How do I find the region of convergence of this infinite series?
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I am given the following question :
Let f(z) = 1/(1+z) [z is a complex number]
(a) Expand f(z) about z=1.
(b) Find the region of convergence.
I am able to solve part (a) using Taylor series expansion, but how should I proceed with part (b)?
Our professor said that the region of convergence is a circle with radius |z-1|<2. How do I arrive at this?
complex-analysis power-series
edited Nov 17 at 7:44
José Carlos Santos
142k20111207
asked Nov 17 at 7:36
Jasmine
111
add a comment |
up vote
0
down vote
favorite
I am given the following question :
Let f(z) = 1/(1+z) [z is a complex number]
(a) Expand f(z) about z=1.
(b) Find the region of convergence.
I am able to solve part (a) using Taylor series expansion, but how should I proceed with part (b)?
Our professor said that the region of convergence is a circle with radius |z-1|<2. How do I arrive at this?
complex-analysis power-series
edited Nov 17 at 7:44
José Carlos Santos
142k20111207
asked Nov 17 at 7:36
Jasmine
111
$1+z=2+(z-1)=2(1+(z-1)/2)$, so when you expand you need $|(z-1)/2|<1$
– Yuriy S
Nov 17 at 7:40
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am given the following question :
Let f(z) = 1/(1+z) [z is a complex number]
(a) Expand f(z) about z=1.
(b) Find the region of convergence.
I am able to solve part (a) using Taylor series expansion, but how should I proceed with part (b)?
Our professor said that the region of convergence is a circle with radius |z-1|<2. How do I arrive at this?
complex-analysis power-series
edited Nov 17 at 7:44
José Carlos Santos
142k20111207
asked Nov 17 at 7:36
Jasmine
111
I am given the following question :
Let f(z) = 1/(1+z) [z is a complex number]
(a) Expand f(z) about z=1.
(b) Find the region of convergence.
I am able to solve part (a) using Taylor series expansion, but how should I proceed with part (b)?
Our professor said that the region of convergence is a circle with radius |z-1|<2. How do I arrive at this?
complex-analysis power-series
complex-analysis power-series
edited Nov 17 at 7:44
José Carlos Santos
142k20111207
asked Nov 17 at 7:36
Jasmine
111
edited Nov 17 at 7:44
José Carlos Santos
142k20111207
asked Nov 17 at 7:36
Jasmine
111
edited Nov 17 at 7:44
José Carlos Santos
142k20111207
edited Nov 17 at 7:44
José Carlos Santos
142k20111207
edited Nov 17 at 7:44
José Carlos Santos
142k20111207
142k20111207
asked Nov 17 at 7:36
Jasmine
111
asked Nov 17 at 7:36
Jasmine
111
asked Nov 17 at 7:36
Jasmine
111
111
$1+z=2+(z-1)=2(1+(z-1)/2)$, so when you expand you need $|(z-1)/2|<1$
– Yuriy S
Nov 17 at 7:40
add a comment |
$1+z=2+(z-1)=2(1+(z-1)/2)$, so when you expand you need $|(z-1)/2|<1$
– Yuriy S
Nov 17 at 7:40
$1+z=2+(z-1)=2(1+(z-1)/2)$, so when you expand you need $|(z-1)/2|<1$
– Yuriy S
Nov 17 at 7:40
$1+z=2+(z-1)=2(1+(z-1)/2)$, so when you expand you need $|(z-1)/2|<1$
– Yuriy S
Nov 17 at 7:40
add a comment |
1 Answer
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Because$$frac1{1+z}=frac1{2+(z-1)}=frac12sum_{n=0}^inftyfrac{(-1)^n}{2^n}(z-1)^ntag1$$when $lvert z-1rvert<2$. Besides, the radius of convergence of $(1)$ is $2$.
answered Nov 17 at 7:43
José Carlos Santos
142k20111207
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Because$$frac1{1+z}=frac1{2+(z-1)}=frac12sum_{n=0}^inftyfrac{(-1)^n}{2^n}(z-1)^ntag1$$when $lvert z-1rvert<2$. Besides, the radius of convergence of $(1)$ is $2$.
answered Nov 17 at 7:43
José Carlos Santos
142k20111207
add a comment |
up vote
0
down vote
Because$$frac1{1+z}=frac1{2+(z-1)}=frac12sum_{n=0}^inftyfrac{(-1)^n}{2^n}(z-1)^ntag1$$when $lvert z-1rvert<2$. Besides, the radius of convergence of $(1)$ is $2$.
answered Nov 17 at 7:43
José Carlos Santos
142k20111207
add a comment |
up vote
0
down vote
up vote
0
down vote
Because$$frac1{1+z}=frac1{2+(z-1)}=frac12sum_{n=0}^inftyfrac{(-1)^n}{2^n}(z-1)^ntag1$$when $lvert z-1rvert<2$. Besides, the radius of convergence of $(1)$ is $2$.
answered Nov 17 at 7:43
José Carlos Santos
142k20111207
Because$$frac1{1+z}=frac1{2+(z-1)}=frac12sum_{n=0}^inftyfrac{(-1)^n}{2^n}(z-1)^ntag1$$when $lvert z-1rvert<2$. Besides, the radius of convergence of $(1)$ is $2$.
answered Nov 17 at 7:43
José Carlos Santos
142k20111207
answered Nov 17 at 7:43
José Carlos Santos
142k20111207
answered Nov 17 at 7:43
José Carlos Santos
142k20111207
answered Nov 17 at 7:43
José Carlos Santos
142k20111207
142k20111207
add a comment |
add a comment |
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$1+z=2+(z-1)=2(1+(z-1)/2)$, so when you expand you need $|(z-1)/2|<1$
– Yuriy S
Nov 17 at 7:40