- #1
Kiefer
- 6
- 0
Use geometric series to find the Laurent series for f (z) = z / (z - 1)(z - 2) in each annulus
(a) Ann(1,0,1)
(b) Ann(1,1,∞)
Ann(a,r,R)
a= center, r=smaller radius, R=larger radius
Ann(1,0,1)=D(1,1)\{0}
My attempt:
f(z)= -1/(z-1) + 2/(z-2)
geometric series: Σ[n=0 to inf] z^n - 1/2 Σ[n=0 to inf] (z/2)^n
a)0<|z-1|<1
2/(z-2)=-2(1/(2-z))=-2(1/(1-(z-1)))=-2Σ[n=0 to inf](z-1)^n
f(z)=-1/(z-1)-2Σ[n=0 to inf](z-1)^n=-(z-1)^(-1)+Σ[n=0 to inf]-2(z-1)^n
b)1<|z-1|<inf
(1/|z-1|)<1
-1/(z-1)=-Σ[n=0 to inf](1/(z-1))^n
2/(z-2)=2/(z-1-1)=2(1/(z-1))(1/(1-1/(z-1)))=2/(z-1)Σ[n=0 to inf](1/(z-1))^n=2Σ[n=0 to inf](z-1)^-(n+1)
f(z)=-Σ[n=0 to inf](1/(z-1))^n+Σ[n=-inf to -2]2(z-1)^-(n+1)
Are my geometric series correct? (I feel good about the second one but am unsure on the first)
Are my summation bounds in the Laurent series correct? I did something funky in the last one, so I'm not sure. Also, I wasn't sure how to find a Laurent series for -1/(z-1) in part a, so I left it as shown.
Any help and corrections are appreciated!
(a) Ann(1,0,1)
(b) Ann(1,1,∞)
Ann(a,r,R)
a= center, r=smaller radius, R=larger radius
Ann(1,0,1)=D(1,1)\{0}
My attempt:
f(z)= -1/(z-1) + 2/(z-2)
geometric series: Σ[n=0 to inf] z^n - 1/2 Σ[n=0 to inf] (z/2)^n
a)0<|z-1|<1
2/(z-2)=-2(1/(2-z))=-2(1/(1-(z-1)))=-2Σ[n=0 to inf](z-1)^n
f(z)=-1/(z-1)-2Σ[n=0 to inf](z-1)^n=-(z-1)^(-1)+Σ[n=0 to inf]-2(z-1)^n
b)1<|z-1|<inf
(1/|z-1|)<1
-1/(z-1)=-Σ[n=0 to inf](1/(z-1))^n
2/(z-2)=2/(z-1-1)=2(1/(z-1))(1/(1-1/(z-1)))=2/(z-1)Σ[n=0 to inf](1/(z-1))^n=2Σ[n=0 to inf](z-1)^-(n+1)
f(z)=-Σ[n=0 to inf](1/(z-1))^n+Σ[n=-inf to -2]2(z-1)^-(n+1)
Are my geometric series correct? (I feel good about the second one but am unsure on the first)
Are my summation bounds in the Laurent series correct? I did something funky in the last one, so I'm not sure. Also, I wasn't sure how to find a Laurent series for -1/(z-1) in part a, so I left it as shown.
Any help and corrections are appreciated!