^2f(z))\big)

=limz1(ddz((z1)2(1(z+2)(z1)2)))=\lim\limits_{z\to 1}\big(\dfrac{d}{dz}((z-1)^2(\dfrac{1}{(z+2)(z-1)^2}))\big)

=limz1(ddz(1z+2))=\lim\limits_{z\to 1}\big(\dfrac{d}{dz}(\dfrac{1}{z+2})\big)

=limz1(1(z+2)2)=\lim\limits_{z\to 1}\big(-\dfrac{1}{(z+2)^2}\big)

=1(1+2)2=19=-\dfrac{1}{(1+2)^2}=-\dfrac{1}{9}


z=2,z=-2, simple pole


rez(f,2)=19rez(f,-2)=\dfrac{1}{9}



z=1,z=1, order 22 pole


rez(f,1)=19rez(f,1)=-\dfrac{1}{9}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment