Compute the integral $int_{|z|=rho}|z-a|^{-4}|dz|$ with $|a|neq rho$
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I need help in computing the integral indicated above. What I've tried so far:
Parametrize the curve indicated by $|z|=rho$ with $gamma = z(t) = rho cos t + isin t$. Then by definition
$$
int_gamma f(z)|dz|=int_gamma f(z(t))|z'(t)| dt
$$
gives the following
begin{align}
int_gamma |z-a|^{-4} |dz| & = int_0^{2pi} |rho cos t+irho sin t-a_1-ia_2|^{-4}rho dt\
& = rhoint_0^{2pi}frac{1}{(rho^2-2a_1cos t-2a_2sin t + a_1^2+a_2^2)^2} dt
end{align}
Where $a=a_1+ia_2$. It's not hard to see how this becomes complicated very easily. I want to know if there is some sort of 'trick' I'm not aware of or something I might be missing.
integration complex-analysis line-integrals
asked Nov 14 at 23:53
D. Brito
345110
add a comment |
up vote
2
down vote
favorite
I need help in computing the integral indicated above. What I've tried so far:
Parametrize the curve indicated by $|z|=rho$ with $gamma = z(t) = rho cos t + isin t$. Then by definition
$$
int_gamma f(z)|dz|=int_gamma f(z(t))|z'(t)| dt
$$
gives the following
begin{align}
int_gamma |z-a|^{-4} |dz| & = int_0^{2pi} |rho cos t+irho sin t-a_1-ia_2|^{-4}rho dt\
& = rhoint_0^{2pi}frac{1}{(rho^2-2a_1cos t-2a_2sin t + a_1^2+a_2^2)^2} dt
end{align}
Where $a=a_1+ia_2$. It's not hard to see how this becomes complicated very easily. I want to know if there is some sort of 'trick' I'm not aware of or something I might be missing.
integration complex-analysis line-integrals
asked Nov 14 at 23:53
D. Brito
345110
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I need help in computing the integral indicated above. What I've tried so far:
Parametrize the curve indicated by $|z|=rho$ with $gamma = z(t) = rho cos t + isin t$. Then by definition
$$
int_gamma f(z)|dz|=int_gamma f(z(t))|z'(t)| dt
$$
gives the following
begin{align}
int_gamma |z-a|^{-4} |dz| & = int_0^{2pi} |rho cos t+irho sin t-a_1-ia_2|^{-4}rho dt\
& = rhoint_0^{2pi}frac{1}{(rho^2-2a_1cos t-2a_2sin t + a_1^2+a_2^2)^2} dt
end{align}
Where $a=a_1+ia_2$. It's not hard to see how this becomes complicated very easily. I want to know if there is some sort of 'trick' I'm not aware of or something I might be missing.
integration complex-analysis line-integrals
asked Nov 14 at 23:53
D. Brito
345110
I need help in computing the integral indicated above. What I've tried so far:
Parametrize the curve indicated by $|z|=rho$ with $gamma = z(t) = rho cos t + isin t$. Then by definition
$$
int_gamma f(z)|dz|=int_gamma f(z(t))|z'(t)| dt
$$
gives the following
begin{align}
int_gamma |z-a|^{-4} |dz| & = int_0^{2pi} |rho cos t+irho sin t-a_1-ia_2|^{-4}rho dt\
& = rhoint_0^{2pi}frac{1}{(rho^2-2a_1cos t-2a_2sin t + a_1^2+a_2^2)^2} dt
end{align}
Where $a=a_1+ia_2$. It's not hard to see how this becomes complicated very easily. I want to know if there is some sort of 'trick' I'm not aware of or something I might be missing.
integration complex-analysis line-integrals
integration complex-analysis line-integrals
asked Nov 14 at 23:53
D. Brito
345110
asked Nov 14 at 23:53
D. Brito
345110
asked Nov 14 at 23:53
D. Brito
345110
asked Nov 14 at 23:53
D. Brito
345110
asked Nov 14 at 23:53
D. Brito
345110
345110
add a comment |
add a comment |
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