What kind of a point does $f(x)=frac{e^{iz}}{z^2+1}$ have at i? [closed]
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How should I approach a problem like this, given a function:
$$f(z)=frac{e^{iz}}{z^2+1}$$
I need to find whether the limit of $|frac{e^{iz}}{z^2+1}|$ at $i$ converges to some number, infinity, or if it does not converge to infinity or any number (IE.the real arguments converge to say 0 from negative side and infinity otherwise).
What would be the most useful things to try here?
complex-analysis
asked Dec 12 '18 at 0:01
DoleDole
914514
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closed as unclear what you're asking by Did, Leucippus, user10354138, Rebellos, rtybase Dec 12 '18 at 9:34
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
How should I approach a problem like this, given a function:
$$f(z)=frac{e^{iz}}{z^2+1}$$
I need to find whether the limit of $|frac{e^{iz}}{z^2+1}|$ at $i$ converges to some number, infinity, or if it does not converge to infinity or any number (IE.the real arguments converge to say 0 from negative side and infinity otherwise).
What would be the most useful things to try here?
complex-analysis
asked Dec 12 '18 at 0:01
DoleDole
914514
$endgroup$
closed as unclear what you're asking by Did, Leucippus, user10354138, Rebellos, rtybase Dec 12 '18 at 9:34
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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You should use the fact that $f(z)$ you described is continuous at $i$.
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– NL1992
Dec 12 '18 at 0:03
2
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Plug in first, does $z=i$ look like a special point?
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– Michael Burr
Dec 12 '18 at 0:04
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@MichaelBurr Sorry, a typo.
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– Dole
Dec 12 '18 at 0:20
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Don't understand the problem here. Plugging in (or taking the limit) $z=i $ is the obvious thing to try right?
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– Winther
Dec 12 '18 at 1:00
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@Winther Yes it goes to infinity, but does it go to somewhere else like zero as well approached from a different direction? To me it seems that it does given I think it might lead to a contradiction if i was a pole.
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– Dole
Dec 12 '18 at 1:02
|
show 1 more comment
$begingroup$
How should I approach a problem like this, given a function:
$$f(z)=frac{e^{iz}}{z^2+1}$$
I need to find whether the limit of $|frac{e^{iz}}{z^2+1}|$ at $i$ converges to some number, infinity, or if it does not converge to infinity or any number (IE.the real arguments converge to say 0 from negative side and infinity otherwise).
What would be the most useful things to try here?
complex-analysis
asked Dec 12 '18 at 0:01
DoleDole
914514
$endgroup$
How should I approach a problem like this, given a function:
$$f(z)=frac{e^{iz}}{z^2+1}$$
I need to find whether the limit of $|frac{e^{iz}}{z^2+1}|$ at $i$ converges to some number, infinity, or if it does not converge to infinity or any number (IE.the real arguments converge to say 0 from negative side and infinity otherwise).
What would be the most useful things to try here?
complex-analysis
complex-analysis
asked Dec 12 '18 at 0:01
DoleDole
914514
asked Dec 12 '18 at 0:01
DoleDole
914514
edited Dec 12 '18 at 0:30
Dole
asked Dec 12 '18 at 0:01
DoleDole
914514
asked Dec 12 '18 at 0:01
DoleDole
914514
asked Dec 12 '18 at 0:01
DoleDole
914514
914514
closed as unclear what you're asking by Did, Leucippus, user10354138, Rebellos, rtybase Dec 12 '18 at 9:34
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Did, Leucippus, user10354138, Rebellos, rtybase Dec 12 '18 at 9:34
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
You should use the fact that $f(z)$ you described is continuous at $i$.
$endgroup$
– NL1992
Dec 12 '18 at 0:03
2
$begingroup$
Plug in first, does $z=i$ look like a special point?
$endgroup$
– Michael Burr
Dec 12 '18 at 0:04
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@MichaelBurr Sorry, a typo.
$endgroup$
– Dole
Dec 12 '18 at 0:20
$begingroup$
Don't understand the problem here. Plugging in (or taking the limit) $z=i $ is the obvious thing to try right?
$endgroup$
– Winther
Dec 12 '18 at 1:00
$begingroup$
@Winther Yes it goes to infinity, but does it go to somewhere else like zero as well approached from a different direction? To me it seems that it does given I think it might lead to a contradiction if i was a pole.
$endgroup$
– Dole
Dec 12 '18 at 1:02
|
show 1 more comment
$begingroup$
You should use the fact that $f(z)$ you described is continuous at $i$.
$endgroup$
– NL1992
Dec 12 '18 at 0:03
2
$begingroup$
Plug in first, does $z=i$ look like a special point?
$endgroup$
– Michael Burr
Dec 12 '18 at 0:04
$begingroup$
@MichaelBurr Sorry, a typo.
$endgroup$
– Dole
Dec 12 '18 at 0:20
$begingroup$
Don't understand the problem here. Plugging in (or taking the limit) $z=i $ is the obvious thing to try right?
$endgroup$
– Winther
Dec 12 '18 at 1:00
$begingroup$
@Winther Yes it goes to infinity, but does it go to somewhere else like zero as well approached from a different direction? To me it seems that it does given I think it might lead to a contradiction if i was a pole.
$endgroup$
– Dole
Dec 12 '18 at 1:02
$begingroup$
You should use the fact that $f(z)$ you described is continuous at $i$.
$endgroup$
– NL1992
Dec 12 '18 at 0:03
$begingroup$
You should use the fact that $f(z)$ you described is continuous at $i$.
$endgroup$
– NL1992
Dec 12 '18 at 0:03
2
2
$begingroup$
Plug in first, does $z=i$ look like a special point?
$endgroup$
– Michael Burr
Dec 12 '18 at 0:04
$begingroup$
Plug in first, does $z=i$ look like a special point?
$endgroup$
– Michael Burr
Dec 12 '18 at 0:04
$begingroup$
@MichaelBurr Sorry, a typo.
$endgroup$
– Dole
Dec 12 '18 at 0:20
$begingroup$
@MichaelBurr Sorry, a typo.
$endgroup$
– Dole
Dec 12 '18 at 0:20
$begingroup$
Don't understand the problem here. Plugging in (or taking the limit) $z=i $ is the obvious thing to try right?
$endgroup$
– Winther
Dec 12 '18 at 1:00
$begingroup$
Don't understand the problem here. Plugging in (or taking the limit) $z=i $ is the obvious thing to try right?
$endgroup$
– Winther
Dec 12 '18 at 1:00
$begingroup$
@Winther Yes it goes to infinity, but does it go to somewhere else like zero as well approached from a different direction? To me it seems that it does given I think it might lead to a contradiction if i was a pole.
$endgroup$
– Dole
Dec 12 '18 at 1:02
$begingroup$
@Winther Yes it goes to infinity, but does it go to somewhere else like zero as well approached from a different direction? To me it seems that it does given I think it might lead to a contradiction if i was a pole.
$endgroup$
– Dole
Dec 12 '18 at 1:02
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show 1 more comment
1 Answer
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$|f(z)|=frac {|e^{iz}|} {|z+i| |z-i|} to infty$ as $z to i$.
answered Dec 12 '18 at 0:33
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$|f(z)|=frac {|e^{iz}|} {|z+i| |z-i|} to infty$ as $z to i$.
answered Dec 12 '18 at 0:33
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
add a comment |
$begingroup$
$|f(z)|=frac {|e^{iz}|} {|z+i| |z-i|} to infty$ as $z to i$.
answered Dec 12 '18 at 0:33
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
add a comment |
$begingroup$
$|f(z)|=frac {|e^{iz}|} {|z+i| |z-i|} to infty$ as $z to i$.
answered Dec 12 '18 at 0:33
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
$|f(z)|=frac {|e^{iz}|} {|z+i| |z-i|} to infty$ as $z to i$.
answered Dec 12 '18 at 0:33
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
answered Dec 12 '18 at 0:33
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
answered Dec 12 '18 at 0:33
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
answered Dec 12 '18 at 0:33
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
57.7k42160
add a comment |
add a comment |
$begingroup$
You should use the fact that $f(z)$ you described is continuous at $i$.
$endgroup$
– NL1992
Dec 12 '18 at 0:03
2
$begingroup$
Plug in first, does $z=i$ look like a special point?
$endgroup$
– Michael Burr
Dec 12 '18 at 0:04
$begingroup$
@MichaelBurr Sorry, a typo.
$endgroup$
– Dole
Dec 12 '18 at 0:20
$begingroup$
Don't understand the problem here. Plugging in (or taking the limit) $z=i $ is the obvious thing to try right?
$endgroup$
– Winther
Dec 12 '18 at 1:00
$begingroup$
@Winther Yes it goes to infinity, but does it go to somewhere else like zero as well approached from a different direction? To me it seems that it does given I think it might lead to a contradiction if i was a pole.
$endgroup$
– Dole
Dec 12 '18 at 1:02