Find the residue of the function $f(z) = frac{z^3}{(z-1)(z^4+2)}$ at $z=0$
Problem: Calculate the residue of the function $f(z) = frac{z^3}{(z-1)(z^4+2)}$ 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
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Problem: Calculate the residue of the function $f(z) = frac{z^3}{(z-1)(z^4+2)}$ 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
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 |
Problem: Calculate the residue of the function $f(z) = frac{z^3}{(z-1)(z^4+2)}$ 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
Problem: Calculate the residue of the function $f(z) = frac{z^3}{(z-1)(z^4+2)}$ 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
complex-analysis residue-calculus
edited Nov 30 at 20:19
asked Jul 24 at 22:01
Fiticous
17218
17218
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 |
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 |
1 Answer
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No, $0$ is not a singularity of $f$. Therefore, the residue at $0$ is $0$.
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1 Answer
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1 Answer
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No, $0$ is not a singularity of $f$. Therefore, the residue at $0$ is $0$.
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No, $0$ is not a singularity of $f$. Therefore, the residue at $0$ is $0$.
add a comment |
No, $0$ is not a singularity of $f$. Therefore, the residue at $0$ is $0$.
No, $0$ is not a singularity of $f$. Therefore, the residue at $0$ is $0$.
answered Jul 24 at 22:04
José Carlos Santos
149k22119221
149k22119221
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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