Mixing
Naive question of taking power of an exponential
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I'm doing a course in complex analysis - and in a question I come to a term that is $[exp(2pi i)]^f $ where $f$ is some fraction. I naively proceed with $[exp(2pi i)]^f=1^f=1$ which leads to an incorrect answer, where instead I should keep $[exp(2pi i)]^f = exp(2 f pi i)ne 1$. This has confused me somewhat (and possibly embarrassingly) - is it incorrect to evaluate the exponential first?
For example, is it $[exp(2pi i)]^frac{1}{4}=exp(frac{pi}{2} i) = i$ or, is it $[exp(2pi i)]^frac{1}{4}=1^frac{1}{4} = 1$?
exponential-function
asked Jan 11 at 9:34
TrockTrock
134
$endgroup$
add a comment |
$begingroup$
I'm doing a course in complex analysis - and in a question I come to a term that is $[exp(2pi i)]^f $ where $f$ is some fraction. I naively proceed with $[exp(2pi i)]^f=1^f=1$ which leads to an incorrect answer, where instead I should keep $[exp(2pi i)]^f = exp(2 f pi i)ne 1$. This has confused me somewhat (and possibly embarrassingly) - is it incorrect to evaluate the exponential first?
For example, is it $[exp(2pi i)]^frac{1}{4}=exp(frac{pi}{2} i) = i$ or, is it $[exp(2pi i)]^frac{1}{4}=1^frac{1}{4} = 1$?
exponential-function
asked Jan 11 at 9:34
TrockTrock
134
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2
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$1^{1/4}$ has four solutions in $mathbb C$.
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– Matteo
Jan 11 at 9:36
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mathoverflow.net/a/94833/13042
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– Giuseppe Negro
Jan 11 at 9:52
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What makes you say that this answer is incorrect ?
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– Yves Daoust
Jan 11 at 9:53
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@GiuseppeNegro thanks for the link, I'm glad I'm not alone in getting muddled!
$endgroup$
– Trock
Jan 11 at 11:29
add a comment |
$begingroup$
I'm doing a course in complex analysis - and in a question I come to a term that is $[exp(2pi i)]^f $ where $f$ is some fraction. I naively proceed with $[exp(2pi i)]^f=1^f=1$ which leads to an incorrect answer, where instead I should keep $[exp(2pi i)]^f = exp(2 f pi i)ne 1$. This has confused me somewhat (and possibly embarrassingly) - is it incorrect to evaluate the exponential first?
For example, is it $[exp(2pi i)]^frac{1}{4}=exp(frac{pi}{2} i) = i$ or, is it $[exp(2pi i)]^frac{1}{4}=1^frac{1}{4} = 1$?
exponential-function
asked Jan 11 at 9:34
TrockTrock
134
$endgroup$
I'm doing a course in complex analysis - and in a question I come to a term that is $[exp(2pi i)]^f $ where $f$ is some fraction. I naively proceed with $[exp(2pi i)]^f=1^f=1$ which leads to an incorrect answer, where instead I should keep $[exp(2pi i)]^f = exp(2 f pi i)ne 1$. This has confused me somewhat (and possibly embarrassingly) - is it incorrect to evaluate the exponential first?
For example, is it $[exp(2pi i)]^frac{1}{4}=exp(frac{pi}{2} i) = i$ or, is it $[exp(2pi i)]^frac{1}{4}=1^frac{1}{4} = 1$?
exponential-function
exponential-function
asked Jan 11 at 9:34
TrockTrock
134
asked Jan 11 at 9:34
TrockTrock
134
asked Jan 11 at 9:34
TrockTrock
134
asked Jan 11 at 9:34
TrockTrock
134
asked Jan 11 at 9:34
TrockTrock
134
134
2
$begingroup$
$1^{1/4}$ has four solutions in $mathbb C$.
$endgroup$
– Matteo
Jan 11 at 9:36
$begingroup$
mathoverflow.net/a/94833/13042
$endgroup$
– Giuseppe Negro
Jan 11 at 9:52
$begingroup$
What makes you say that this answer is incorrect ?
$endgroup$
– Yves Daoust
Jan 11 at 9:53
$begingroup$
@GiuseppeNegro thanks for the link, I'm glad I'm not alone in getting muddled!
$endgroup$
– Trock
Jan 11 at 11:29
add a comment |
2
$begingroup$
$1^{1/4}$ has four solutions in $mathbb C$.
$endgroup$
– Matteo
Jan 11 at 9:36
$begingroup$
mathoverflow.net/a/94833/13042
$endgroup$
– Giuseppe Negro
Jan 11 at 9:52
$begingroup$
What makes you say that this answer is incorrect ?
$endgroup$
– Yves Daoust
Jan 11 at 9:53
$begingroup$
@GiuseppeNegro thanks for the link, I'm glad I'm not alone in getting muddled!
$endgroup$
– Trock
Jan 11 at 11:29
2
2
$begingroup$
$1^{1/4}$ has four solutions in $mathbb C$.
$endgroup$
– Matteo
Jan 11 at 9:36
$begingroup$
$1^{1/4}$ has four solutions in $mathbb C$.
$endgroup$
– Matteo
Jan 11 at 9:36
$begingroup$
mathoverflow.net/a/94833/13042
$endgroup$
– Giuseppe Negro
Jan 11 at 9:52
$begingroup$
mathoverflow.net/a/94833/13042
$endgroup$
– Giuseppe Negro
Jan 11 at 9:52
$begingroup$
What makes you say that this answer is incorrect ?
$endgroup$
– Yves Daoust
Jan 11 at 9:53
$begingroup$
What makes you say that this answer is incorrect ?
$endgroup$
– Yves Daoust
Jan 11 at 9:53
$begingroup$
@GiuseppeNegro thanks for the link, I'm glad I'm not alone in getting muddled!
$endgroup$
– Trock
Jan 11 at 11:29
$begingroup$
@GiuseppeNegro thanks for the link, I'm glad I'm not alone in getting muddled!
$endgroup$
– Trock
Jan 11 at 11:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Raising complex numbers to a power is problematic because there is an undeterminacy. Notice that
$$e^{i2pi}=e^{i2kpi}$$ for every integer $k$ so that
$$(e^{i2pi})^f$$ could take infinitely many values
$$e^{i2kpi f}.$$
When $f$ is irrational, the expression has no meaning and remains undefined.
When $f$ is rational, you can choose to consider all $q$ distinct values of
$$e^{i2kpi p/q},$$ assuming the fraction irreducible.
You can also choose a particular "branch", by constraining the argument to certain range, such as $[0,2pi)$. In this case
$$(e^{i2pi})^f=(e^{i,0})^f=e^{i,0f}=1.$$
answered Jan 11 at 10:00
Yves DaoustYves Daoust
127k673226
$endgroup$
add a comment |
$begingroup$
Look what happens if you consider
$$left[e^{2 ipi}right]^f = left[e^{i(2pi + 2k pi)}right]^f, kinmathbb Z.$$
answered Jan 11 at 9:49
MatteoMatteo
63439
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Raising complex numbers to a power is problematic because there is an undeterminacy. Notice that
$$e^{i2pi}=e^{i2kpi}$$ for every integer $k$ so that
$$(e^{i2pi})^f$$ could take infinitely many values
$$e^{i2kpi f}.$$
When $f$ is irrational, the expression has no meaning and remains undefined.
When $f$ is rational, you can choose to consider all $q$ distinct values of
$$e^{i2kpi p/q},$$ assuming the fraction irreducible.
You can also choose a particular "branch", by constraining the argument to certain range, such as $[0,2pi)$. In this case
$$(e^{i2pi})^f=(e^{i,0})^f=e^{i,0f}=1.$$
answered Jan 11 at 10:00
Yves DaoustYves Daoust
127k673226
$endgroup$
add a comment |
$begingroup$
Raising complex numbers to a power is problematic because there is an undeterminacy. Notice that
$$e^{i2pi}=e^{i2kpi}$$ for every integer $k$ so that
$$(e^{i2pi})^f$$ could take infinitely many values
$$e^{i2kpi f}.$$
When $f$ is irrational, the expression has no meaning and remains undefined.
When $f$ is rational, you can choose to consider all $q$ distinct values of
$$e^{i2kpi p/q},$$ assuming the fraction irreducible.
You can also choose a particular "branch", by constraining the argument to certain range, such as $[0,2pi)$. In this case
$$(e^{i2pi})^f=(e^{i,0})^f=e^{i,0f}=1.$$
answered Jan 11 at 10:00
Yves DaoustYves Daoust
127k673226
$endgroup$
add a comment |
$begingroup$
Raising complex numbers to a power is problematic because there is an undeterminacy. Notice that
$$e^{i2pi}=e^{i2kpi}$$ for every integer $k$ so that
$$(e^{i2pi})^f$$ could take infinitely many values
$$e^{i2kpi f}.$$
When $f$ is irrational, the expression has no meaning and remains undefined.
When $f$ is rational, you can choose to consider all $q$ distinct values of
$$e^{i2kpi p/q},$$ assuming the fraction irreducible.
You can also choose a particular "branch", by constraining the argument to certain range, such as $[0,2pi)$. In this case
$$(e^{i2pi})^f=(e^{i,0})^f=e^{i,0f}=1.$$
answered Jan 11 at 10:00
Yves DaoustYves Daoust
127k673226
$endgroup$
Raising complex numbers to a power is problematic because there is an undeterminacy. Notice that
$$e^{i2pi}=e^{i2kpi}$$ for every integer $k$ so that
$$(e^{i2pi})^f$$ could take infinitely many values
$$e^{i2kpi f}.$$
When $f$ is irrational, the expression has no meaning and remains undefined.
When $f$ is rational, you can choose to consider all $q$ distinct values of
$$e^{i2kpi p/q},$$ assuming the fraction irreducible.
You can also choose a particular "branch", by constraining the argument to certain range, such as $[0,2pi)$. In this case
$$(e^{i2pi})^f=(e^{i,0})^f=e^{i,0f}=1.$$
answered Jan 11 at 10:00
Yves DaoustYves Daoust
127k673226
edited Jan 11 at 10:06
answered Jan 11 at 10:00
Yves DaoustYves Daoust
127k673226
answered Jan 11 at 10:00
Yves DaoustYves Daoust
127k673226
answered Jan 11 at 10:00
Yves DaoustYves Daoust
127k673226
127k673226
add a comment |
add a comment |
$begingroup$
Look what happens if you consider
$$left[e^{2 ipi}right]^f = left[e^{i(2pi + 2k pi)}right]^f, kinmathbb Z.$$
answered Jan 11 at 9:49
MatteoMatteo
63439
$endgroup$
add a comment |
$begingroup$
Look what happens if you consider
$$left[e^{2 ipi}right]^f = left[e^{i(2pi + 2k pi)}right]^f, kinmathbb Z.$$
answered Jan 11 at 9:49
MatteoMatteo
63439
$endgroup$
add a comment |
$begingroup$
Look what happens if you consider
$$left[e^{2 ipi}right]^f = left[e^{i(2pi + 2k pi)}right]^f, kinmathbb Z.$$
answered Jan 11 at 9:49
MatteoMatteo
63439
$endgroup$
Look what happens if you consider
$$left[e^{2 ipi}right]^f = left[e^{i(2pi + 2k pi)}right]^f, kinmathbb Z.$$
answered Jan 11 at 9:49
MatteoMatteo
63439
edited Jan 11 at 9:56
answered Jan 11 at 9:49
MatteoMatteo
63439
answered Jan 11 at 9:49
MatteoMatteo
63439
answered Jan 11 at 9:49
MatteoMatteo
63439
63439
add a comment |
add a comment |
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2
$begingroup$
$1^{1/4}$ has four solutions in $mathbb C$.
$endgroup$
– Matteo
Jan 11 at 9:36
$begingroup$
mathoverflow.net/a/94833/13042
$endgroup$
– Giuseppe Negro
Jan 11 at 9:52
$begingroup$
What makes you say that this answer is incorrect ?
$endgroup$
– Yves Daoust
Jan 11 at 9:53
$begingroup$
@GiuseppeNegro thanks for the link, I'm glad I'm not alone in getting muddled!
$endgroup$
– Trock
Jan 11 at 11:29