What type of singularity does $frac{e^{iaz}-e^{-z}}{z}$ have at $z=0$?
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What type of singularity does $frac{e^{iaz}-e^{-z}}{z}$ have at $z=0$? It is not a pole since the numerator is zero when $z=0$. The derivatives of this function always have a $z$ at the denominator, so we can't take the limit as $zto 0$. Is it
a removable singularity or an essential singularity? Why?
complex-analysis analysis
asked Dec 8 '18 at 1:19
user398843user398843
643215
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add a comment |
$begingroup$
What type of singularity does $frac{e^{iaz}-e^{-z}}{z}$ have at $z=0$? It is not a pole since the numerator is zero when $z=0$. The derivatives of this function always have a $z$ at the denominator, so we can't take the limit as $zto 0$. Is it
a removable singularity or an essential singularity? Why?
complex-analysis analysis
asked Dec 8 '18 at 1:19
user398843user398843
643215
$endgroup$
$begingroup$
Don't think it has a singularity atall, unless I've made a bit of a blunder. It appears to me to have a perfectly welldefined limit at z=0: -1+ia .
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– AmbretteOrrisey
Dec 8 '18 at 2:09
add a comment |
$begingroup$
What type of singularity does $frac{e^{iaz}-e^{-z}}{z}$ have at $z=0$? It is not a pole since the numerator is zero when $z=0$. The derivatives of this function always have a $z$ at the denominator, so we can't take the limit as $zto 0$. Is it
a removable singularity or an essential singularity? Why?
complex-analysis analysis
asked Dec 8 '18 at 1:19
user398843user398843
643215
$endgroup$
What type of singularity does $frac{e^{iaz}-e^{-z}}{z}$ have at $z=0$? It is not a pole since the numerator is zero when $z=0$. The derivatives of this function always have a $z$ at the denominator, so we can't take the limit as $zto 0$. Is it
a removable singularity or an essential singularity? Why?
complex-analysis analysis
complex-analysis analysis
asked Dec 8 '18 at 1:19
user398843user398843
643215
asked Dec 8 '18 at 1:19
user398843user398843
643215
edited Dec 8 '18 at 1:42
user398843
asked Dec 8 '18 at 1:19
user398843user398843
643215
asked Dec 8 '18 at 1:19
user398843user398843
643215
asked Dec 8 '18 at 1:19
user398843user398843
643215
643215
$begingroup$
Don't think it has a singularity atall, unless I've made a bit of a blunder. It appears to me to have a perfectly welldefined limit at z=0: -1+ia .
$endgroup$
– AmbretteOrrisey
Dec 8 '18 at 2:09
add a comment |
$begingroup$
Don't think it has a singularity atall, unless I've made a bit of a blunder. It appears to me to have a perfectly welldefined limit at z=0: -1+ia .
$endgroup$
– AmbretteOrrisey
Dec 8 '18 at 2:09
$begingroup$
Don't think it has a singularity atall, unless I've made a bit of a blunder. It appears to me to have a perfectly welldefined limit at z=0: -1+ia .
$endgroup$
– AmbretteOrrisey
Dec 8 '18 at 2:09
$begingroup$
Don't think it has a singularity atall, unless I've made a bit of a blunder. It appears to me to have a perfectly welldefined limit at z=0: -1+ia .
$endgroup$
– AmbretteOrrisey
Dec 8 '18 at 2:09
add a comment |
1 Answer
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This is a removable singularity. This can be seen by the power series representation of this function. Using
$$e^z=sum_{n=0}^inftyfrac{z^n}{n!}$$
we can deduce that
$$frac{e^{iaz}-e^{-z}}{z}=frac{1}{z}sum_{n=0}^inftyfrac{(iaz)^n-(-z)^n}{n!}=sum_{n=0}^inftyfrac{(ia)(iaz)^n+(-z)^n}{(n+1)!}$$
which is a continuous function. Thus, the singularity is removable.
answered Dec 8 '18 at 2:05
Josh B.Josh B.
8748
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add a comment |
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1 Answer
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$begingroup$
This is a removable singularity. This can be seen by the power series representation of this function. Using
$$e^z=sum_{n=0}^inftyfrac{z^n}{n!}$$
we can deduce that
$$frac{e^{iaz}-e^{-z}}{z}=frac{1}{z}sum_{n=0}^inftyfrac{(iaz)^n-(-z)^n}{n!}=sum_{n=0}^inftyfrac{(ia)(iaz)^n+(-z)^n}{(n+1)!}$$
which is a continuous function. Thus, the singularity is removable.
answered Dec 8 '18 at 2:05
Josh B.Josh B.
8748
$endgroup$
add a comment |
$begingroup$
This is a removable singularity. This can be seen by the power series representation of this function. Using
$$e^z=sum_{n=0}^inftyfrac{z^n}{n!}$$
we can deduce that
$$frac{e^{iaz}-e^{-z}}{z}=frac{1}{z}sum_{n=0}^inftyfrac{(iaz)^n-(-z)^n}{n!}=sum_{n=0}^inftyfrac{(ia)(iaz)^n+(-z)^n}{(n+1)!}$$
which is a continuous function. Thus, the singularity is removable.
answered Dec 8 '18 at 2:05
Josh B.Josh B.
8748
$endgroup$
add a comment |
$begingroup$
This is a removable singularity. This can be seen by the power series representation of this function. Using
$$e^z=sum_{n=0}^inftyfrac{z^n}{n!}$$
we can deduce that
$$frac{e^{iaz}-e^{-z}}{z}=frac{1}{z}sum_{n=0}^inftyfrac{(iaz)^n-(-z)^n}{n!}=sum_{n=0}^inftyfrac{(ia)(iaz)^n+(-z)^n}{(n+1)!}$$
which is a continuous function. Thus, the singularity is removable.
answered Dec 8 '18 at 2:05
Josh B.Josh B.
8748
$endgroup$
This is a removable singularity. This can be seen by the power series representation of this function. Using
$$e^z=sum_{n=0}^inftyfrac{z^n}{n!}$$
we can deduce that
$$frac{e^{iaz}-e^{-z}}{z}=frac{1}{z}sum_{n=0}^inftyfrac{(iaz)^n-(-z)^n}{n!}=sum_{n=0}^inftyfrac{(ia)(iaz)^n+(-z)^n}{(n+1)!}$$
which is a continuous function. Thus, the singularity is removable.
answered Dec 8 '18 at 2:05
Josh B.Josh B.
8748
answered Dec 8 '18 at 2:05
Josh B.Josh B.
8748
answered Dec 8 '18 at 2:05
Josh B.Josh B.
8748
answered Dec 8 '18 at 2:05
Josh B.Josh B.
8748
8748
add a comment |
add a comment |
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$begingroup$
Don't think it has a singularity atall, unless I've made a bit of a blunder. It appears to me to have a perfectly welldefined limit at z=0: -1+ia .
$endgroup$
– AmbretteOrrisey
Dec 8 '18 at 2:09