Is $intlimits_0^R e^{-(1+i)r^2}dr$ the same as $intlimits_0^R e^{-({1over2}+{iover2})r^2}dr$
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In a corrected exercise there is this equality
$Gamma_R^3$ is the line ${re^{ipi/8}~: ~rin[0,R]}$
$-sqrt2(re^{ipi/8})^2=-sqrt2(r^2e^{ipi/4})=-sqrt2r^2({sqrt2over2}+i{sqrt2over2})$ so I don't get how this is right. Mybe I'm missing something...
Thanks
integration complex-integration
asked Jan 2 at 16:10
John CataldoJohn Cataldo
1,1881316
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add a comment |
$begingroup$
In a corrected exercise there is this equality
$Gamma_R^3$ is the line ${re^{ipi/8}~: ~rin[0,R]}$
$-sqrt2(re^{ipi/8})^2=-sqrt2(r^2e^{ipi/4})=-sqrt2r^2({sqrt2over2}+i{sqrt2over2})$ so I don't get how this is right. Mybe I'm missing something...
Thanks
integration complex-integration
asked Jan 2 at 16:10
John CataldoJohn Cataldo
1,1881316
$endgroup$
add a comment |
$begingroup$
In a corrected exercise there is this equality
$Gamma_R^3$ is the line ${re^{ipi/8}~: ~rin[0,R]}$
$-sqrt2(re^{ipi/8})^2=-sqrt2(r^2e^{ipi/4})=-sqrt2r^2({sqrt2over2}+i{sqrt2over2})$ so I don't get how this is right. Mybe I'm missing something...
Thanks
integration complex-integration
asked Jan 2 at 16:10
John CataldoJohn Cataldo
1,1881316
$endgroup$
In a corrected exercise there is this equality
$Gamma_R^3$ is the line ${re^{ipi/8}~: ~rin[0,R]}$
$-sqrt2(re^{ipi/8})^2=-sqrt2(r^2e^{ipi/4})=-sqrt2r^2({sqrt2over2}+i{sqrt2over2})$ so I don't get how this is right. Mybe I'm missing something...
Thanks
integration complex-integration
integration complex-integration
asked Jan 2 at 16:10
John CataldoJohn Cataldo
1,1881316
asked Jan 2 at 16:10
John CataldoJohn Cataldo
1,1881316
asked Jan 2 at 16:10
John CataldoJohn Cataldo
1,1881316
asked Jan 2 at 16:10
John CataldoJohn Cataldo
1,1881316
asked Jan 2 at 16:10
John CataldoJohn Cataldo
1,1881316
1,1881316
add a comment |
add a comment |
1 Answer
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$begingroup$
Continuing from your calculation,$$-sqrt2(re^{ipi/8})^2=-sqrt2(r^2e^{ipi/4})=-sqrt2r^2({sqrt2over2}+i{sqrt2over2})=-r^2(1+i)sqrt2cdotfrac{sqrt2}2=-r^2(1+i)$$since $sqrt2cdotfrac{sqrt2}2=1$.
answered Jan 2 at 16:14
John DoeJohn Doe
11.2k11239
$endgroup$
$begingroup$
Thanks yeah just a stupid mistake...
$endgroup$
– John Cataldo
Jan 2 at 16:16
1
$begingroup$
That's ok, often it takes longer than it should to spot simple things like that because you're expecting something more complicated!
$endgroup$
– John Doe
Jan 2 at 16:19
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Continuing from your calculation,$$-sqrt2(re^{ipi/8})^2=-sqrt2(r^2e^{ipi/4})=-sqrt2r^2({sqrt2over2}+i{sqrt2over2})=-r^2(1+i)sqrt2cdotfrac{sqrt2}2=-r^2(1+i)$$since $sqrt2cdotfrac{sqrt2}2=1$.
answered Jan 2 at 16:14
John DoeJohn Doe
11.2k11239
$endgroup$
$begingroup$
Thanks yeah just a stupid mistake...
$endgroup$
– John Cataldo
Jan 2 at 16:16
1
$begingroup$
That's ok, often it takes longer than it should to spot simple things like that because you're expecting something more complicated!
$endgroup$
– John Doe
Jan 2 at 16:19
add a comment |
$begingroup$
Continuing from your calculation,$$-sqrt2(re^{ipi/8})^2=-sqrt2(r^2e^{ipi/4})=-sqrt2r^2({sqrt2over2}+i{sqrt2over2})=-r^2(1+i)sqrt2cdotfrac{sqrt2}2=-r^2(1+i)$$since $sqrt2cdotfrac{sqrt2}2=1$.
answered Jan 2 at 16:14
John DoeJohn Doe
11.2k11239
$endgroup$
$begingroup$
Thanks yeah just a stupid mistake...
$endgroup$
– John Cataldo
Jan 2 at 16:16
1
$begingroup$
That's ok, often it takes longer than it should to spot simple things like that because you're expecting something more complicated!
$endgroup$
– John Doe
Jan 2 at 16:19
add a comment |
$begingroup$
Continuing from your calculation,$$-sqrt2(re^{ipi/8})^2=-sqrt2(r^2e^{ipi/4})=-sqrt2r^2({sqrt2over2}+i{sqrt2over2})=-r^2(1+i)sqrt2cdotfrac{sqrt2}2=-r^2(1+i)$$since $sqrt2cdotfrac{sqrt2}2=1$.
answered Jan 2 at 16:14
John DoeJohn Doe
11.2k11239
$endgroup$
Continuing from your calculation,$$-sqrt2(re^{ipi/8})^2=-sqrt2(r^2e^{ipi/4})=-sqrt2r^2({sqrt2over2}+i{sqrt2over2})=-r^2(1+i)sqrt2cdotfrac{sqrt2}2=-r^2(1+i)$$since $sqrt2cdotfrac{sqrt2}2=1$.
answered Jan 2 at 16:14
John DoeJohn Doe
11.2k11239
answered Jan 2 at 16:14
John DoeJohn Doe
11.2k11239
answered Jan 2 at 16:14
John DoeJohn Doe
11.2k11239
answered Jan 2 at 16:14
John DoeJohn Doe
11.2k11239
11.2k11239
$begingroup$
Thanks yeah just a stupid mistake...
$endgroup$
– John Cataldo
Jan 2 at 16:16
1
$begingroup$
That's ok, often it takes longer than it should to spot simple things like that because you're expecting something more complicated!
$endgroup$
– John Doe
Jan 2 at 16:19
add a comment |
$begingroup$
Thanks yeah just a stupid mistake...
$endgroup$
– John Cataldo
Jan 2 at 16:16
1
$begingroup$
That's ok, often it takes longer than it should to spot simple things like that because you're expecting something more complicated!
$endgroup$
– John Doe
Jan 2 at 16:19
$begingroup$
Thanks yeah just a stupid mistake...
$endgroup$
– John Cataldo
Jan 2 at 16:16
$begingroup$
Thanks yeah just a stupid mistake...
$endgroup$
– John Cataldo
Jan 2 at 16:16
1
1
$begingroup$
That's ok, often it takes longer than it should to spot simple things like that because you're expecting something more complicated!
$endgroup$
– John Doe
Jan 2 at 16:19
$begingroup$
That's ok, often it takes longer than it should to spot simple things like that because you're expecting something more complicated!
$endgroup$
– John Doe
Jan 2 at 16:19
add a comment |
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