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Find the residue of the function $f(z) = fracz^3(z-1)(z^4+2)$ at $z=0$
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Problem: Calculate the residue of the function $f(z) = fracz^3(z-1)(z^4+2)$ at $z=0$
I am confused on where to even start on this question since there is no pole at z=0. Is z=0 even a singularity of this function?
complex-analysis residue-calculus
asked Jul 24 at 22:01
Fiticous
84
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up vote
1
down vote
favorite
Problem: Calculate the residue of the function $f(z) = fracz^3(z-1)(z^4+2)$ at $z=0$
I am confused on where to even start on this question since there is no pole at z=0. Is z=0 even a singularity of this function?
complex-analysis residue-calculus
asked Jul 24 at 22:01
Fiticous
84
Suppose you were a mathematician in the early 19th century and you had defined the concept of a residue at the poles of meromorphic functions. How would you have generalized your definition in a consistent way such that it also applies to points where the function doesn't have a pole?
– Count Iblis
Jul 24 at 22:07
1
@CountIblis I am guessing your wanting me to notice that if a function $f$ doesn't have a pole at $z_0$, then the coefficient $a_-1$ in the Laurent expansion $f(z) = sum_n=-infty^infty a_n(z - z_0)^n$ is always going to be zero?
– Fiticous
Jul 24 at 22:17
That's right! :)
– Count Iblis
Jul 24 at 23:12
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Problem: Calculate the residue of the function $f(z) = fracz^3(z-1)(z^4+2)$ at $z=0$
I am confused on where to even start on this question since there is no pole at z=0. Is z=0 even a singularity of this function?
complex-analysis residue-calculus
asked Jul 24 at 22:01
Fiticous
84
Problem: Calculate the residue of the function $f(z) = fracz^3(z-1)(z^4+2)$ at $z=0$
I am confused on where to even start on this question since there is no pole at z=0. Is z=0 even a singularity of this function?
complex-analysis residue-calculus
asked Jul 24 at 22:01
Fiticous
84
asked Jul 24 at 22:01
Fiticous
84
asked Jul 24 at 22:01
Fiticous
84
asked Jul 24 at 22:01
Fiticous
84
84
Suppose you were a mathematician in the early 19th century and you had defined the concept of a residue at the poles of meromorphic functions. How would you have generalized your definition in a consistent way such that it also applies to points where the function doesn't have a pole?
– Count Iblis
Jul 24 at 22:07
1
@CountIblis I am guessing your wanting me to notice that if a function $f$ doesn't have a pole at $z_0$, then the coefficient $a_-1$ in the Laurent expansion $f(z) = sum_n=-infty^infty a_n(z - z_0)^n$ is always going to be zero?
– Fiticous
Jul 24 at 22:17
That's right! :)
– Count Iblis
Jul 24 at 23:12
add a comment |Â
Suppose you were a mathematician in the early 19th century and you had defined the concept of a residue at the poles of meromorphic functions. How would you have generalized your definition in a consistent way such that it also applies to points where the function doesn't have a pole?
– Count Iblis
Jul 24 at 22:07
1
@CountIblis I am guessing your wanting me to notice that if a function $f$ doesn't have a pole at $z_0$, then the coefficient $a_-1$ in the Laurent expansion $f(z) = sum_n=-infty^infty a_n(z - z_0)^n$ is always going to be zero?
– Fiticous
Jul 24 at 22:17
That's right! :)
– Count Iblis
Jul 24 at 23:12
Suppose you were a mathematician in the early 19th century and you had defined the concept of a residue at the poles of meromorphic functions. How would you have generalized your definition in a consistent way such that it also applies to points where the function doesn't have a pole?
– Count Iblis
Jul 24 at 22:07
Suppose you were a mathematician in the early 19th century and you had defined the concept of a residue at the poles of meromorphic functions. How would you have generalized your definition in a consistent way such that it also applies to points where the function doesn't have a pole?
– Count Iblis
Jul 24 at 22:07
1
1
@CountIblis I am guessing your wanting me to notice that if a function $f$ doesn't have a pole at $z_0$, then the coefficient $a_-1$ in the Laurent expansion $f(z) = sum_n=-infty^infty a_n(z - z_0)^n$ is always going to be zero?
– Fiticous
Jul 24 at 22:17
@CountIblis I am guessing your wanting me to notice that if a function $f$ doesn't have a pole at $z_0$, then the coefficient $a_-1$ in the Laurent expansion $f(z) = sum_n=-infty^infty a_n(z - z_0)^n$ is always going to be zero?
– Fiticous
Jul 24 at 22:17
That's right! :)
– Count Iblis
Jul 24 at 23:12
That's right! :)
– Count Iblis
Jul 24 at 23:12
add a comment |Â
1 Answer
1
active
oldest
votes
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1
down vote
accepted
No, $0$ is not a singularity of $f$. Therefore, the residue at $0$ is $0$.
answered Jul 24 at 22:04
José Carlos Santos
113k1697176
Thanks, that makes sense, should have been obvious.
– Fiticous
Jul 24 at 22:07
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
accepted
No, $0$ is not a singularity of $f$. Therefore, the residue at $0$ is $0$.
answered Jul 24 at 22:04
José Carlos Santos
113k1697176
Thanks, that makes sense, should have been obvious.
– Fiticous
Jul 24 at 22:07
add a comment |Â
up vote
1
down vote
accepted
No, $0$ is not a singularity of $f$. Therefore, the residue at $0$ is $0$.
answered Jul 24 at 22:04
José Carlos Santos
113k1697176
Thanks, that makes sense, should have been obvious.
– Fiticous
Jul 24 at 22:07
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
No, $0$ is not a singularity of $f$. Therefore, the residue at $0$ is $0$.
answered Jul 24 at 22:04
José Carlos Santos
113k1697176
No, $0$ is not a singularity of $f$. Therefore, the residue at $0$ is $0$.
answered Jul 24 at 22:04
José Carlos Santos
113k1697176
answered Jul 24 at 22:04
José Carlos Santos
113k1697176
answered Jul 24 at 22:04
José Carlos Santos
113k1697176
answered Jul 24 at 22:04
José Carlos Santos
113k1697176
113k1697176
Thanks, that makes sense, should have been obvious.
– Fiticous
Jul 24 at 22:07
add a comment |Â
Thanks, that makes sense, should have been obvious.
– Fiticous
Jul 24 at 22:07
Thanks, that makes sense, should have been obvious.
– Fiticous
Jul 24 at 22:07
Thanks, that makes sense, should have been obvious.
– Fiticous
Jul 24 at 22:07
add a comment |Â
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Suppose you were a mathematician in the early 19th century and you had defined the concept of a residue at the poles of meromorphic functions. How would you have generalized your definition in a consistent way such that it also applies to points where the function doesn't have a pole?
– Count Iblis
Jul 24 at 22:07
1
@CountIblis I am guessing your wanting me to notice that if a function $f$ doesn't have a pole at $z_0$, then the coefficient $a_-1$ in the Laurent expansion $f(z) = sum_n=-infty^infty a_n(z - z_0)^n$ is always going to be zero?
– Fiticous
Jul 24 at 22:17
That's right! :)
– Count Iblis
Jul 24 at 23:12