result of this integral? $int_{|z|=3} frac{e^z}{(z-1)(z-2)}$












2












$begingroup$


I think I can solve it by taking:
$z(t)=3(cos t + i sin t)=3e^{it}$
and by using



$$int_c f(z)dz = int_c frac{dz}{dt}cdot f(z(t))$$
we have:



$displaystyleint_{|z|=3} frac{e^{3e^{it}}}{(3e^{it}-1)(3e^{it}-2)}cdot3ie^{it}$
but I can't continue this approach to find a solution without using residue theorem






complex-analysis complex-integration







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The solution is pretty trivial using the residue theorem. Is that tool available to you at this point in your coursework?
    $endgroup$
    – Eevee Trainer
    Jan 1 at 20:48










  • $begingroup$
    no; indeed I should solve it without residue theorem
    $endgroup$
    – mohamadreza
    Jan 1 at 20:51










  • $begingroup$
    I have doubts concerning the equality $e^{it}=(sin t+i cos t)$. Should it not be $e^{it}=(cos t + i sin t)$ or am I missing something?
    $endgroup$
    – mrtaurho
    Jan 1 at 20:54










  • $begingroup$
    you are right I have a typing mistake unfortunately
    $endgroup$
    – mohamadreza
    Jan 1 at 20:56










  • $begingroup$
    The second and third integral should involve a $dt$.
    $endgroup$
    – Fakemistake
    Jan 1 at 21:30
















2












$begingroup$


I think I can solve it by taking:
$z(t)=3(cos t + i sin t)=3e^{it}$
and by using



$$int_c f(z)dz = int_c frac{dz}{dt}cdot f(z(t))$$
we have:



$displaystyleint_{|z|=3} frac{e^{3e^{it}}}{(3e^{it}-1)(3e^{it}-2)}cdot3ie^{it}$
but I can't continue this approach to find a solution without using residue theorem






complex-analysis complex-integration







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The solution is pretty trivial using the residue theorem. Is that tool available to you at this point in your coursework?
    $endgroup$
    – Eevee Trainer
    Jan 1 at 20:48










  • $begingroup$
    no; indeed I should solve it without residue theorem
    $endgroup$
    – mohamadreza
    Jan 1 at 20:51










  • $begingroup$
    I have doubts concerning the equality $e^{it}=(sin t+i cos t)$. Should it not be $e^{it}=(cos t + i sin t)$ or am I missing something?
    $endgroup$
    – mrtaurho
    Jan 1 at 20:54










  • $begingroup$
    you are right I have a typing mistake unfortunately
    $endgroup$
    – mohamadreza
    Jan 1 at 20:56










  • $begingroup$
    The second and third integral should involve a $dt$.
    $endgroup$
    – Fakemistake
    Jan 1 at 21:30














2












2








2





$begingroup$


I think I can solve it by taking:
$z(t)=3(cos t + i sin t)=3e^{it}$
and by using



$$int_c f(z)dz = int_c frac{dz}{dt}cdot f(z(t))$$
we have:



$displaystyleint_{|z|=3} frac{e^{3e^{it}}}{(3e^{it}-1)(3e^{it}-2)}cdot3ie^{it}$
but I can't continue this approach to find a solution without using residue theorem






complex-analysis complex-integration







share|cite|improve this question











$endgroup$




I think I can solve it by taking:
$z(t)=3(cos t + i sin t)=3e^{it}$
and by using



$$int_c f(z)dz = int_c frac{dz}{dt}cdot f(z(t))$$
we have:



$displaystyleint_{|z|=3} frac{e^{3e^{it}}}{(3e^{it}-1)(3e^{it}-2)}cdot3ie^{it}$
but I can't continue this approach to find a solution without using residue theorem






complex-analysis complex-integration




complex-analysis complex-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 20:55







mohamadreza

















asked Jan 1 at 20:44









mohamadrezamohamadreza

275




275








  • 2




    $begingroup$
    The solution is pretty trivial using the residue theorem. Is that tool available to you at this point in your coursework?
    $endgroup$
    – Eevee Trainer
    Jan 1 at 20:48










  • $begingroup$
    no; indeed I should solve it without residue theorem
    $endgroup$
    – mohamadreza
    Jan 1 at 20:51










  • $begingroup$
    I have doubts concerning the equality $e^{it}=(sin t+i cos t)$. Should it not be $e^{it}=(cos t + i sin t)$ or am I missing something?
    $endgroup$
    – mrtaurho
    Jan 1 at 20:54










  • $begingroup$
    you are right I have a typing mistake unfortunately
    $endgroup$
    – mohamadreza
    Jan 1 at 20:56










  • $begingroup$
    The second and third integral should involve a $dt$.
    $endgroup$
    – Fakemistake
    Jan 1 at 21:30














  • 2




    $begingroup$
    The solution is pretty trivial using the residue theorem. Is that tool available to you at this point in your coursework?
    $endgroup$
    – Eevee Trainer
    Jan 1 at 20:48










  • $begingroup$
    no; indeed I should solve it without residue theorem
    $endgroup$
    – mohamadreza
    Jan 1 at 20:51










  • $begingroup$
    I have doubts concerning the equality $e^{it}=(sin t+i cos t)$. Should it not be $e^{it}=(cos t + i sin t)$ or am I missing something?
    $endgroup$
    – mrtaurho
    Jan 1 at 20:54










  • $begingroup$
    you are right I have a typing mistake unfortunately
    $endgroup$
    – mohamadreza
    Jan 1 at 20:56










  • $begingroup$
    The second and third integral should involve a $dt$.
    $endgroup$
    – Fakemistake
    Jan 1 at 21:30








2




2




$begingroup$
The solution is pretty trivial using the residue theorem. Is that tool available to you at this point in your coursework?
$endgroup$
– Eevee Trainer
Jan 1 at 20:48




$begingroup$
The solution is pretty trivial using the residue theorem. Is that tool available to you at this point in your coursework?
$endgroup$
– Eevee Trainer
Jan 1 at 20:48












$begingroup$
no; indeed I should solve it without residue theorem
$endgroup$
– mohamadreza
Jan 1 at 20:51




$begingroup$
no; indeed I should solve it without residue theorem
$endgroup$
– mohamadreza
Jan 1 at 20:51












$begingroup$
I have doubts concerning the equality $e^{it}=(sin t+i cos t)$. Should it not be $e^{it}=(cos t + i sin t)$ or am I missing something?
$endgroup$
– mrtaurho
Jan 1 at 20:54




$begingroup$
I have doubts concerning the equality $e^{it}=(sin t+i cos t)$. Should it not be $e^{it}=(cos t + i sin t)$ or am I missing something?
$endgroup$
– mrtaurho
Jan 1 at 20:54












$begingroup$
you are right I have a typing mistake unfortunately
$endgroup$
– mohamadreza
Jan 1 at 20:56




$begingroup$
you are right I have a typing mistake unfortunately
$endgroup$
– mohamadreza
Jan 1 at 20:56












$begingroup$
The second and third integral should involve a $dt$.
$endgroup$
– Fakemistake
Jan 1 at 21:30




$begingroup$
The second and third integral should involve a $dt$.
$endgroup$
– Fakemistake
Jan 1 at 21:30










1 Answer
1






active

oldest

votes


















4












$begingroup$

Hint



$$frac{1}{(z-1)(z-2)}=frac{1}{z-2}-frac{1}{z-1}$$



Thus
$$int_{|z|=3} frac{e^z}{(z-1)(z-2)}dz=int_{|z|=3} frac{e^z}{z-2}dz-int_{|z|=3} frac{e^z}{z-1}dz$$



The answer follows easily from Cauchy Integral Formula.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you.I got the answer using your hint
    $endgroup$
    – mohamadreza
    Jan 1 at 20:58










  • $begingroup$
    @mohamadreza Then accept his answer to show that it helped you ^^
    $endgroup$
    – mrtaurho
    Jan 1 at 21:05












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Hint



$$frac{1}{(z-1)(z-2)}=frac{1}{z-2}-frac{1}{z-1}$$



Thus
$$int_{|z|=3} frac{e^z}{(z-1)(z-2)}dz=int_{|z|=3} frac{e^z}{z-2}dz-int_{|z|=3} frac{e^z}{z-1}dz$$



The answer follows easily from Cauchy Integral Formula.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you.I got the answer using your hint
    $endgroup$
    – mohamadreza
    Jan 1 at 20:58










  • $begingroup$
    @mohamadreza Then accept his answer to show that it helped you ^^
    $endgroup$
    – mrtaurho
    Jan 1 at 21:05
















4












$begingroup$

Hint



$$frac{1}{(z-1)(z-2)}=frac{1}{z-2}-frac{1}{z-1}$$



Thus
$$int_{|z|=3} frac{e^z}{(z-1)(z-2)}dz=int_{|z|=3} frac{e^z}{z-2}dz-int_{|z|=3} frac{e^z}{z-1}dz$$



The answer follows easily from Cauchy Integral Formula.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you.I got the answer using your hint
    $endgroup$
    – mohamadreza
    Jan 1 at 20:58










  • $begingroup$
    @mohamadreza Then accept his answer to show that it helped you ^^
    $endgroup$
    – mrtaurho
    Jan 1 at 21:05














4












4








4





$begingroup$

Hint



$$frac{1}{(z-1)(z-2)}=frac{1}{z-2}-frac{1}{z-1}$$



Thus
$$int_{|z|=3} frac{e^z}{(z-1)(z-2)}dz=int_{|z|=3} frac{e^z}{z-2}dz-int_{|z|=3} frac{e^z}{z-1}dz$$



The answer follows easily from Cauchy Integral Formula.






share|cite|improve this answer









$endgroup$



Hint



$$frac{1}{(z-1)(z-2)}=frac{1}{z-2}-frac{1}{z-1}$$



Thus
$$int_{|z|=3} frac{e^z}{(z-1)(z-2)}dz=int_{|z|=3} frac{e^z}{z-2}dz-int_{|z|=3} frac{e^z}{z-1}dz$$



The answer follows easily from Cauchy Integral Formula.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 1 at 20:53









N. S.N. S.

105k7114210




105k7114210












  • $begingroup$
    thank you.I got the answer using your hint
    $endgroup$
    – mohamadreza
    Jan 1 at 20:58










  • $begingroup$
    @mohamadreza Then accept his answer to show that it helped you ^^
    $endgroup$
    – mrtaurho
    Jan 1 at 21:05


















  • $begingroup$
    thank you.I got the answer using your hint
    $endgroup$
    – mohamadreza
    Jan 1 at 20:58










  • $begingroup$
    @mohamadreza Then accept his answer to show that it helped you ^^
    $endgroup$
    – mrtaurho
    Jan 1 at 21:05
















$begingroup$
thank you.I got the answer using your hint
$endgroup$
– mohamadreza
Jan 1 at 20:58




$begingroup$
thank you.I got the answer using your hint
$endgroup$
– mohamadreza
Jan 1 at 20:58












$begingroup$
@mohamadreza Then accept his answer to show that it helped you ^^
$endgroup$
– mrtaurho
Jan 1 at 21:05




$begingroup$
@mohamadreza Then accept his answer to show that it helped you ^^
$endgroup$
– mrtaurho
Jan 1 at 21:05


















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