Mixing
What's $i^e$, and why is there an imaginary part?
$begingroup$
I am not sure if this is a straightforward question, since I am not familiar with complex analysis.
$i^e = e^{(i*π/2)*e} = (e^{iπ})^{e/2} = (-1)^{e/2}$
I think that's somewhat right, but when I put this into a calculator an imaginary part also comes out.
Is this because of $e$?
complex-analysis complex-numbers
edited Jan 15 at 19:34
David G. Stork
11k41432
asked Jan 15 at 19:29
Casimir RönnlöfCasimir Rönnlöf
1374
$endgroup$
add a comment |
$begingroup$
I am not sure if this is a straightforward question, since I am not familiar with complex analysis.
$i^e = e^{(i*π/2)*e} = (e^{iπ})^{e/2} = (-1)^{e/2}$
I think that's somewhat right, but when I put this into a calculator an imaginary part also comes out.
Is this because of $e$?
complex-analysis complex-numbers
edited Jan 15 at 19:34
David G. Stork
11k41432
asked Jan 15 at 19:29
Casimir RönnlöfCasimir Rönnlöf
1374
$endgroup$
2
$begingroup$
How would you expect $(-1)^{e/2}$ to be real? Not even $(-1)^{1/2}$ is real.
$endgroup$
– Arthur
Jan 15 at 19:31
$begingroup$
A continuous version of $(-1)^t$ goes around the unit circle at a certain speed. It goes once around when $t$ increases by $2$. And $(-1)^t$ is non-real whenever $t$ is non-integer.
$endgroup$
– GEdgar
Jan 15 at 19:37
add a comment |
$begingroup$
I am not sure if this is a straightforward question, since I am not familiar with complex analysis.
$i^e = e^{(i*π/2)*e} = (e^{iπ})^{e/2} = (-1)^{e/2}$
I think that's somewhat right, but when I put this into a calculator an imaginary part also comes out.
Is this because of $e$?
complex-analysis complex-numbers
edited Jan 15 at 19:34
David G. Stork
11k41432
asked Jan 15 at 19:29
Casimir RönnlöfCasimir Rönnlöf
1374
$endgroup$
I am not sure if this is a straightforward question, since I am not familiar with complex analysis.
$i^e = e^{(i*π/2)*e} = (e^{iπ})^{e/2} = (-1)^{e/2}$
I think that's somewhat right, but when I put this into a calculator an imaginary part also comes out.
Is this because of $e$?
complex-analysis complex-numbers
complex-analysis complex-numbers
edited Jan 15 at 19:34
David G. Stork
11k41432
asked Jan 15 at 19:29
Casimir RönnlöfCasimir Rönnlöf
1374
edited Jan 15 at 19:34
David G. Stork
11k41432
asked Jan 15 at 19:29
Casimir RönnlöfCasimir Rönnlöf
1374
edited Jan 15 at 19:34
David G. Stork
11k41432
edited Jan 15 at 19:34
David G. Stork
11k41432
edited Jan 15 at 19:34
David G. Stork
11k41432
11k41432
asked Jan 15 at 19:29
Casimir RönnlöfCasimir Rönnlöf
1374
asked Jan 15 at 19:29
Casimir RönnlöfCasimir Rönnlöf
1374
asked Jan 15 at 19:29
Casimir RönnlöfCasimir Rönnlöf
1374
1374
2
$begingroup$
How would you expect $(-1)^{e/2}$ to be real? Not even $(-1)^{1/2}$ is real.
$endgroup$
– Arthur
Jan 15 at 19:31
$begingroup$
A continuous version of $(-1)^t$ goes around the unit circle at a certain speed. It goes once around when $t$ increases by $2$. And $(-1)^t$ is non-real whenever $t$ is non-integer.
$endgroup$
– GEdgar
Jan 15 at 19:37
add a comment |
2
$begingroup$
How would you expect $(-1)^{e/2}$ to be real? Not even $(-1)^{1/2}$ is real.
$endgroup$
– Arthur
Jan 15 at 19:31
$begingroup$
A continuous version of $(-1)^t$ goes around the unit circle at a certain speed. It goes once around when $t$ increases by $2$. And $(-1)^t$ is non-real whenever $t$ is non-integer.
$endgroup$
– GEdgar
Jan 15 at 19:37
2
2
$begingroup$
How would you expect $(-1)^{e/2}$ to be real? Not even $(-1)^{1/2}$ is real.
$endgroup$
– Arthur
Jan 15 at 19:31
$begingroup$
How would you expect $(-1)^{e/2}$ to be real? Not even $(-1)^{1/2}$ is real.
$endgroup$
– Arthur
Jan 15 at 19:31
$begingroup$
A continuous version of $(-1)^t$ goes around the unit circle at a certain speed. It goes once around when $t$ increases by $2$. And $(-1)^t$ is non-real whenever $t$ is non-integer.
$endgroup$
– GEdgar
Jan 15 at 19:37
$begingroup$
A continuous version of $(-1)^t$ goes around the unit circle at a certain speed. It goes once around when $t$ increases by $2$. And $(-1)^t$ is non-real whenever $t$ is non-integer.
$endgroup$
– GEdgar
Jan 15 at 19:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$i^e = e^{(i*π/2)*e} = (e^{iπ})^{e/2} = (-1)^{e/2}$
You still have $(-1)$ to a power of a multiple of $frac 12$ and that will create an imaginary part.
$(e^{ifrac {eπ}{2}}) = cos frac {eπ}{2} + isin frac {eπ}{2}$
answered Jan 15 at 19:33
Doug MDoug M
45.2k31854
$endgroup$
add a comment |
$begingroup$
Complex exponentiation is in general multi-valued. Since $i=e^z$ iff $z=frac{pi i (4n+1)}{2}$ for some integer $n$, $i^e$ can be any value of the form $expfrac{pi e i (4n+1)}{2}=cosfrac{pi e (4n+1)}{2}+isinfrac{pi e (4n+1)}{2}$.
answered Jan 15 at 19:39
J.G.J.G.
27.3k22843
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$i^e = e^{(i*π/2)*e} = (e^{iπ})^{e/2} = (-1)^{e/2}$
You still have $(-1)$ to a power of a multiple of $frac 12$ and that will create an imaginary part.
$(e^{ifrac {eπ}{2}}) = cos frac {eπ}{2} + isin frac {eπ}{2}$
answered Jan 15 at 19:33
Doug MDoug M
45.2k31854
$endgroup$
add a comment |
$begingroup$
$i^e = e^{(i*π/2)*e} = (e^{iπ})^{e/2} = (-1)^{e/2}$
You still have $(-1)$ to a power of a multiple of $frac 12$ and that will create an imaginary part.
$(e^{ifrac {eπ}{2}}) = cos frac {eπ}{2} + isin frac {eπ}{2}$
answered Jan 15 at 19:33
Doug MDoug M
45.2k31854
$endgroup$
add a comment |
$begingroup$
$i^e = e^{(i*π/2)*e} = (e^{iπ})^{e/2} = (-1)^{e/2}$
You still have $(-1)$ to a power of a multiple of $frac 12$ and that will create an imaginary part.
$(e^{ifrac {eπ}{2}}) = cos frac {eπ}{2} + isin frac {eπ}{2}$
answered Jan 15 at 19:33
Doug MDoug M
45.2k31854
$endgroup$
$i^e = e^{(i*π/2)*e} = (e^{iπ})^{e/2} = (-1)^{e/2}$
You still have $(-1)$ to a power of a multiple of $frac 12$ and that will create an imaginary part.
$(e^{ifrac {eπ}{2}}) = cos frac {eπ}{2} + isin frac {eπ}{2}$
answered Jan 15 at 19:33
Doug MDoug M
45.2k31854
answered Jan 15 at 19:33
Doug MDoug M
45.2k31854
answered Jan 15 at 19:33
Doug MDoug M
45.2k31854
answered Jan 15 at 19:33
Doug MDoug M
45.2k31854
45.2k31854
add a comment |
add a comment |
$begingroup$
Complex exponentiation is in general multi-valued. Since $i=e^z$ iff $z=frac{pi i (4n+1)}{2}$ for some integer $n$, $i^e$ can be any value of the form $expfrac{pi e i (4n+1)}{2}=cosfrac{pi e (4n+1)}{2}+isinfrac{pi e (4n+1)}{2}$.
answered Jan 15 at 19:39
J.G.J.G.
27.3k22843
$endgroup$
add a comment |
$begingroup$
Complex exponentiation is in general multi-valued. Since $i=e^z$ iff $z=frac{pi i (4n+1)}{2}$ for some integer $n$, $i^e$ can be any value of the form $expfrac{pi e i (4n+1)}{2}=cosfrac{pi e (4n+1)}{2}+isinfrac{pi e (4n+1)}{2}$.
answered Jan 15 at 19:39
J.G.J.G.
27.3k22843
$endgroup$
add a comment |
$begingroup$
Complex exponentiation is in general multi-valued. Since $i=e^z$ iff $z=frac{pi i (4n+1)}{2}$ for some integer $n$, $i^e$ can be any value of the form $expfrac{pi e i (4n+1)}{2}=cosfrac{pi e (4n+1)}{2}+isinfrac{pi e (4n+1)}{2}$.
answered Jan 15 at 19:39
J.G.J.G.
27.3k22843
$endgroup$
Complex exponentiation is in general multi-valued. Since $i=e^z$ iff $z=frac{pi i (4n+1)}{2}$ for some integer $n$, $i^e$ can be any value of the form $expfrac{pi e i (4n+1)}{2}=cosfrac{pi e (4n+1)}{2}+isinfrac{pi e (4n+1)}{2}$.
answered Jan 15 at 19:39
J.G.J.G.
27.3k22843
answered Jan 15 at 19:39
J.G.J.G.
27.3k22843
answered Jan 15 at 19:39
J.G.J.G.
27.3k22843
answered Jan 15 at 19:39
J.G.J.G.
27.3k22843
27.3k22843
add a comment |
add a comment |
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2
$begingroup$
How would you expect $(-1)^{e/2}$ to be real? Not even $(-1)^{1/2}$ is real.
$endgroup$
– Arthur
Jan 15 at 19:31
$begingroup$
A continuous version of $(-1)^t$ goes around the unit circle at a certain speed. It goes once around when $t$ increases by $2$. And $(-1)^t$ is non-real whenever $t$ is non-integer.
$endgroup$
– GEdgar
Jan 15 at 19:37