Laurent series $f(z)=dfrac{sqrt{z}}{z+i}$











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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.






proof-verification alternative-proof laurent-series







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  • 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















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







share|cite|improve this question


















  • 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













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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









Yadati Kiran

240110




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  • 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




    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















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