Question about integrating a Laurent series












0














I want to expand the function $ f(z) = frac {z-1}{z^2 -2z -3} $



in $ 0 < |z+1| < 4$



Then I want to use the result to evaluate this integral $ int_C frac {z-1}{(z+1)(z-3)} dz $



in $ C : |z+1|=2$



Now I approach this in 2 different ways, by extracting the residue from the expansion or by integrating the whole expansion



Laurent expansion:



$ f(z) = frac {z-1}{z^2 -2z -3} = frac {z-1}{(z+1)(z-3)} = (frac {z-1}{z+1} )(frac {1}{z-3}) = frac {z-1}{z+1} (frac {1}{(z+1)-4}) = frac {z-1}{z+1} (frac{-1}{4-(z+1)}) = frac {1-z}{4(z+1)} sum_{n=0}^{infty} (frac {z+1}{4})^n = frac {1-z}{4(z+1)} [1 + frac {z+1}{4} + frac {(z+1)^2}{16} + frac {(z+1)^3}{64} + ..... ] = frac {1-z}{4(z+1)} + frac {1-z}{16} + frac{(1-z)(z+1)}{64} + frac {(1-z)(z+1)^2}{256} + ..... $



Now I use 2 approaches, first one is to take the residue and work out the integral using the residue theorem and the other is to just integrate the whole expansion



Residue theorem approach:



From the expansion, the residue for $ C : |z+1|=2$ is equal to $ frac{1}{2}$



Then, $int_C frac {z-1}{(z+1)(z-3)} dz = 2 pi i sum Res = 2 pi i ( frac {1}{2}) = pi i$



Now the second approach:



$int_C frac {z-1}{(z+1)(z-3)} dz = int_C frac {1-z}{4(z+1)} sum_{n=0}^{infty} (frac {z+1}{4})^n dz$



$ C: |z+1| = 2$ then, $ z+ 1 = 2 e^{i theta}$ , $ dz = 2 i z d theta$



$ int_C frac {z-1}{(z+1)(z-3)} dz = int_C frac {1-z}{4(z+1)} + frac {1-z}{16} + frac{(1-z)(z+1)}{64} + frac {(1-z)(z+1)^2}{256} + ..... dz = 2i int_0^{2 pi} frac{1-e^{i theta}}{4} + frac {(1-e^{i theta})(e^{i theta})}{8} + frac {(1-e^{i theta})(e^{2i theta})}{16} + frac {(1-e^{i theta})(e^{3i theta})}{32} + .... d theta$



= $2i [ frac {1}{4} theta - frac {e^{i theta}}{4i} + frac {e^{i theta}}{8i} - frac {e^{2i theta}}{16i} + frac {e^{2i theta}}{32i} - frac {e^{3i theta}}{48i} + frac {e^{3i theta}}{98i} - frac {e^{4i theta}}{128i} + ..... ]$ as $ theta $ goes from $0$ to $ 2 pi$



Finally $ = 2i [ frac { pi}{2} - frac {1}{8i} - frac {1}{32i} - ... ] = pi i - frac {1}{4} - frac {1}{16} - ....$



Now the problem is that with the residue theorem I got the answer $ pi i $ , but by integrating the whole expansion, keeping in mind that the radius is $2$ yields $ pi i $ minus a bunch of terms, is there a reason why?










share|cite|improve this question






















  • What you computed is not the Laurent series (because the coefficients depend on $z$).
    – Martin R
    Jan 4 at 13:52










  • How can I do it right if I may ask?
    – khaled014z
    Jan 4 at 15:45










  • Use $z-1=(z+1)-2$ and split the sum in two. Then simplify and combine terms with the same power of $z+1$
    – N74
    Jan 4 at 18:33
















0














I want to expand the function $ f(z) = frac {z-1}{z^2 -2z -3} $



in $ 0 < |z+1| < 4$



Then I want to use the result to evaluate this integral $ int_C frac {z-1}{(z+1)(z-3)} dz $



in $ C : |z+1|=2$



Now I approach this in 2 different ways, by extracting the residue from the expansion or by integrating the whole expansion



Laurent expansion:



$ f(z) = frac {z-1}{z^2 -2z -3} = frac {z-1}{(z+1)(z-3)} = (frac {z-1}{z+1} )(frac {1}{z-3}) = frac {z-1}{z+1} (frac {1}{(z+1)-4}) = frac {z-1}{z+1} (frac{-1}{4-(z+1)}) = frac {1-z}{4(z+1)} sum_{n=0}^{infty} (frac {z+1}{4})^n = frac {1-z}{4(z+1)} [1 + frac {z+1}{4} + frac {(z+1)^2}{16} + frac {(z+1)^3}{64} + ..... ] = frac {1-z}{4(z+1)} + frac {1-z}{16} + frac{(1-z)(z+1)}{64} + frac {(1-z)(z+1)^2}{256} + ..... $



Now I use 2 approaches, first one is to take the residue and work out the integral using the residue theorem and the other is to just integrate the whole expansion



Residue theorem approach:



From the expansion, the residue for $ C : |z+1|=2$ is equal to $ frac{1}{2}$



Then, $int_C frac {z-1}{(z+1)(z-3)} dz = 2 pi i sum Res = 2 pi i ( frac {1}{2}) = pi i$



Now the second approach:



$int_C frac {z-1}{(z+1)(z-3)} dz = int_C frac {1-z}{4(z+1)} sum_{n=0}^{infty} (frac {z+1}{4})^n dz$



$ C: |z+1| = 2$ then, $ z+ 1 = 2 e^{i theta}$ , $ dz = 2 i z d theta$



$ int_C frac {z-1}{(z+1)(z-3)} dz = int_C frac {1-z}{4(z+1)} + frac {1-z}{16} + frac{(1-z)(z+1)}{64} + frac {(1-z)(z+1)^2}{256} + ..... dz = 2i int_0^{2 pi} frac{1-e^{i theta}}{4} + frac {(1-e^{i theta})(e^{i theta})}{8} + frac {(1-e^{i theta})(e^{2i theta})}{16} + frac {(1-e^{i theta})(e^{3i theta})}{32} + .... d theta$



= $2i [ frac {1}{4} theta - frac {e^{i theta}}{4i} + frac {e^{i theta}}{8i} - frac {e^{2i theta}}{16i} + frac {e^{2i theta}}{32i} - frac {e^{3i theta}}{48i} + frac {e^{3i theta}}{98i} - frac {e^{4i theta}}{128i} + ..... ]$ as $ theta $ goes from $0$ to $ 2 pi$



Finally $ = 2i [ frac { pi}{2} - frac {1}{8i} - frac {1}{32i} - ... ] = pi i - frac {1}{4} - frac {1}{16} - ....$



Now the problem is that with the residue theorem I got the answer $ pi i $ , but by integrating the whole expansion, keeping in mind that the radius is $2$ yields $ pi i $ minus a bunch of terms, is there a reason why?










share|cite|improve this question






















  • What you computed is not the Laurent series (because the coefficients depend on $z$).
    – Martin R
    Jan 4 at 13:52










  • How can I do it right if I may ask?
    – khaled014z
    Jan 4 at 15:45










  • Use $z-1=(z+1)-2$ and split the sum in two. Then simplify and combine terms with the same power of $z+1$
    – N74
    Jan 4 at 18:33














0












0








0







I want to expand the function $ f(z) = frac {z-1}{z^2 -2z -3} $



in $ 0 < |z+1| < 4$



Then I want to use the result to evaluate this integral $ int_C frac {z-1}{(z+1)(z-3)} dz $



in $ C : |z+1|=2$



Now I approach this in 2 different ways, by extracting the residue from the expansion or by integrating the whole expansion



Laurent expansion:



$ f(z) = frac {z-1}{z^2 -2z -3} = frac {z-1}{(z+1)(z-3)} = (frac {z-1}{z+1} )(frac {1}{z-3}) = frac {z-1}{z+1} (frac {1}{(z+1)-4}) = frac {z-1}{z+1} (frac{-1}{4-(z+1)}) = frac {1-z}{4(z+1)} sum_{n=0}^{infty} (frac {z+1}{4})^n = frac {1-z}{4(z+1)} [1 + frac {z+1}{4} + frac {(z+1)^2}{16} + frac {(z+1)^3}{64} + ..... ] = frac {1-z}{4(z+1)} + frac {1-z}{16} + frac{(1-z)(z+1)}{64} + frac {(1-z)(z+1)^2}{256} + ..... $



Now I use 2 approaches, first one is to take the residue and work out the integral using the residue theorem and the other is to just integrate the whole expansion



Residue theorem approach:



From the expansion, the residue for $ C : |z+1|=2$ is equal to $ frac{1}{2}$



Then, $int_C frac {z-1}{(z+1)(z-3)} dz = 2 pi i sum Res = 2 pi i ( frac {1}{2}) = pi i$



Now the second approach:



$int_C frac {z-1}{(z+1)(z-3)} dz = int_C frac {1-z}{4(z+1)} sum_{n=0}^{infty} (frac {z+1}{4})^n dz$



$ C: |z+1| = 2$ then, $ z+ 1 = 2 e^{i theta}$ , $ dz = 2 i z d theta$



$ int_C frac {z-1}{(z+1)(z-3)} dz = int_C frac {1-z}{4(z+1)} + frac {1-z}{16} + frac{(1-z)(z+1)}{64} + frac {(1-z)(z+1)^2}{256} + ..... dz = 2i int_0^{2 pi} frac{1-e^{i theta}}{4} + frac {(1-e^{i theta})(e^{i theta})}{8} + frac {(1-e^{i theta})(e^{2i theta})}{16} + frac {(1-e^{i theta})(e^{3i theta})}{32} + .... d theta$



= $2i [ frac {1}{4} theta - frac {e^{i theta}}{4i} + frac {e^{i theta}}{8i} - frac {e^{2i theta}}{16i} + frac {e^{2i theta}}{32i} - frac {e^{3i theta}}{48i} + frac {e^{3i theta}}{98i} - frac {e^{4i theta}}{128i} + ..... ]$ as $ theta $ goes from $0$ to $ 2 pi$



Finally $ = 2i [ frac { pi}{2} - frac {1}{8i} - frac {1}{32i} - ... ] = pi i - frac {1}{4} - frac {1}{16} - ....$



Now the problem is that with the residue theorem I got the answer $ pi i $ , but by integrating the whole expansion, keeping in mind that the radius is $2$ yields $ pi i $ minus a bunch of terms, is there a reason why?










share|cite|improve this question













I want to expand the function $ f(z) = frac {z-1}{z^2 -2z -3} $



in $ 0 < |z+1| < 4$



Then I want to use the result to evaluate this integral $ int_C frac {z-1}{(z+1)(z-3)} dz $



in $ C : |z+1|=2$



Now I approach this in 2 different ways, by extracting the residue from the expansion or by integrating the whole expansion



Laurent expansion:



$ f(z) = frac {z-1}{z^2 -2z -3} = frac {z-1}{(z+1)(z-3)} = (frac {z-1}{z+1} )(frac {1}{z-3}) = frac {z-1}{z+1} (frac {1}{(z+1)-4}) = frac {z-1}{z+1} (frac{-1}{4-(z+1)}) = frac {1-z}{4(z+1)} sum_{n=0}^{infty} (frac {z+1}{4})^n = frac {1-z}{4(z+1)} [1 + frac {z+1}{4} + frac {(z+1)^2}{16} + frac {(z+1)^3}{64} + ..... ] = frac {1-z}{4(z+1)} + frac {1-z}{16} + frac{(1-z)(z+1)}{64} + frac {(1-z)(z+1)^2}{256} + ..... $



Now I use 2 approaches, first one is to take the residue and work out the integral using the residue theorem and the other is to just integrate the whole expansion



Residue theorem approach:



From the expansion, the residue for $ C : |z+1|=2$ is equal to $ frac{1}{2}$



Then, $int_C frac {z-1}{(z+1)(z-3)} dz = 2 pi i sum Res = 2 pi i ( frac {1}{2}) = pi i$



Now the second approach:



$int_C frac {z-1}{(z+1)(z-3)} dz = int_C frac {1-z}{4(z+1)} sum_{n=0}^{infty} (frac {z+1}{4})^n dz$



$ C: |z+1| = 2$ then, $ z+ 1 = 2 e^{i theta}$ , $ dz = 2 i z d theta$



$ int_C frac {z-1}{(z+1)(z-3)} dz = int_C frac {1-z}{4(z+1)} + frac {1-z}{16} + frac{(1-z)(z+1)}{64} + frac {(1-z)(z+1)^2}{256} + ..... dz = 2i int_0^{2 pi} frac{1-e^{i theta}}{4} + frac {(1-e^{i theta})(e^{i theta})}{8} + frac {(1-e^{i theta})(e^{2i theta})}{16} + frac {(1-e^{i theta})(e^{3i theta})}{32} + .... d theta$



= $2i [ frac {1}{4} theta - frac {e^{i theta}}{4i} + frac {e^{i theta}}{8i} - frac {e^{2i theta}}{16i} + frac {e^{2i theta}}{32i} - frac {e^{3i theta}}{48i} + frac {e^{3i theta}}{98i} - frac {e^{4i theta}}{128i} + ..... ]$ as $ theta $ goes from $0$ to $ 2 pi$



Finally $ = 2i [ frac { pi}{2} - frac {1}{8i} - frac {1}{32i} - ... ] = pi i - frac {1}{4} - frac {1}{16} - ....$



Now the problem is that with the residue theorem I got the answer $ pi i $ , but by integrating the whole expansion, keeping in mind that the radius is $2$ yields $ pi i $ minus a bunch of terms, is there a reason why?







complex-analysis residue-calculus laurent-series






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asked Jan 4 at 13:26









khaled014zkhaled014z

1107




1107












  • What you computed is not the Laurent series (because the coefficients depend on $z$).
    – Martin R
    Jan 4 at 13:52










  • How can I do it right if I may ask?
    – khaled014z
    Jan 4 at 15:45










  • Use $z-1=(z+1)-2$ and split the sum in two. Then simplify and combine terms with the same power of $z+1$
    – N74
    Jan 4 at 18:33


















  • What you computed is not the Laurent series (because the coefficients depend on $z$).
    – Martin R
    Jan 4 at 13:52










  • How can I do it right if I may ask?
    – khaled014z
    Jan 4 at 15:45










  • Use $z-1=(z+1)-2$ and split the sum in two. Then simplify and combine terms with the same power of $z+1$
    – N74
    Jan 4 at 18:33
















What you computed is not the Laurent series (because the coefficients depend on $z$).
– Martin R
Jan 4 at 13:52




What you computed is not the Laurent series (because the coefficients depend on $z$).
– Martin R
Jan 4 at 13:52












How can I do it right if I may ask?
– khaled014z
Jan 4 at 15:45




How can I do it right if I may ask?
– khaled014z
Jan 4 at 15:45












Use $z-1=(z+1)-2$ and split the sum in two. Then simplify and combine terms with the same power of $z+1$
– N74
Jan 4 at 18:33




Use $z-1=(z+1)-2$ and split the sum in two. Then simplify and combine terms with the same power of $z+1$
– N74
Jan 4 at 18:33










1 Answer
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$$f(z)=frac{z-1}{(z+1)(z-3)}=frac{(z+1)-2}{(z+1)(z+1-4)}$$Let $u=z+1$.



$$f(u)=frac1u(u-2)frac1{u-4}$$For $|u|in(0,4)$, we can expand the third term



$$begin{align}f(u)&=frac1u(2-u)frac14sum_{n=0}^infty left(frac u4right)^n\&=frac14sum_{n=0}^inftyleft(frac{2u^{n-1}}{4^n}-frac{u^n}{4^n}right)\&=frac1{2u}+frac14sum_{n=0}^inftyleft(-frac{2u^n}{4^{n+1}}right)end{align}$$



This means that the Laurent series for $f(z) $ is $$f(z)=frac1{2(z+1)}+frac14sum_{n=0}^inftyleft(-frac{2(z+1)^n}{4^{n+1}}right)=frac{1}{2(z+1)}-frac18-frac{(z+1)}{32}-frac{(z+1)^2}{512}-cdots$$



Now if you integrate the whole expansion, you should get $pi i$ again.






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    $$f(z)=frac{z-1}{(z+1)(z-3)}=frac{(z+1)-2}{(z+1)(z+1-4)}$$Let $u=z+1$.



    $$f(u)=frac1u(u-2)frac1{u-4}$$For $|u|in(0,4)$, we can expand the third term



    $$begin{align}f(u)&=frac1u(2-u)frac14sum_{n=0}^infty left(frac u4right)^n\&=frac14sum_{n=0}^inftyleft(frac{2u^{n-1}}{4^n}-frac{u^n}{4^n}right)\&=frac1{2u}+frac14sum_{n=0}^inftyleft(-frac{2u^n}{4^{n+1}}right)end{align}$$



    This means that the Laurent series for $f(z) $ is $$f(z)=frac1{2(z+1)}+frac14sum_{n=0}^inftyleft(-frac{2(z+1)^n}{4^{n+1}}right)=frac{1}{2(z+1)}-frac18-frac{(z+1)}{32}-frac{(z+1)^2}{512}-cdots$$



    Now if you integrate the whole expansion, you should get $pi i$ again.






    share|cite|improve this answer


























      0














      $$f(z)=frac{z-1}{(z+1)(z-3)}=frac{(z+1)-2}{(z+1)(z+1-4)}$$Let $u=z+1$.



      $$f(u)=frac1u(u-2)frac1{u-4}$$For $|u|in(0,4)$, we can expand the third term



      $$begin{align}f(u)&=frac1u(2-u)frac14sum_{n=0}^infty left(frac u4right)^n\&=frac14sum_{n=0}^inftyleft(frac{2u^{n-1}}{4^n}-frac{u^n}{4^n}right)\&=frac1{2u}+frac14sum_{n=0}^inftyleft(-frac{2u^n}{4^{n+1}}right)end{align}$$



      This means that the Laurent series for $f(z) $ is $$f(z)=frac1{2(z+1)}+frac14sum_{n=0}^inftyleft(-frac{2(z+1)^n}{4^{n+1}}right)=frac{1}{2(z+1)}-frac18-frac{(z+1)}{32}-frac{(z+1)^2}{512}-cdots$$



      Now if you integrate the whole expansion, you should get $pi i$ again.






      share|cite|improve this answer
























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        0






        $$f(z)=frac{z-1}{(z+1)(z-3)}=frac{(z+1)-2}{(z+1)(z+1-4)}$$Let $u=z+1$.



        $$f(u)=frac1u(u-2)frac1{u-4}$$For $|u|in(0,4)$, we can expand the third term



        $$begin{align}f(u)&=frac1u(2-u)frac14sum_{n=0}^infty left(frac u4right)^n\&=frac14sum_{n=0}^inftyleft(frac{2u^{n-1}}{4^n}-frac{u^n}{4^n}right)\&=frac1{2u}+frac14sum_{n=0}^inftyleft(-frac{2u^n}{4^{n+1}}right)end{align}$$



        This means that the Laurent series for $f(z) $ is $$f(z)=frac1{2(z+1)}+frac14sum_{n=0}^inftyleft(-frac{2(z+1)^n}{4^{n+1}}right)=frac{1}{2(z+1)}-frac18-frac{(z+1)}{32}-frac{(z+1)^2}{512}-cdots$$



        Now if you integrate the whole expansion, you should get $pi i$ again.






        share|cite|improve this answer












        $$f(z)=frac{z-1}{(z+1)(z-3)}=frac{(z+1)-2}{(z+1)(z+1-4)}$$Let $u=z+1$.



        $$f(u)=frac1u(u-2)frac1{u-4}$$For $|u|in(0,4)$, we can expand the third term



        $$begin{align}f(u)&=frac1u(2-u)frac14sum_{n=0}^infty left(frac u4right)^n\&=frac14sum_{n=0}^inftyleft(frac{2u^{n-1}}{4^n}-frac{u^n}{4^n}right)\&=frac1{2u}+frac14sum_{n=0}^inftyleft(-frac{2u^n}{4^{n+1}}right)end{align}$$



        This means that the Laurent series for $f(z) $ is $$f(z)=frac1{2(z+1)}+frac14sum_{n=0}^inftyleft(-frac{2(z+1)^n}{4^{n+1}}right)=frac{1}{2(z+1)}-frac18-frac{(z+1)}{32}-frac{(z+1)^2}{512}-cdots$$



        Now if you integrate the whole expansion, you should get $pi i$ again.







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        share|cite|improve this answer










        answered Jan 4 at 21:22









        John DoeJohn Doe

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        10.9k11238






























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