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Clarification regarding Winding Number Theorem
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In the course notes for my class, we have an example of calculating a complex integral about a path. The specific example uses the Winding number but I am having some difficulty understanding.
For a path $alpha : [t_1, t_2]to U subseteq mathbf C$ and a point $ainmathbf C$ which does not lie on the path $alpha$, we define the winding number $eta(alpha, a)$ of $alpha$ about $a$ as follows. We write $alpha(t) = a + r(t)e^itheta(t)$ where $r(t) = |alpha(t) - a|$ and $theta(t)$ is chosen continuously with $0letheta(t_1)le2pi$ (it can be shown that the map $theta(t)$ exists and is uniquely determined), and then we set $$eta(alpha, a) = dfractheta(t_2) - theta(t_1)2pi.$$ If $alpha$ is a loop then we have $alpha(t_1) = alpha(t_2)$ and so $e^itheta(t_1) = e^itheta(t_2)$ and hence $theta(t_2) - theta(t_1)$ will be a multiple of $2pi$. Thus for a loop $alpha$, we have $eta(alpha, a)inmathbf Z$.
Specifically, how does one go about choosing a function $theta(t)$
For example, given the integral $int_alpha frac1z$ where $alpha(t) = (1+ frac 38t)e^ipi t$. Here $t_1 = 0, t_2 = frac83$ The answer is $lnfrac21 + 2pi i frac 43$. I am not sure how this is evaluated.
Edit: In the same question, there is another integral $int_alpha frac1z+1$ along the same path, how does $theta(t)$ and $r(t)$ change?
integration complex-analysis winding-number
asked Jul 30 at 1:48
CyclicGroupOfOrder2
62
add a comment |Â
up vote
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In the course notes for my class, we have an example of calculating a complex integral about a path. The specific example uses the Winding number but I am having some difficulty understanding.
For a path $alpha : [t_1, t_2]to U subseteq mathbf C$ and a point $ainmathbf C$ which does not lie on the path $alpha$, we define the winding number $eta(alpha, a)$ of $alpha$ about $a$ as follows. We write $alpha(t) = a + r(t)e^itheta(t)$ where $r(t) = |alpha(t) - a|$ and $theta(t)$ is chosen continuously with $0letheta(t_1)le2pi$ (it can be shown that the map $theta(t)$ exists and is uniquely determined), and then we set $$eta(alpha, a) = dfractheta(t_2) - theta(t_1)2pi.$$ If $alpha$ is a loop then we have $alpha(t_1) = alpha(t_2)$ and so $e^itheta(t_1) = e^itheta(t_2)$ and hence $theta(t_2) - theta(t_1)$ will be a multiple of $2pi$. Thus for a loop $alpha$, we have $eta(alpha, a)inmathbf Z$.
Specifically, how does one go about choosing a function $theta(t)$
For example, given the integral $int_alpha frac1z$ where $alpha(t) = (1+ frac 38t)e^ipi t$. Here $t_1 = 0, t_2 = frac83$ The answer is $lnfrac21 + 2pi i frac 43$. I am not sure how this is evaluated.
Edit: In the same question, there is another integral $int_alpha frac1z+1$ along the same path, how does $theta(t)$ and $r(t)$ change?
integration complex-analysis winding-number
asked Jul 30 at 1:48
CyclicGroupOfOrder2
62
What are $t_1$ and $t_2$ in your example?
– Lord Shark the Unknown
Jul 30 at 2:04
Oh whoops, t1 is 0 and t2 is 8/3. I've edited it into the question, thanks.
– CyclicGroupOfOrder2
Jul 30 at 2:06
add a comment |Â
up vote
1
down vote
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up vote
1
down vote
favorite
In the course notes for my class, we have an example of calculating a complex integral about a path. The specific example uses the Winding number but I am having some difficulty understanding.
For a path $alpha : [t_1, t_2]to U subseteq mathbf C$ and a point $ainmathbf C$ which does not lie on the path $alpha$, we define the winding number $eta(alpha, a)$ of $alpha$ about $a$ as follows. We write $alpha(t) = a + r(t)e^itheta(t)$ where $r(t) = |alpha(t) - a|$ and $theta(t)$ is chosen continuously with $0letheta(t_1)le2pi$ (it can be shown that the map $theta(t)$ exists and is uniquely determined), and then we set $$eta(alpha, a) = dfractheta(t_2) - theta(t_1)2pi.$$ If $alpha$ is a loop then we have $alpha(t_1) = alpha(t_2)$ and so $e^itheta(t_1) = e^itheta(t_2)$ and hence $theta(t_2) - theta(t_1)$ will be a multiple of $2pi$. Thus for a loop $alpha$, we have $eta(alpha, a)inmathbf Z$.
Specifically, how does one go about choosing a function $theta(t)$
For example, given the integral $int_alpha frac1z$ where $alpha(t) = (1+ frac 38t)e^ipi t$. Here $t_1 = 0, t_2 = frac83$ The answer is $lnfrac21 + 2pi i frac 43$. I am not sure how this is evaluated.
Edit: In the same question, there is another integral $int_alpha frac1z+1$ along the same path, how does $theta(t)$ and $r(t)$ change?
integration complex-analysis winding-number
asked Jul 30 at 1:48
CyclicGroupOfOrder2
62
In the course notes for my class, we have an example of calculating a complex integral about a path. The specific example uses the Winding number but I am having some difficulty understanding.
For a path $alpha : [t_1, t_2]to U subseteq mathbf C$ and a point $ainmathbf C$ which does not lie on the path $alpha$, we define the winding number $eta(alpha, a)$ of $alpha$ about $a$ as follows. We write $alpha(t) = a + r(t)e^itheta(t)$ where $r(t) = |alpha(t) - a|$ and $theta(t)$ is chosen continuously with $0letheta(t_1)le2pi$ (it can be shown that the map $theta(t)$ exists and is uniquely determined), and then we set $$eta(alpha, a) = dfractheta(t_2) - theta(t_1)2pi.$$ If $alpha$ is a loop then we have $alpha(t_1) = alpha(t_2)$ and so $e^itheta(t_1) = e^itheta(t_2)$ and hence $theta(t_2) - theta(t_1)$ will be a multiple of $2pi$. Thus for a loop $alpha$, we have $eta(alpha, a)inmathbf Z$.
Specifically, how does one go about choosing a function $theta(t)$
For example, given the integral $int_alpha frac1z$ where $alpha(t) = (1+ frac 38t)e^ipi t$. Here $t_1 = 0, t_2 = frac83$ The answer is $lnfrac21 + 2pi i frac 43$. I am not sure how this is evaluated.
Edit: In the same question, there is another integral $int_alpha frac1z+1$ along the same path, how does $theta(t)$ and $r(t)$ change?
integration complex-analysis winding-number
asked Jul 30 at 1:48
CyclicGroupOfOrder2
62
edited Jul 30 at 2:20
asked Jul 30 at 1:48
CyclicGroupOfOrder2
62
asked Jul 30 at 1:48
CyclicGroupOfOrder2
62
asked Jul 30 at 1:48
CyclicGroupOfOrder2
62
62
What are $t_1$ and $t_2$ in your example?
– Lord Shark the Unknown
Jul 30 at 2:04
Oh whoops, t1 is 0 and t2 is 8/3. I've edited it into the question, thanks.
– CyclicGroupOfOrder2
Jul 30 at 2:06
add a comment |Â
What are $t_1$ and $t_2$ in your example?
– Lord Shark the Unknown
Jul 30 at 2:04
Oh whoops, t1 is 0 and t2 is 8/3. I've edited it into the question, thanks.
– CyclicGroupOfOrder2
Jul 30 at 2:06
What are $t_1$ and $t_2$ in your example?
– Lord Shark the Unknown
Jul 30 at 2:04
What are $t_1$ and $t_2$ in your example?
– Lord Shark the Unknown
Jul 30 at 2:04
Oh whoops, t1 is 0 and t2 is 8/3. I've edited it into the question, thanks.
– CyclicGroupOfOrder2
Jul 30 at 2:06
Oh whoops, t1 is 0 and t2 is 8/3. I've edited it into the question, thanks.
– CyclicGroupOfOrder2
Jul 30 at 2:06
add a comment |Â
1 Answer
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up vote
1
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On your path, $theta(t)=pi t$ is an allowable value. But to evaluate
$int_alphafracdzz$ you need to take a branch $ell(t)$ for $lnalpha(t)$ on your curve. In general that would be $ln|alpha(t)|+itheta(t)$, and here would be $ln(1+3t/8)+ipi t$. The value of
the integral is then $ell(t_2)-ell(t_1)$, which should come out
to your given value.
answered Jul 30 at 2:18
Lord Shark the Unknown
84.5k950111
Thanks, could you also take a look at my edit. Much appreciated
– CyclicGroupOfOrder2
Jul 30 at 2:26
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
On your path, $theta(t)=pi t$ is an allowable value. But to evaluate
$int_alphafracdzz$ you need to take a branch $ell(t)$ for $lnalpha(t)$ on your curve. In general that would be $ln|alpha(t)|+itheta(t)$, and here would be $ln(1+3t/8)+ipi t$. The value of
the integral is then $ell(t_2)-ell(t_1)$, which should come out
to your given value.
answered Jul 30 at 2:18
Lord Shark the Unknown
84.5k950111
Thanks, could you also take a look at my edit. Much appreciated
– CyclicGroupOfOrder2
Jul 30 at 2:26
add a comment |Â
up vote
1
down vote
On your path, $theta(t)=pi t$ is an allowable value. But to evaluate
$int_alphafracdzz$ you need to take a branch $ell(t)$ for $lnalpha(t)$ on your curve. In general that would be $ln|alpha(t)|+itheta(t)$, and here would be $ln(1+3t/8)+ipi t$. The value of
the integral is then $ell(t_2)-ell(t_1)$, which should come out
to your given value.
answered Jul 30 at 2:18
Lord Shark the Unknown
84.5k950111
Thanks, could you also take a look at my edit. Much appreciated
– CyclicGroupOfOrder2
Jul 30 at 2:26
add a comment |Â
up vote
1
down vote
up vote
1
down vote
On your path, $theta(t)=pi t$ is an allowable value. But to evaluate
$int_alphafracdzz$ you need to take a branch $ell(t)$ for $lnalpha(t)$ on your curve. In general that would be $ln|alpha(t)|+itheta(t)$, and here would be $ln(1+3t/8)+ipi t$. The value of
the integral is then $ell(t_2)-ell(t_1)$, which should come out
to your given value.
answered Jul 30 at 2:18
Lord Shark the Unknown
84.5k950111
On your path, $theta(t)=pi t$ is an allowable value. But to evaluate
$int_alphafracdzz$ you need to take a branch $ell(t)$ for $lnalpha(t)$ on your curve. In general that would be $ln|alpha(t)|+itheta(t)$, and here would be $ln(1+3t/8)+ipi t$. The value of
the integral is then $ell(t_2)-ell(t_1)$, which should come out
to your given value.
answered Jul 30 at 2:18
Lord Shark the Unknown
84.5k950111
answered Jul 30 at 2:18
Lord Shark the Unknown
84.5k950111
answered Jul 30 at 2:18
Lord Shark the Unknown
84.5k950111
answered Jul 30 at 2:18
Lord Shark the Unknown
84.5k950111
84.5k950111
Thanks, could you also take a look at my edit. Much appreciated
– CyclicGroupOfOrder2
Jul 30 at 2:26
add a comment |Â
Thanks, could you also take a look at my edit. Much appreciated
– CyclicGroupOfOrder2
Jul 30 at 2:26
Thanks, could you also take a look at my edit. Much appreciated
– CyclicGroupOfOrder2
Jul 30 at 2:26
Thanks, could you also take a look at my edit. Much appreciated
– CyclicGroupOfOrder2
Jul 30 at 2:26
add a comment |Â
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What are $t_1$ and $t_2$ in your example?
– Lord Shark the Unknown
Jul 30 at 2:04
Oh whoops, t1 is 0 and t2 is 8/3. I've edited it into the question, thanks.
– CyclicGroupOfOrder2
Jul 30 at 2:06