find order of pole
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$f(z) = frac12-e^z$
I found that $f$ is holomorphic on $mathbbC-A$, where $A$ is the discrete set $A = log(2) + 2pi ik, k in mathbbZ $. If I now take $z_k = log(2) + 2pi ik$, then $z_k$ is a pole, since $lim_zto z_k |f(x)| = infty$.
Trying to find the order of the pole $z_k$, I got stuck in the computation of $lim_zto z_k fracz-z_k2-e^z$.
I know that the solution should be $lim_zto z_k fracz-z_k2-e^z = lim_zto z_k frac1-e^z = lim_zto z_k -frac12$. So the pole is of order 1. But I don't see how I can get there.
Thanks for any hints.
complex-analysis complex-numbers
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up vote
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$f(z) = frac12-e^z$
I found that $f$ is holomorphic on $mathbbC-A$, where $A$ is the discrete set $A = log(2) + 2pi ik, k in mathbbZ $. If I now take $z_k = log(2) + 2pi ik$, then $z_k$ is a pole, since $lim_zto z_k |f(x)| = infty$.
Trying to find the order of the pole $z_k$, I got stuck in the computation of $lim_zto z_k fracz-z_k2-e^z$.
I know that the solution should be $lim_zto z_k fracz-z_k2-e^z = lim_zto z_k frac1-e^z = lim_zto z_k -frac12$. So the pole is of order 1. But I don't see how I can get there.
Thanks for any hints.
complex-analysis complex-numbers
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$f(z) = frac12-e^z$
I found that $f$ is holomorphic on $mathbbC-A$, where $A$ is the discrete set $A = log(2) + 2pi ik, k in mathbbZ $. If I now take $z_k = log(2) + 2pi ik$, then $z_k$ is a pole, since $lim_zto z_k |f(x)| = infty$.
Trying to find the order of the pole $z_k$, I got stuck in the computation of $lim_zto z_k fracz-z_k2-e^z$.
I know that the solution should be $lim_zto z_k fracz-z_k2-e^z = lim_zto z_k frac1-e^z = lim_zto z_k -frac12$. So the pole is of order 1. But I don't see how I can get there.
Thanks for any hints.
complex-analysis complex-numbers
$f(z) = frac12-e^z$
I found that $f$ is holomorphic on $mathbbC-A$, where $A$ is the discrete set $A = log(2) + 2pi ik, k in mathbbZ $. If I now take $z_k = log(2) + 2pi ik$, then $z_k$ is a pole, since $lim_zto z_k |f(x)| = infty$.
Trying to find the order of the pole $z_k$, I got stuck in the computation of $lim_zto z_k fracz-z_k2-e^z$.
I know that the solution should be $lim_zto z_k fracz-z_k2-e^z = lim_zto z_k frac1-e^z = lim_zto z_k -frac12$. So the pole is of order 1. But I don't see how I can get there.
Thanks for any hints.
complex-analysis complex-numbers
edited Aug 14 at 12:04
Bernard
111k635103
111k635103
asked Aug 14 at 11:53
Livpez.
826
826
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2 Answers
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active
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up vote
4
down vote
Just notice that, taken any pole $z_k$, $2 = e^z_k$, hence the limit
$$
lim_z to z_k frac e^z - 2z - z_k
=
lim_z to z_k frac e^z - e^z_kz - z_k
=
fracddz[e^z](z_k) = e^z_k
$$
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up vote
0
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Hint: The denominator $g(z)=2-e^z$ has only simple zeros because $g'(z)=-e^z$ is never zero.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Just notice that, taken any pole $z_k$, $2 = e^z_k$, hence the limit
$$
lim_z to z_k frac e^z - 2z - z_k
=
lim_z to z_k frac e^z - e^z_kz - z_k
=
fracddz[e^z](z_k) = e^z_k
$$
add a comment |Â
up vote
4
down vote
Just notice that, taken any pole $z_k$, $2 = e^z_k$, hence the limit
$$
lim_z to z_k frac e^z - 2z - z_k
=
lim_z to z_k frac e^z - e^z_kz - z_k
=
fracddz[e^z](z_k) = e^z_k
$$
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Just notice that, taken any pole $z_k$, $2 = e^z_k$, hence the limit
$$
lim_z to z_k frac e^z - 2z - z_k
=
lim_z to z_k frac e^z - e^z_kz - z_k
=
fracddz[e^z](z_k) = e^z_k
$$
Just notice that, taken any pole $z_k$, $2 = e^z_k$, hence the limit
$$
lim_z to z_k frac e^z - 2z - z_k
=
lim_z to z_k frac e^z - e^z_kz - z_k
=
fracddz[e^z](z_k) = e^z_k
$$
edited Aug 18 at 20:58
answered Aug 14 at 12:11
JayTuma
1,335118
1,335118
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint: The denominator $g(z)=2-e^z$ has only simple zeros because $g'(z)=-e^z$ is never zero.
add a comment |Â
up vote
0
down vote
Hint: The denominator $g(z)=2-e^z$ has only simple zeros because $g'(z)=-e^z$ is never zero.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: The denominator $g(z)=2-e^z$ has only simple zeros because $g'(z)=-e^z$ is never zero.
Hint: The denominator $g(z)=2-e^z$ has only simple zeros because $g'(z)=-e^z$ is never zero.
answered Aug 14 at 12:10
lhf
156k9161367
156k9161367
add a comment |Â
add a comment |Â
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