Mixing
Compute: $int_{partial B(0,r)} frac{(z-a)dz}{(z+a)}$
I want to calculate the complex integral:
$$int_{partial B(0,1)} frac{(z-a)dz}{(z+a)}$$
When $|a|<1$ and $|a|>1$.
In the first case I believe the Cauchy integral formula can be applied with $f(z)=z-a$. I get $-4pi ia$ as an answer then. Is there any way of easily computing the second case, other than going through the long process of computing the integral?
complex-analysis
asked Nov 21 '18 at 1:12
Dole
906514
add a comment |
I want to calculate the complex integral:
$$int_{partial B(0,1)} frac{(z-a)dz}{(z+a)}$$
When $|a|<1$ and $|a|>1$.
In the first case I believe the Cauchy integral formula can be applied with $f(z)=z-a$. I get $-4pi ia$ as an answer then. Is there any way of easily computing the second case, other than going through the long process of computing the integral?
complex-analysis
asked Nov 21 '18 at 1:12
Dole
906514
Well for $|a|>1$ you can expand the denominator in terms of powers $z/a$ and use the general formula for powers of $z$ whereat no negative power appears, so it must be zero. Doesn't seem like a long process.
– Diger
Nov 21 '18 at 1:40
add a comment |
I want to calculate the complex integral:
$$int_{partial B(0,1)} frac{(z-a)dz}{(z+a)}$$
When $|a|<1$ and $|a|>1$.
In the first case I believe the Cauchy integral formula can be applied with $f(z)=z-a$. I get $-4pi ia$ as an answer then. Is there any way of easily computing the second case, other than going through the long process of computing the integral?
complex-analysis
asked Nov 21 '18 at 1:12
Dole
906514
I want to calculate the complex integral:
$$int_{partial B(0,1)} frac{(z-a)dz}{(z+a)}$$
When $|a|<1$ and $|a|>1$.
In the first case I believe the Cauchy integral formula can be applied with $f(z)=z-a$. I get $-4pi ia$ as an answer then. Is there any way of easily computing the second case, other than going through the long process of computing the integral?
complex-analysis
complex-analysis
asked Nov 21 '18 at 1:12
Dole
906514
asked Nov 21 '18 at 1:12
Dole
906514
asked Nov 21 '18 at 1:12
Dole
906514
asked Nov 21 '18 at 1:12
Dole
906514
asked Nov 21 '18 at 1:12
Dole
906514
906514
Well for $|a|>1$ you can expand the denominator in terms of powers $z/a$ and use the general formula for powers of $z$ whereat no negative power appears, so it must be zero. Doesn't seem like a long process.
– Diger
Nov 21 '18 at 1:40
add a comment |
Well for $|a|>1$ you can expand the denominator in terms of powers $z/a$ and use the general formula for powers of $z$ whereat no negative power appears, so it must be zero. Doesn't seem like a long process.
– Diger
Nov 21 '18 at 1:40
Well for $|a|>1$ you can expand the denominator in terms of powers $z/a$ and use the general formula for powers of $z$ whereat no negative power appears, so it must be zero. Doesn't seem like a long process.
– Diger
Nov 21 '18 at 1:40
Well for $|a|>1$ you can expand the denominator in terms of powers $z/a$ and use the general formula for powers of $z$ whereat no negative power appears, so it must be zero. Doesn't seem like a long process.
– Diger
Nov 21 '18 at 1:40
add a comment |
1 Answer
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When $|a|>1$, the integrand $frac{z-a}{z+a}$ is holomorphic on $B(0,1)$, therefore by Cauchy’s integral theorem the integral equals zero.
answered Nov 21 '18 at 3:46
Szeto
6,4362926
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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oldest
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When $|a|>1$, the integrand $frac{z-a}{z+a}$ is holomorphic on $B(0,1)$, therefore by Cauchy’s integral theorem the integral equals zero.
answered Nov 21 '18 at 3:46
Szeto
6,4362926
add a comment |
When $|a|>1$, the integrand $frac{z-a}{z+a}$ is holomorphic on $B(0,1)$, therefore by Cauchy’s integral theorem the integral equals zero.
answered Nov 21 '18 at 3:46
Szeto
6,4362926
add a comment |
When $|a|>1$, the integrand $frac{z-a}{z+a}$ is holomorphic on $B(0,1)$, therefore by Cauchy’s integral theorem the integral equals zero.
answered Nov 21 '18 at 3:46
Szeto
6,4362926
When $|a|>1$, the integrand $frac{z-a}{z+a}$ is holomorphic on $B(0,1)$, therefore by Cauchy’s integral theorem the integral equals zero.
answered Nov 21 '18 at 3:46
Szeto
6,4362926
answered Nov 21 '18 at 3:46
Szeto
6,4362926
answered Nov 21 '18 at 3:46
Szeto
6,4362926
answered Nov 21 '18 at 3:46
Szeto
6,4362926
6,4362926
add a comment |
add a comment |
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Well for $|a|>1$ you can expand the denominator in terms of powers $z/a$ and use the general formula for powers of $z$ whereat no negative power appears, so it must be zero. Doesn't seem like a long process.
– Diger
Nov 21 '18 at 1:40