Find the residue of the function $f(z) = e^{(z^2)/(z^n)}$ at all its poles.
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I'm currently working on this problem, and I think there is a trick to it because of the $z^2$ in the exponent of e. I fail to see it though.
Find the residue of the function $f(z) = e^{(z^2)/(z^n)}$ at all its poles.
real-analysis complex-analysis analysis
asked Jan 4 at 15:48
user628226user628226
575
$endgroup$
add a comment |
$begingroup$
I'm currently working on this problem, and I think there is a trick to it because of the $z^2$ in the exponent of e. I fail to see it though.
Find the residue of the function $f(z) = e^{(z^2)/(z^n)}$ at all its poles.
real-analysis complex-analysis analysis
asked Jan 4 at 15:48
user628226user628226
575
$endgroup$
$begingroup$
What do we know about $n$? Is $n$ an integer?
$endgroup$
– Omnomnomnom
Jan 4 at 15:51
2
$begingroup$
What poles? This has no poles. $0$ is an essential singularity (if $n$ is an integer $>2$), or a branch point if $n$ is not an integer.
$endgroup$
– Robert Israel
Jan 4 at 15:56
add a comment |
$begingroup$
I'm currently working on this problem, and I think there is a trick to it because of the $z^2$ in the exponent of e. I fail to see it though.
Find the residue of the function $f(z) = e^{(z^2)/(z^n)}$ at all its poles.
real-analysis complex-analysis analysis
asked Jan 4 at 15:48
user628226user628226
575
$endgroup$
I'm currently working on this problem, and I think there is a trick to it because of the $z^2$ in the exponent of e. I fail to see it though.
Find the residue of the function $f(z) = e^{(z^2)/(z^n)}$ at all its poles.
real-analysis complex-analysis analysis
real-analysis complex-analysis analysis
asked Jan 4 at 15:48
user628226user628226
575
asked Jan 4 at 15:48
user628226user628226
575
asked Jan 4 at 15:48
user628226user628226
575
asked Jan 4 at 15:48
user628226user628226
575
asked Jan 4 at 15:48
user628226user628226
575
575
$begingroup$
What do we know about $n$? Is $n$ an integer?
$endgroup$
– Omnomnomnom
Jan 4 at 15:51
2
$begingroup$
What poles? This has no poles. $0$ is an essential singularity (if $n$ is an integer $>2$), or a branch point if $n$ is not an integer.
$endgroup$
– Robert Israel
Jan 4 at 15:56
add a comment |
$begingroup$
What do we know about $n$? Is $n$ an integer?
$endgroup$
– Omnomnomnom
Jan 4 at 15:51
2
$begingroup$
What poles? This has no poles. $0$ is an essential singularity (if $n$ is an integer $>2$), or a branch point if $n$ is not an integer.
$endgroup$
– Robert Israel
Jan 4 at 15:56
$begingroup$
What do we know about $n$? Is $n$ an integer?
$endgroup$
– Omnomnomnom
Jan 4 at 15:51
$begingroup$
What do we know about $n$? Is $n$ an integer?
$endgroup$
– Omnomnomnom
Jan 4 at 15:51
2
2
$begingroup$
What poles? This has no poles. $0$ is an essential singularity (if $n$ is an integer $>2$), or a branch point if $n$ is not an integer.
$endgroup$
– Robert Israel
Jan 4 at 15:56
$begingroup$
What poles? This has no poles. $0$ is an essential singularity (if $n$ is an integer $>2$), or a branch point if $n$ is not an integer.
$endgroup$
– Robert Israel
Jan 4 at 15:56
add a comment |
1 Answer
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$begingroup$
I am assuming that $n$ is supposed to be an integer.
Hint: $f$ only fails to be complex-differentiable at $z = 0$ (which is a singularity, but not a pole). Using the series $e^x = sum_k frac 1{k!}x^k$, we can express $f(z)$ as a Laurent series centered at $z = 0$. Using the Laurent series, we can find the desired residue.
answered Jan 4 at 15:55
OmnomnomnomOmnomnomnom
129k793187
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1 Answer
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1 Answer
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$begingroup$
I am assuming that $n$ is supposed to be an integer.
Hint: $f$ only fails to be complex-differentiable at $z = 0$ (which is a singularity, but not a pole). Using the series $e^x = sum_k frac 1{k!}x^k$, we can express $f(z)$ as a Laurent series centered at $z = 0$. Using the Laurent series, we can find the desired residue.
answered Jan 4 at 15:55
OmnomnomnomOmnomnomnom
129k793187
$endgroup$
add a comment |
$begingroup$
I am assuming that $n$ is supposed to be an integer.
Hint: $f$ only fails to be complex-differentiable at $z = 0$ (which is a singularity, but not a pole). Using the series $e^x = sum_k frac 1{k!}x^k$, we can express $f(z)$ as a Laurent series centered at $z = 0$. Using the Laurent series, we can find the desired residue.
answered Jan 4 at 15:55
OmnomnomnomOmnomnomnom
129k793187
$endgroup$
add a comment |
$begingroup$
I am assuming that $n$ is supposed to be an integer.
Hint: $f$ only fails to be complex-differentiable at $z = 0$ (which is a singularity, but not a pole). Using the series $e^x = sum_k frac 1{k!}x^k$, we can express $f(z)$ as a Laurent series centered at $z = 0$. Using the Laurent series, we can find the desired residue.
answered Jan 4 at 15:55
OmnomnomnomOmnomnomnom
129k793187
$endgroup$
I am assuming that $n$ is supposed to be an integer.
Hint: $f$ only fails to be complex-differentiable at $z = 0$ (which is a singularity, but not a pole). Using the series $e^x = sum_k frac 1{k!}x^k$, we can express $f(z)$ as a Laurent series centered at $z = 0$. Using the Laurent series, we can find the desired residue.
answered Jan 4 at 15:55
OmnomnomnomOmnomnomnom
129k793187
answered Jan 4 at 15:55
OmnomnomnomOmnomnomnom
129k793187
answered Jan 4 at 15:55
OmnomnomnomOmnomnomnom
129k793187
answered Jan 4 at 15:55
OmnomnomnomOmnomnomnom
129k793187
129k793187
add a comment |
add a comment |
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$begingroup$
What do we know about $n$? Is $n$ an integer?
$endgroup$
– Omnomnomnom
Jan 4 at 15:51
2
$begingroup$
What poles? This has no poles. $0$ is an essential singularity (if $n$ is an integer $>2$), or a branch point if $n$ is not an integer.
$endgroup$
– Robert Israel
Jan 4 at 15:56