$int_{phi} z^2e^{z^3}sin z$ dz












0












$begingroup$


$int_{phi} z^2e^{z^3}sin z$ dz where $phi in [0, 2pi]$ and $ phi (t) =e^{4it}$.



I want to use Cauchy's Integral Formuła for Derivtives, but I can't simplify this expression. I use substitution $ z=e^{4it}$ and $dz=ie^{4it}$ dt.






integration complex-analysis







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    $z^2e^{z^3}sin(z)$ is holomorphic in and on the circle $z=e^{i4t}$, $tin [0,pi/2]$. The integral is $0$ by Cauchy's Integral Theorem.
    $endgroup$
    – Mark Viola
    Jan 19 at 23:08












  • $begingroup$
    Do should I use any any another way?
    $endgroup$
    – Kamilo
    Jan 19 at 23:19






  • 2




    $begingroup$
    Why would you want to proceed any other way when the conclusion is immediate here?
    $endgroup$
    – Mark Viola
    Jan 20 at 1:03
















0












$begingroup$


$int_{phi} z^2e^{z^3}sin z$ dz where $phi in [0, 2pi]$ and $ phi (t) =e^{4it}$.



I want to use Cauchy's Integral Formuła for Derivtives, but I can't simplify this expression. I use substitution $ z=e^{4it}$ and $dz=ie^{4it}$ dt.






integration complex-analysis







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    $z^2e^{z^3}sin(z)$ is holomorphic in and on the circle $z=e^{i4t}$, $tin [0,pi/2]$. The integral is $0$ by Cauchy's Integral Theorem.
    $endgroup$
    – Mark Viola
    Jan 19 at 23:08












  • $begingroup$
    Do should I use any any another way?
    $endgroup$
    – Kamilo
    Jan 19 at 23:19






  • 2




    $begingroup$
    Why would you want to proceed any other way when the conclusion is immediate here?
    $endgroup$
    – Mark Viola
    Jan 20 at 1:03














0












0








0





$begingroup$


$int_{phi} z^2e^{z^3}sin z$ dz where $phi in [0, 2pi]$ and $ phi (t) =e^{4it}$.



I want to use Cauchy's Integral Formuła for Derivtives, but I can't simplify this expression. I use substitution $ z=e^{4it}$ and $dz=ie^{4it}$ dt.






integration complex-analysis







share|cite|improve this question











$endgroup$




$int_{phi} z^2e^{z^3}sin z$ dz where $phi in [0, 2pi]$ and $ phi (t) =e^{4it}$.



I want to use Cauchy's Integral Formuła for Derivtives, but I can't simplify this expression. I use substitution $ z=e^{4it}$ and $dz=ie^{4it}$ dt.






integration complex-analysis




integration complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 20 at 6:28









stone-zeng

210210




210210










asked Jan 19 at 23:01









KamiloKamilo

43




43








  • 5




    $begingroup$
    $z^2e^{z^3}sin(z)$ is holomorphic in and on the circle $z=e^{i4t}$, $tin [0,pi/2]$. The integral is $0$ by Cauchy's Integral Theorem.
    $endgroup$
    – Mark Viola
    Jan 19 at 23:08












  • $begingroup$
    Do should I use any any another way?
    $endgroup$
    – Kamilo
    Jan 19 at 23:19






  • 2




    $begingroup$
    Why would you want to proceed any other way when the conclusion is immediate here?
    $endgroup$
    – Mark Viola
    Jan 20 at 1:03














  • 5




    $begingroup$
    $z^2e^{z^3}sin(z)$ is holomorphic in and on the circle $z=e^{i4t}$, $tin [0,pi/2]$. The integral is $0$ by Cauchy's Integral Theorem.
    $endgroup$
    – Mark Viola
    Jan 19 at 23:08












  • $begingroup$
    Do should I use any any another way?
    $endgroup$
    – Kamilo
    Jan 19 at 23:19






  • 2




    $begingroup$
    Why would you want to proceed any other way when the conclusion is immediate here?
    $endgroup$
    – Mark Viola
    Jan 20 at 1:03








5




5




$begingroup$
$z^2e^{z^3}sin(z)$ is holomorphic in and on the circle $z=e^{i4t}$, $tin [0,pi/2]$. The integral is $0$ by Cauchy's Integral Theorem.
$endgroup$
– Mark Viola
Jan 19 at 23:08






$begingroup$
$z^2e^{z^3}sin(z)$ is holomorphic in and on the circle $z=e^{i4t}$, $tin [0,pi/2]$. The integral is $0$ by Cauchy's Integral Theorem.
$endgroup$
– Mark Viola
Jan 19 at 23:08














$begingroup$
Do should I use any any another way?
$endgroup$
– Kamilo
Jan 19 at 23:19




$begingroup$
Do should I use any any another way?
$endgroup$
– Kamilo
Jan 19 at 23:19




2




2




$begingroup$
Why would you want to proceed any other way when the conclusion is immediate here?
$endgroup$
– Mark Viola
Jan 20 at 1:03




$begingroup$
Why would you want to proceed any other way when the conclusion is immediate here?
$endgroup$
– Mark Viola
Jan 20 at 1:03










0






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