Laurent series $f(z)=frac{sqrt{z}}{z+i}$
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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.
complex-analysis proof-verification alternative-proof laurent-series
edited Nov 23 at 17:45
Martin Sleziak
44.6k7115270
asked Nov 15 at 7:27
Yadati Kiran
1,634519
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.
complex-analysis proof-verification alternative-proof laurent-series
edited Nov 23 at 17:45
Martin Sleziak
44.6k7115270
asked Nov 15 at 7:27
Yadati Kiran
1,634519
2
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
Nov 15 at 7:29
Oh yeah! I get it. Thank you Sir @Kavi Rama Murthy
– Yadati Kiran
Nov 15 at 7:31
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.
complex-analysis proof-verification alternative-proof laurent-series
edited Nov 23 at 17:45
Martin Sleziak
44.6k7115270
asked Nov 15 at 7:27
Yadati Kiran
1,634519
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.
complex-analysis proof-verification alternative-proof laurent-series
complex-analysis proof-verification alternative-proof laurent-series
edited Nov 23 at 17:45
Martin Sleziak
44.6k7115270
asked Nov 15 at 7:27
Yadati Kiran
1,634519
edited Nov 23 at 17:45
Martin Sleziak
44.6k7115270
asked Nov 15 at 7:27
Yadati Kiran
1,634519
edited Nov 23 at 17:45
Martin Sleziak
44.6k7115270
edited Nov 23 at 17:45
Martin Sleziak
44.6k7115270
edited Nov 23 at 17:45
Martin Sleziak
44.6k7115270
44.6k7115270
asked Nov 15 at 7:27
Yadati Kiran
1,634519
asked Nov 15 at 7:27
Yadati Kiran
1,634519
asked Nov 15 at 7:27
Yadati Kiran
1,634519
1,634519
2
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
Nov 15 at 7:29
Oh yeah! I get it. Thank you Sir @Kavi Rama Murthy
– Yadati Kiran
Nov 15 at 7:31
add a comment |
2
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
Nov 15 at 7:29
Oh yeah! I get it. Thank you Sir @Kavi Rama Murthy
– Yadati Kiran
Nov 15 at 7:31
2
2
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
Nov 15 at 7:29
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
Nov 15 at 7:29
Oh yeah! I get it. Thank you Sir @Kavi Rama Murthy
– Yadati Kiran
Nov 15 at 7:31
Oh yeah! I get it. Thank you Sir @Kavi Rama Murthy
– Yadati Kiran
Nov 15 at 7:31
add a comment |
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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
Nov 15 at 7:29
Oh yeah! I get it. Thank you Sir @Kavi Rama Murthy
– Yadati Kiran
Nov 15 at 7:31