Laurent series $f(z)=dfrac{sqrt{z}}{z+i}$
up vote
0
down vote
favorite
I came across to calculate an integral for which I had to find the Laurent series of $f(z)=dfrac{sqrt{z}}{z+i}$.
I see that at $z=-i$ the function has a simple pole and the Laurent series I got is as follows:
for $0<|z|<1$ we have $dfrac{sqrt{z}}{i},left(dfrac{1}{1+dfrac{z}{i}} right)=dfrac{sqrt{z}}{i},left(1-dfrac{z}{i}+dfrac{z^2}{i^2}-cdots right)$
and for $|z|>1$ we have $dfrac{sqrt{z}}{z},left(dfrac{1}{1+dfrac{i}{z}} right)=dfrac{sqrt{z}}{z},left(1-dfrac{i}{z}+dfrac{i^2}{z^2}-cdots right)$.
Thus $f(z)=dfrac{sqrt{z}}{i},left(1-dfrac{z}{i}+dfrac{z^2}{i^2}-cdots right)+dfrac{sqrt{z}}{z},left(1-dfrac{i}{z}+dfrac{i^2}{z^2}-cdots right)$.
I am not particularly sure about the answer. Any suggestions or hints would be appreciated.
proof-verification alternative-proof laurent-series
add a comment |
up vote
0
down vote
favorite
I came across to calculate an integral for which I had to find the Laurent series of $f(z)=dfrac{sqrt{z}}{z+i}$.
I see that at $z=-i$ the function has a simple pole and the Laurent series I got is as follows:
for $0<|z|<1$ we have $dfrac{sqrt{z}}{i},left(dfrac{1}{1+dfrac{z}{i}} right)=dfrac{sqrt{z}}{i},left(1-dfrac{z}{i}+dfrac{z^2}{i^2}-cdots right)$
and for $|z|>1$ we have $dfrac{sqrt{z}}{z},left(dfrac{1}{1+dfrac{i}{z}} right)=dfrac{sqrt{z}}{z},left(1-dfrac{i}{z}+dfrac{i^2}{z^2}-cdots right)$.
Thus $f(z)=dfrac{sqrt{z}}{i},left(1-dfrac{z}{i}+dfrac{z^2}{i^2}-cdots right)+dfrac{sqrt{z}}{z},left(1-dfrac{i}{z}+dfrac{i^2}{z^2}-cdots right)$.
I am not particularly sure about the answer. Any suggestions or hints would be appreciated.
proof-verification alternative-proof laurent-series
1
There is no continuous branch of $sqrt z$ in a neighborhood of $0$ so the question does not make sense. Besides you cannot have $sqrt z$ in the terms of any Laurent series.
– Kavi Rama Murthy
yesterday
Oh yeah! I get it. Thank you Sir @Kavi Rama Murthy
– Yadati Kiran
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I came across to calculate an integral for which I had to find the Laurent series of $f(z)=dfrac{sqrt{z}}{z+i}$.
I see that at $z=-i$ the function has a simple pole and the Laurent series I got is as follows:
for $0<|z|<1$ we have $dfrac{sqrt{z}}{i},left(dfrac{1}{1+dfrac{z}{i}} right)=dfrac{sqrt{z}}{i},left(1-dfrac{z}{i}+dfrac{z^2}{i^2}-cdots right)$
and for $|z|>1$ we have $dfrac{sqrt{z}}{z},left(dfrac{1}{1+dfrac{i}{z}} right)=dfrac{sqrt{z}}{z},left(1-dfrac{i}{z}+dfrac{i^2}{z^2}-cdots right)$.
Thus $f(z)=dfrac{sqrt{z}}{i},left(1-dfrac{z}{i}+dfrac{z^2}{i^2}-cdots right)+dfrac{sqrt{z}}{z},left(1-dfrac{i}{z}+dfrac{i^2}{z^2}-cdots right)$.
I am not particularly sure about the answer. Any suggestions or hints would be appreciated.
proof-verification alternative-proof laurent-series
I came across to calculate an integral for which I had to find the Laurent series of $f(z)=dfrac{sqrt{z}}{z+i}$.
I see that at $z=-i$ the function has a simple pole and the Laurent series I got is as follows:
for $0<|z|<1$ we have $dfrac{sqrt{z}}{i},left(dfrac{1}{1+dfrac{z}{i}} right)=dfrac{sqrt{z}}{i},left(1-dfrac{z}{i}+dfrac{z^2}{i^2}-cdots right)$
and for $|z|>1$ we have $dfrac{sqrt{z}}{z},left(dfrac{1}{1+dfrac{i}{z}} right)=dfrac{sqrt{z}}{z},left(1-dfrac{i}{z}+dfrac{i^2}{z^2}-cdots right)$.
Thus $f(z)=dfrac{sqrt{z}}{i},left(1-dfrac{z}{i}+dfrac{z^2}{i^2}-cdots right)+dfrac{sqrt{z}}{z},left(1-dfrac{i}{z}+dfrac{i^2}{z^2}-cdots right)$.
I am not particularly sure about the answer. Any suggestions or hints would be appreciated.
proof-verification alternative-proof laurent-series
proof-verification alternative-proof laurent-series
asked yesterday
Yadati Kiran
240110
240110
1
There is no continuous branch of $sqrt z$ in a neighborhood of $0$ so the question does not make sense. Besides you cannot have $sqrt z$ in the terms of any Laurent series.
– Kavi Rama Murthy
yesterday
Oh yeah! I get it. Thank you Sir @Kavi Rama Murthy
– Yadati Kiran
yesterday
add a comment |
1
There is no continuous branch of $sqrt z$ in a neighborhood of $0$ so the question does not make sense. Besides you cannot have $sqrt z$ in the terms of any Laurent series.
– Kavi Rama Murthy
yesterday
Oh yeah! I get it. Thank you Sir @Kavi Rama Murthy
– Yadati Kiran
yesterday
1
1
There is no continuous branch of $sqrt z$ in a neighborhood of $0$ so the question does not make sense. Besides you cannot have $sqrt z$ in the terms of any Laurent series.
– Kavi Rama Murthy
yesterday
There is no continuous branch of $sqrt z$ in a neighborhood of $0$ so the question does not make sense. Besides you cannot have $sqrt z$ in the terms of any Laurent series.
– Kavi Rama Murthy
yesterday
Oh yeah! I get it. Thank you Sir @Kavi Rama Murthy
– Yadati Kiran
yesterday
Oh yeah! I get it. Thank you Sir @Kavi Rama Murthy
– Yadati Kiran
yesterday
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999362%2flaurent-series-fz-dfrac-sqrtzzi%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
There is no continuous branch of $sqrt z$ in a neighborhood of $0$ so the question does not make sense. Besides you cannot have $sqrt z$ in the terms of any Laurent series.
– Kavi Rama Murthy
yesterday
Oh yeah! I get it. Thank you Sir @Kavi Rama Murthy
– Yadati Kiran
yesterday