Why is integral not equal to zero even though the path is closed?
I have an integral $$int{z^{i}}dz$$ the path is $e^{it}$ where $t$ is between $0$ and $2pi$. Why is it not equal to zero even though the path is closed.
complex-analysis contour-integration complex-integration
edited Nov 20 at 18:13
Snookie
1,30017
asked Nov 20 at 12:03
ryszard eggink
308110
add a comment |
I have an integral $$int{z^{i}}dz$$ the path is $e^{it}$ where $t$ is between $0$ and $2pi$. Why is it not equal to zero even though the path is closed.
complex-analysis contour-integration complex-integration
edited Nov 20 at 18:13
Snookie
1,30017
asked Nov 20 at 12:03
ryszard eggink
308110
add a comment |
I have an integral $$int{z^{i}}dz$$ the path is $e^{it}$ where $t$ is between $0$ and $2pi$. Why is it not equal to zero even though the path is closed.
complex-analysis contour-integration complex-integration
edited Nov 20 at 18:13
Snookie
1,30017
asked Nov 20 at 12:03
ryszard eggink
308110
I have an integral $$int{z^{i}}dz$$ the path is $e^{it}$ where $t$ is between $0$ and $2pi$. Why is it not equal to zero even though the path is closed.
complex-analysis contour-integration complex-integration
complex-analysis contour-integration complex-integration
edited Nov 20 at 18:13
Snookie
1,30017
asked Nov 20 at 12:03
ryszard eggink
308110
edited Nov 20 at 18:13
Snookie
1,30017
asked Nov 20 at 12:03
ryszard eggink
308110
edited Nov 20 at 18:13
Snookie
1,30017
edited Nov 20 at 18:13
Snookie
1,30017
edited Nov 20 at 18:13
Snookie
1,30017
1,30017
asked Nov 20 at 12:03
ryszard eggink
308110
asked Nov 20 at 12:03
ryszard eggink
308110
asked Nov 20 at 12:03
ryszard eggink
308110
308110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You need to be careful about the definition of $z^i$ as it is singular at $0$. If you insert a branch cut somewhere then the contour is no longer closed. If you do not, you cannot apply Cauchy's theorem.
answered Nov 20 at 12:10
Richard Martin
1,63618
add a comment |
You seem to believe that $z^i$ has a well-defined meaning, at least as a function from $mathbb C$ into itself. It hasn't. The number $z^i$ is any number of the form $e^{ilog z}$, where $log z$ is any logarihtm of $z$ (and, in particular, $z^i$ has no meaning if $z=0$). And you will find no way of choosing, for each $zinmathbb{C}setminus{0}$, a number $z^i$ in such a way that the function that you get is continuous, and much less analytic.
answered Nov 20 at 12:29
José Carlos Santos
149k22117219
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You need to be careful about the definition of $z^i$ as it is singular at $0$. If you insert a branch cut somewhere then the contour is no longer closed. If you do not, you cannot apply Cauchy's theorem.
answered Nov 20 at 12:10
Richard Martin
1,63618
add a comment |
You need to be careful about the definition of $z^i$ as it is singular at $0$. If you insert a branch cut somewhere then the contour is no longer closed. If you do not, you cannot apply Cauchy's theorem.
answered Nov 20 at 12:10
Richard Martin
1,63618
add a comment |
You need to be careful about the definition of $z^i$ as it is singular at $0$. If you insert a branch cut somewhere then the contour is no longer closed. If you do not, you cannot apply Cauchy's theorem.
answered Nov 20 at 12:10
Richard Martin
1,63618
You need to be careful about the definition of $z^i$ as it is singular at $0$. If you insert a branch cut somewhere then the contour is no longer closed. If you do not, you cannot apply Cauchy's theorem.
answered Nov 20 at 12:10
Richard Martin
1,63618
answered Nov 20 at 12:10
Richard Martin
1,63618
answered Nov 20 at 12:10
Richard Martin
1,63618
answered Nov 20 at 12:10
Richard Martin
1,63618
1,63618
add a comment |
add a comment |
You seem to believe that $z^i$ has a well-defined meaning, at least as a function from $mathbb C$ into itself. It hasn't. The number $z^i$ is any number of the form $e^{ilog z}$, where $log z$ is any logarihtm of $z$ (and, in particular, $z^i$ has no meaning if $z=0$). And you will find no way of choosing, for each $zinmathbb{C}setminus{0}$, a number $z^i$ in such a way that the function that you get is continuous, and much less analytic.
answered Nov 20 at 12:29
José Carlos Santos
149k22117219
add a comment |
You seem to believe that $z^i$ has a well-defined meaning, at least as a function from $mathbb C$ into itself. It hasn't. The number $z^i$ is any number of the form $e^{ilog z}$, where $log z$ is any logarihtm of $z$ (and, in particular, $z^i$ has no meaning if $z=0$). And you will find no way of choosing, for each $zinmathbb{C}setminus{0}$, a number $z^i$ in such a way that the function that you get is continuous, and much less analytic.
answered Nov 20 at 12:29
José Carlos Santos
149k22117219
add a comment |
You seem to believe that $z^i$ has a well-defined meaning, at least as a function from $mathbb C$ into itself. It hasn't. The number $z^i$ is any number of the form $e^{ilog z}$, where $log z$ is any logarihtm of $z$ (and, in particular, $z^i$ has no meaning if $z=0$). And you will find no way of choosing, for each $zinmathbb{C}setminus{0}$, a number $z^i$ in such a way that the function that you get is continuous, and much less analytic.
answered Nov 20 at 12:29
José Carlos Santos
149k22117219
You seem to believe that $z^i$ has a well-defined meaning, at least as a function from $mathbb C$ into itself. It hasn't. The number $z^i$ is any number of the form $e^{ilog z}$, where $log z$ is any logarihtm of $z$ (and, in particular, $z^i$ has no meaning if $z=0$). And you will find no way of choosing, for each $zinmathbb{C}setminus{0}$, a number $z^i$ in such a way that the function that you get is continuous, and much less analytic.
answered Nov 20 at 12:29
José Carlos Santos
149k22117219
edited Nov 20 at 18:06
answered Nov 20 at 12:29
José Carlos Santos
149k22117219
answered Nov 20 at 12:29
José Carlos Santos
149k22117219
answered Nov 20 at 12:29
José Carlos Santos
149k22117219
149k22117219
add a comment |
add a comment |
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