Compute $int_{|z|=R} frac{|dz|}{|z-a|^4}$ when $R>0, |a|neq R$
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I want to compute $int_{|z|=R} frac{|dz|}{|z-a|^4}$ when $R>0, |a|neq R$.
I did so by parametrizing the circle, but this way required many pages of tedious calculations.
Do you have a more straightforward way to do it?
Let me mention that my issue here is that we integrate by $|dz|$ not by $dz$, so all the known theorems I know fail to be applied -at least in this form of the integral.
Thanks
complex-integration
asked Oct 24 '17 at 5:22
perlmanperlman
8110
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add a comment |
$begingroup$
I want to compute $int_{|z|=R} frac{|dz|}{|z-a|^4}$ when $R>0, |a|neq R$.
I did so by parametrizing the circle, but this way required many pages of tedious calculations.
Do you have a more straightforward way to do it?
Let me mention that my issue here is that we integrate by $|dz|$ not by $dz$, so all the known theorems I know fail to be applied -at least in this form of the integral.
Thanks
complex-integration
asked Oct 24 '17 at 5:22
perlmanperlman
8110
$endgroup$
add a comment |
$begingroup$
I want to compute $int_{|z|=R} frac{|dz|}{|z-a|^4}$ when $R>0, |a|neq R$.
I did so by parametrizing the circle, but this way required many pages of tedious calculations.
Do you have a more straightforward way to do it?
Let me mention that my issue here is that we integrate by $|dz|$ not by $dz$, so all the known theorems I know fail to be applied -at least in this form of the integral.
Thanks
complex-integration
asked Oct 24 '17 at 5:22
perlmanperlman
8110
$endgroup$
I want to compute $int_{|z|=R} frac{|dz|}{|z-a|^4}$ when $R>0, |a|neq R$.
I did so by parametrizing the circle, but this way required many pages of tedious calculations.
Do you have a more straightforward way to do it?
Let me mention that my issue here is that we integrate by $|dz|$ not by $dz$, so all the known theorems I know fail to be applied -at least in this form of the integral.
Thanks
complex-integration
complex-integration
asked Oct 24 '17 at 5:22
perlmanperlman
8110
asked Oct 24 '17 at 5:22
perlmanperlman
8110
edited Oct 24 '17 at 5:28
perlman
asked Oct 24 '17 at 5:22
perlmanperlman
8110
asked Oct 24 '17 at 5:22
perlmanperlman
8110
asked Oct 24 '17 at 5:22
perlmanperlman
8110
8110
add a comment |
add a comment |
2 Answers
2
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$begingroup$
Try This
$f^n(z)$ = $frac {n!}{2(pi)n}$ $int$ $frac {f(w) dw}{(w-z)^{n+1}}$
answered Oct 24 '17 at 5:41
Math ModelMath Model
445
$endgroup$
$begingroup$
Can you make it a bit more clear? Specifically, how do I treat the $|dz|$?
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– perlman
Oct 24 '17 at 5:47
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I just assumed the absolute value sign was meant to be around the function. I don't think I've specifically dealt with a case you're talking about, because I tried googling it, and |dz| turns out to be some kind of arc length.
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– Math Model
Oct 24 '17 at 5:57
add a comment |
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After parametrising $z= R e^{itheta}$ and after a lot of simplifications you will get
$$I = int_{0}^{2pi} dtheta frac{R}{[(Rcostheta-Re(a))^2+(Rsintheta-Im(a))^2]^2_{}}.$$
To proceed further : go back to complex variable $Z = e^{itheta}$, rewrite it in terms of $costheta=frac{1}{2}(z+frac{1}{z})$ and $sintheta=frac{1}{2i}(z-frac{1}{z})$. The resulting integral can be evaluated using complex integration methods.
answered Oct 24 '17 at 6:34
SunyamSunyam
167214
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try This
$f^n(z)$ = $frac {n!}{2(pi)n}$ $int$ $frac {f(w) dw}{(w-z)^{n+1}}$
answered Oct 24 '17 at 5:41
Math ModelMath Model
445
$endgroup$
$begingroup$
Can you make it a bit more clear? Specifically, how do I treat the $|dz|$?
$endgroup$
– perlman
Oct 24 '17 at 5:47
$begingroup$
I just assumed the absolute value sign was meant to be around the function. I don't think I've specifically dealt with a case you're talking about, because I tried googling it, and |dz| turns out to be some kind of arc length.
$endgroup$
– Math Model
Oct 24 '17 at 5:57
add a comment |
$begingroup$
Try This
$f^n(z)$ = $frac {n!}{2(pi)n}$ $int$ $frac {f(w) dw}{(w-z)^{n+1}}$
answered Oct 24 '17 at 5:41
Math ModelMath Model
445
$endgroup$
$begingroup$
Can you make it a bit more clear? Specifically, how do I treat the $|dz|$?
$endgroup$
– perlman
Oct 24 '17 at 5:47
$begingroup$
I just assumed the absolute value sign was meant to be around the function. I don't think I've specifically dealt with a case you're talking about, because I tried googling it, and |dz| turns out to be some kind of arc length.
$endgroup$
– Math Model
Oct 24 '17 at 5:57
add a comment |
$begingroup$
Try This
$f^n(z)$ = $frac {n!}{2(pi)n}$ $int$ $frac {f(w) dw}{(w-z)^{n+1}}$
answered Oct 24 '17 at 5:41
Math ModelMath Model
445
$endgroup$
Try This
$f^n(z)$ = $frac {n!}{2(pi)n}$ $int$ $frac {f(w) dw}{(w-z)^{n+1}}$
answered Oct 24 '17 at 5:41
Math ModelMath Model
445
edited Oct 24 '17 at 5:55
answered Oct 24 '17 at 5:41
Math ModelMath Model
445
answered Oct 24 '17 at 5:41
Math ModelMath Model
445
answered Oct 24 '17 at 5:41
Math ModelMath Model
445
445
$begingroup$
Can you make it a bit more clear? Specifically, how do I treat the $|dz|$?
$endgroup$
– perlman
Oct 24 '17 at 5:47
$begingroup$
I just assumed the absolute value sign was meant to be around the function. I don't think I've specifically dealt with a case you're talking about, because I tried googling it, and |dz| turns out to be some kind of arc length.
$endgroup$
– Math Model
Oct 24 '17 at 5:57
add a comment |
$begingroup$
Can you make it a bit more clear? Specifically, how do I treat the $|dz|$?
$endgroup$
– perlman
Oct 24 '17 at 5:47
$begingroup$
I just assumed the absolute value sign was meant to be around the function. I don't think I've specifically dealt with a case you're talking about, because I tried googling it, and |dz| turns out to be some kind of arc length.
$endgroup$
– Math Model
Oct 24 '17 at 5:57
$begingroup$
Can you make it a bit more clear? Specifically, how do I treat the $|dz|$?
$endgroup$
– perlman
Oct 24 '17 at 5:47
$begingroup$
Can you make it a bit more clear? Specifically, how do I treat the $|dz|$?
$endgroup$
– perlman
Oct 24 '17 at 5:47
$begingroup$
I just assumed the absolute value sign was meant to be around the function. I don't think I've specifically dealt with a case you're talking about, because I tried googling it, and |dz| turns out to be some kind of arc length.
$endgroup$
– Math Model
Oct 24 '17 at 5:57
$begingroup$
I just assumed the absolute value sign was meant to be around the function. I don't think I've specifically dealt with a case you're talking about, because I tried googling it, and |dz| turns out to be some kind of arc length.
$endgroup$
– Math Model
Oct 24 '17 at 5:57
add a comment |
$begingroup$
After parametrising $z= R e^{itheta}$ and after a lot of simplifications you will get
$$I = int_{0}^{2pi} dtheta frac{R}{[(Rcostheta-Re(a))^2+(Rsintheta-Im(a))^2]^2_{}}.$$
To proceed further : go back to complex variable $Z = e^{itheta}$, rewrite it in terms of $costheta=frac{1}{2}(z+frac{1}{z})$ and $sintheta=frac{1}{2i}(z-frac{1}{z})$. The resulting integral can be evaluated using complex integration methods.
answered Oct 24 '17 at 6:34
SunyamSunyam
167214
$endgroup$
add a comment |
$begingroup$
After parametrising $z= R e^{itheta}$ and after a lot of simplifications you will get
$$I = int_{0}^{2pi} dtheta frac{R}{[(Rcostheta-Re(a))^2+(Rsintheta-Im(a))^2]^2_{}}.$$
To proceed further : go back to complex variable $Z = e^{itheta}$, rewrite it in terms of $costheta=frac{1}{2}(z+frac{1}{z})$ and $sintheta=frac{1}{2i}(z-frac{1}{z})$. The resulting integral can be evaluated using complex integration methods.
answered Oct 24 '17 at 6:34
SunyamSunyam
167214
$endgroup$
add a comment |
$begingroup$
After parametrising $z= R e^{itheta}$ and after a lot of simplifications you will get
$$I = int_{0}^{2pi} dtheta frac{R}{[(Rcostheta-Re(a))^2+(Rsintheta-Im(a))^2]^2_{}}.$$
To proceed further : go back to complex variable $Z = e^{itheta}$, rewrite it in terms of $costheta=frac{1}{2}(z+frac{1}{z})$ and $sintheta=frac{1}{2i}(z-frac{1}{z})$. The resulting integral can be evaluated using complex integration methods.
answered Oct 24 '17 at 6:34
SunyamSunyam
167214
$endgroup$
After parametrising $z= R e^{itheta}$ and after a lot of simplifications you will get
$$I = int_{0}^{2pi} dtheta frac{R}{[(Rcostheta-Re(a))^2+(Rsintheta-Im(a))^2]^2_{}}.$$
To proceed further : go back to complex variable $Z = e^{itheta}$, rewrite it in terms of $costheta=frac{1}{2}(z+frac{1}{z})$ and $sintheta=frac{1}{2i}(z-frac{1}{z})$. The resulting integral can be evaluated using complex integration methods.
answered Oct 24 '17 at 6:34
SunyamSunyam
167214
answered Oct 24 '17 at 6:34
SunyamSunyam
167214
answered Oct 24 '17 at 6:34
SunyamSunyam
167214
answered Oct 24 '17 at 6:34
SunyamSunyam
167214
167214
add a comment |
add a comment |
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