Find order of pole $frac{e^z -1}{z^2 +4}$, about $z=2i$.












2












$begingroup$


$$frac{e^z -1}{z^2 +4},quadtext{about $z=2i$.}$$
The textbook I'm reading isn't specific about these case, only gives basic examples.
Basically to find pole I'd have to expand a Laurent series about some point, in this case $z=2i$, I did $$ frac{e^{2i} e^{z-2i} -1}{(z-2i)^2 +4-(4z-4)}$$ and from there just expanded $e^{z-2i}$ as a series, just see which powers (without caring about constants) of $z-2i$ was left after canceling, giving me a pole of order 2. Answer however says of order 1.






complex-analysis laurent-series singularity







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$endgroup$








  • 1




    $begingroup$
    $$frac{e^z-1}{z^2+4}=frac{frac{e^z-1}{z+2i}}{z-2i}=frac{h(z)}{z-2i}$$ with $h(2i) neq 0$.
    $endgroup$
    – Chinnapparaj R
    Jan 22 at 6:37
















2












$begingroup$


$$frac{e^z -1}{z^2 +4},quadtext{about $z=2i$.}$$
The textbook I'm reading isn't specific about these case, only gives basic examples.
Basically to find pole I'd have to expand a Laurent series about some point, in this case $z=2i$, I did $$ frac{e^{2i} e^{z-2i} -1}{(z-2i)^2 +4-(4z-4)}$$ and from there just expanded $e^{z-2i}$ as a series, just see which powers (without caring about constants) of $z-2i$ was left after canceling, giving me a pole of order 2. Answer however says of order 1.






complex-analysis laurent-series singularity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $$frac{e^z-1}{z^2+4}=frac{frac{e^z-1}{z+2i}}{z-2i}=frac{h(z)}{z-2i}$$ with $h(2i) neq 0$.
    $endgroup$
    – Chinnapparaj R
    Jan 22 at 6:37














2












2








2





$begingroup$


$$frac{e^z -1}{z^2 +4},quadtext{about $z=2i$.}$$
The textbook I'm reading isn't specific about these case, only gives basic examples.
Basically to find pole I'd have to expand a Laurent series about some point, in this case $z=2i$, I did $$ frac{e^{2i} e^{z-2i} -1}{(z-2i)^2 +4-(4z-4)}$$ and from there just expanded $e^{z-2i}$ as a series, just see which powers (without caring about constants) of $z-2i$ was left after canceling, giving me a pole of order 2. Answer however says of order 1.






complex-analysis laurent-series singularity







share|cite|improve this question











$endgroup$




$$frac{e^z -1}{z^2 +4},quadtext{about $z=2i$.}$$
The textbook I'm reading isn't specific about these case, only gives basic examples.
Basically to find pole I'd have to expand a Laurent series about some point, in this case $z=2i$, I did $$ frac{e^{2i} e^{z-2i} -1}{(z-2i)^2 +4-(4z-4)}$$ and from there just expanded $e^{z-2i}$ as a series, just see which powers (without caring about constants) of $z-2i$ was left after canceling, giving me a pole of order 2. Answer however says of order 1.






complex-analysis laurent-series singularity




complex-analysis laurent-series singularity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 6:53









El borito

666216




666216










asked Jan 22 at 6:34









MinYoung KimMinYoung Kim

907




907








  • 1




    $begingroup$
    $$frac{e^z-1}{z^2+4}=frac{frac{e^z-1}{z+2i}}{z-2i}=frac{h(z)}{z-2i}$$ with $h(2i) neq 0$.
    $endgroup$
    – Chinnapparaj R
    Jan 22 at 6:37














  • 1




    $begingroup$
    $$frac{e^z-1}{z^2+4}=frac{frac{e^z-1}{z+2i}}{z-2i}=frac{h(z)}{z-2i}$$ with $h(2i) neq 0$.
    $endgroup$
    – Chinnapparaj R
    Jan 22 at 6:37








1




1




$begingroup$
$$frac{e^z-1}{z^2+4}=frac{frac{e^z-1}{z+2i}}{z-2i}=frac{h(z)}{z-2i}$$ with $h(2i) neq 0$.
$endgroup$
– Chinnapparaj R
Jan 22 at 6:37




$begingroup$
$$frac{e^z-1}{z^2+4}=frac{frac{e^z-1}{z+2i}}{z-2i}=frac{h(z)}{z-2i}$$ with $h(2i) neq 0$.
$endgroup$
– Chinnapparaj R
Jan 22 at 6:37










1 Answer
1






active

oldest

votes


















3












$begingroup$

$z^2+4$ has in $2i$ a simple zero and $e^{2i}-1 ne 0$, hence $2i$ is a simple pole of $displaystylefrac{e^z -1}{z^2 +4}$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you again Fred. If the numerator was $= 0$, then what would we have to do? Say $frac{z^3-1}{(z-1)^3}$ at z = 1
    $endgroup$
    – MinYoung Kim
    Jan 22 at 6:44












  • $begingroup$
    nevermind, I understand why you cared about the numerator. in this case, it would e $frac{0}{0^3}$ and have one of the pole order is removable singularity, so pole would be order of 2.
    $endgroup$
    – MinYoung Kim
    Jan 22 at 7:08










  • $begingroup$
    $frac{z^3-1}{(z-1)^3}=frac{z^2+z+1}{(z-1)^2}$, hence $frac{z^3-1}{(z-1)^3}$ has in $1$ a pole of order $2$.
    $endgroup$
    – Fred
    Jan 22 at 8:48











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1 Answer
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1 Answer
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active

oldest

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3












$begingroup$

$z^2+4$ has in $2i$ a simple zero and $e^{2i}-1 ne 0$, hence $2i$ is a simple pole of $displaystylefrac{e^z -1}{z^2 +4}$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you again Fred. If the numerator was $= 0$, then what would we have to do? Say $frac{z^3-1}{(z-1)^3}$ at z = 1
    $endgroup$
    – MinYoung Kim
    Jan 22 at 6:44












  • $begingroup$
    nevermind, I understand why you cared about the numerator. in this case, it would e $frac{0}{0^3}$ and have one of the pole order is removable singularity, so pole would be order of 2.
    $endgroup$
    – MinYoung Kim
    Jan 22 at 7:08










  • $begingroup$
    $frac{z^3-1}{(z-1)^3}=frac{z^2+z+1}{(z-1)^2}$, hence $frac{z^3-1}{(z-1)^3}$ has in $1$ a pole of order $2$.
    $endgroup$
    – Fred
    Jan 22 at 8:48
















3












$begingroup$

$z^2+4$ has in $2i$ a simple zero and $e^{2i}-1 ne 0$, hence $2i$ is a simple pole of $displaystylefrac{e^z -1}{z^2 +4}$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you again Fred. If the numerator was $= 0$, then what would we have to do? Say $frac{z^3-1}{(z-1)^3}$ at z = 1
    $endgroup$
    – MinYoung Kim
    Jan 22 at 6:44












  • $begingroup$
    nevermind, I understand why you cared about the numerator. in this case, it would e $frac{0}{0^3}$ and have one of the pole order is removable singularity, so pole would be order of 2.
    $endgroup$
    – MinYoung Kim
    Jan 22 at 7:08










  • $begingroup$
    $frac{z^3-1}{(z-1)^3}=frac{z^2+z+1}{(z-1)^2}$, hence $frac{z^3-1}{(z-1)^3}$ has in $1$ a pole of order $2$.
    $endgroup$
    – Fred
    Jan 22 at 8:48














3












3








3





$begingroup$

$z^2+4$ has in $2i$ a simple zero and $e^{2i}-1 ne 0$, hence $2i$ is a simple pole of $displaystylefrac{e^z -1}{z^2 +4}$.






share|cite|improve this answer











$endgroup$



$z^2+4$ has in $2i$ a simple zero and $e^{2i}-1 ne 0$, hence $2i$ is a simple pole of $displaystylefrac{e^z -1}{z^2 +4}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 22 at 7:54









El borito

666216




666216










answered Jan 22 at 6:39









FredFred

47.8k1849




47.8k1849








  • 1




    $begingroup$
    Thank you again Fred. If the numerator was $= 0$, then what would we have to do? Say $frac{z^3-1}{(z-1)^3}$ at z = 1
    $endgroup$
    – MinYoung Kim
    Jan 22 at 6:44












  • $begingroup$
    nevermind, I understand why you cared about the numerator. in this case, it would e $frac{0}{0^3}$ and have one of the pole order is removable singularity, so pole would be order of 2.
    $endgroup$
    – MinYoung Kim
    Jan 22 at 7:08










  • $begingroup$
    $frac{z^3-1}{(z-1)^3}=frac{z^2+z+1}{(z-1)^2}$, hence $frac{z^3-1}{(z-1)^3}$ has in $1$ a pole of order $2$.
    $endgroup$
    – Fred
    Jan 22 at 8:48














  • 1




    $begingroup$
    Thank you again Fred. If the numerator was $= 0$, then what would we have to do? Say $frac{z^3-1}{(z-1)^3}$ at z = 1
    $endgroup$
    – MinYoung Kim
    Jan 22 at 6:44












  • $begingroup$
    nevermind, I understand why you cared about the numerator. in this case, it would e $frac{0}{0^3}$ and have one of the pole order is removable singularity, so pole would be order of 2.
    $endgroup$
    – MinYoung Kim
    Jan 22 at 7:08










  • $begingroup$
    $frac{z^3-1}{(z-1)^3}=frac{z^2+z+1}{(z-1)^2}$, hence $frac{z^3-1}{(z-1)^3}$ has in $1$ a pole of order $2$.
    $endgroup$
    – Fred
    Jan 22 at 8:48








1




1




$begingroup$
Thank you again Fred. If the numerator was $= 0$, then what would we have to do? Say $frac{z^3-1}{(z-1)^3}$ at z = 1
$endgroup$
– MinYoung Kim
Jan 22 at 6:44






$begingroup$
Thank you again Fred. If the numerator was $= 0$, then what would we have to do? Say $frac{z^3-1}{(z-1)^3}$ at z = 1
$endgroup$
– MinYoung Kim
Jan 22 at 6:44














$begingroup$
nevermind, I understand why you cared about the numerator. in this case, it would e $frac{0}{0^3}$ and have one of the pole order is removable singularity, so pole would be order of 2.
$endgroup$
– MinYoung Kim
Jan 22 at 7:08




$begingroup$
nevermind, I understand why you cared about the numerator. in this case, it would e $frac{0}{0^3}$ and have one of the pole order is removable singularity, so pole would be order of 2.
$endgroup$
– MinYoung Kim
Jan 22 at 7:08












$begingroup$
$frac{z^3-1}{(z-1)^3}=frac{z^2+z+1}{(z-1)^2}$, hence $frac{z^3-1}{(z-1)^3}$ has in $1$ a pole of order $2$.
$endgroup$
– Fred
Jan 22 at 8:48




$begingroup$
$frac{z^3-1}{(z-1)^3}=frac{z^2+z+1}{(z-1)^2}$, hence $frac{z^3-1}{(z-1)^3}$ has in $1$ a pole of order $2$.
$endgroup$
– Fred
Jan 22 at 8:48


















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