Good treatment of EulerâÂÂs formula
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I need help because I'm a bit confused about the correct way to treat Euler's formula.
If I ask Wolfram Alpha (WA) about $(i^i)^4$ the answer is $e^(-2 pi)$, therefore, I thought, ok this must be equal to $(e^i 2 pi)^i$ but, this is not true, WA says that is equal 1...but, it says that its general form is $e^-2pi n$ but that can not be ever 1...so my head explodes here.
Note: I will appreciate a link to a good explanation about that kind of things.
Thank you
complex-numbers
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up vote
1
down vote
favorite
I need help because I'm a bit confused about the correct way to treat Euler's formula.
If I ask Wolfram Alpha (WA) about $(i^i)^4$ the answer is $e^(-2 pi)$, therefore, I thought, ok this must be equal to $(e^i 2 pi)^i$ but, this is not true, WA says that is equal 1...but, it says that its general form is $e^-2pi n$ but that can not be ever 1...so my head explodes here.
Note: I will appreciate a link to a good explanation about that kind of things.
Thank you
complex-numbers
2
It's dangerous to write $a^b$ unless either $a$ is a positive real or $b$ is an integer.
â Lord Shark the Unknown
Aug 24 at 8:21
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need help because I'm a bit confused about the correct way to treat Euler's formula.
If I ask Wolfram Alpha (WA) about $(i^i)^4$ the answer is $e^(-2 pi)$, therefore, I thought, ok this must be equal to $(e^i 2 pi)^i$ but, this is not true, WA says that is equal 1...but, it says that its general form is $e^-2pi n$ but that can not be ever 1...so my head explodes here.
Note: I will appreciate a link to a good explanation about that kind of things.
Thank you
complex-numbers
I need help because I'm a bit confused about the correct way to treat Euler's formula.
If I ask Wolfram Alpha (WA) about $(i^i)^4$ the answer is $e^(-2 pi)$, therefore, I thought, ok this must be equal to $(e^i 2 pi)^i$ but, this is not true, WA says that is equal 1...but, it says that its general form is $e^-2pi n$ but that can not be ever 1...so my head explodes here.
Note: I will appreciate a link to a good explanation about that kind of things.
Thank you
complex-numbers
edited Aug 24 at 8:30
InertialObserver
37719
37719
asked Aug 24 at 8:20
Perico Cruel
245
245
2
It's dangerous to write $a^b$ unless either $a$ is a positive real or $b$ is an integer.
â Lord Shark the Unknown
Aug 24 at 8:21
add a comment |Â
2
It's dangerous to write $a^b$ unless either $a$ is a positive real or $b$ is an integer.
â Lord Shark the Unknown
Aug 24 at 8:21
2
2
It's dangerous to write $a^b$ unless either $a$ is a positive real or $b$ is an integer.
â Lord Shark the Unknown
Aug 24 at 8:21
It's dangerous to write $a^b$ unless either $a$ is a positive real or $b$ is an integer.
â Lord Shark the Unknown
Aug 24 at 8:21
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
The result is obtained from
$$i=e^fracipi2+2 k pi i$$
then
$$i^i=e^-fracpi2-2k pi$$
and
$$(i^i)^4=e^-2pi-8k pi$$
1
You mean $i=e^fracipi2+2kpi i$ etc.
â Lord Shark the Unknown
Aug 24 at 8:36
@LordSharktheUnknown Opssss yes of course! I fix that. Thanks
â gimusi
Aug 24 at 8:39
But, if I do $i=e^fracipi2+2 k pi i$ then $i^4=e^2pi i-8k pi i$ and $(i^4)^i=e^-2pi+8k pi$...but is not true, because $(i^i)^4 != (i^4)^i$
â Perico Cruel
Aug 24 at 9:24
What I mean is: Is that true: $((e^(i*(2/pi) + 2*k*pi*i)) ^i)^4 = ((e^(i*(2/pi) + 2*k*pi*i)) ^4)^i$ ?
â Perico Cruel
Aug 24 at 9:30
@PericoCruel If we consider the multivalued $i^4=e^ileft(fracpi2+2kpiright)$ then $(i^4)^i=e^-2pi-8k pi=(i^i)^4$.
â gimusi
Aug 24 at 10:29
add a comment |Â
up vote
2
down vote
The complex exponential does not satisfy $a^bc=(a^b)^c$.
Of course the question is how you define the complex exponential in the first place. In gimusi's answer, it is a multi-valued function. Personally, I prefer to define it as single-valued (which means it is not defined or discontinuous somewhere).
Some information is on Wikipedia at https://en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers
More application-oriented textbooks (such as Wunsch, Complex Variables with Applications or Brown-Churchill, Complex Variables and Applications) define $a^b$ as multi-valued. More theory-oriented textbooks (Lang, Complex Analysis or Stein-Shakarchi, Complex Analysis) tend to make $a^b$ single-valued, but only defined on a subset of the complex numbers, and dependent on a choice of a branch of the logarithm.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The result is obtained from
$$i=e^fracipi2+2 k pi i$$
then
$$i^i=e^-fracpi2-2k pi$$
and
$$(i^i)^4=e^-2pi-8k pi$$
1
You mean $i=e^fracipi2+2kpi i$ etc.
â Lord Shark the Unknown
Aug 24 at 8:36
@LordSharktheUnknown Opssss yes of course! I fix that. Thanks
â gimusi
Aug 24 at 8:39
But, if I do $i=e^fracipi2+2 k pi i$ then $i^4=e^2pi i-8k pi i$ and $(i^4)^i=e^-2pi+8k pi$...but is not true, because $(i^i)^4 != (i^4)^i$
â Perico Cruel
Aug 24 at 9:24
What I mean is: Is that true: $((e^(i*(2/pi) + 2*k*pi*i)) ^i)^4 = ((e^(i*(2/pi) + 2*k*pi*i)) ^4)^i$ ?
â Perico Cruel
Aug 24 at 9:30
@PericoCruel If we consider the multivalued $i^4=e^ileft(fracpi2+2kpiright)$ then $(i^4)^i=e^-2pi-8k pi=(i^i)^4$.
â gimusi
Aug 24 at 10:29
add a comment |Â
up vote
2
down vote
accepted
The result is obtained from
$$i=e^fracipi2+2 k pi i$$
then
$$i^i=e^-fracpi2-2k pi$$
and
$$(i^i)^4=e^-2pi-8k pi$$
1
You mean $i=e^fracipi2+2kpi i$ etc.
â Lord Shark the Unknown
Aug 24 at 8:36
@LordSharktheUnknown Opssss yes of course! I fix that. Thanks
â gimusi
Aug 24 at 8:39
But, if I do $i=e^fracipi2+2 k pi i$ then $i^4=e^2pi i-8k pi i$ and $(i^4)^i=e^-2pi+8k pi$...but is not true, because $(i^i)^4 != (i^4)^i$
â Perico Cruel
Aug 24 at 9:24
What I mean is: Is that true: $((e^(i*(2/pi) + 2*k*pi*i)) ^i)^4 = ((e^(i*(2/pi) + 2*k*pi*i)) ^4)^i$ ?
â Perico Cruel
Aug 24 at 9:30
@PericoCruel If we consider the multivalued $i^4=e^ileft(fracpi2+2kpiright)$ then $(i^4)^i=e^-2pi-8k pi=(i^i)^4$.
â gimusi
Aug 24 at 10:29
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The result is obtained from
$$i=e^fracipi2+2 k pi i$$
then
$$i^i=e^-fracpi2-2k pi$$
and
$$(i^i)^4=e^-2pi-8k pi$$
The result is obtained from
$$i=e^fracipi2+2 k pi i$$
then
$$i^i=e^-fracpi2-2k pi$$
and
$$(i^i)^4=e^-2pi-8k pi$$
edited Aug 24 at 8:41
answered Aug 24 at 8:26
gimusi
69.7k73686
69.7k73686
1
You mean $i=e^fracipi2+2kpi i$ etc.
â Lord Shark the Unknown
Aug 24 at 8:36
@LordSharktheUnknown Opssss yes of course! I fix that. Thanks
â gimusi
Aug 24 at 8:39
But, if I do $i=e^fracipi2+2 k pi i$ then $i^4=e^2pi i-8k pi i$ and $(i^4)^i=e^-2pi+8k pi$...but is not true, because $(i^i)^4 != (i^4)^i$
â Perico Cruel
Aug 24 at 9:24
What I mean is: Is that true: $((e^(i*(2/pi) + 2*k*pi*i)) ^i)^4 = ((e^(i*(2/pi) + 2*k*pi*i)) ^4)^i$ ?
â Perico Cruel
Aug 24 at 9:30
@PericoCruel If we consider the multivalued $i^4=e^ileft(fracpi2+2kpiright)$ then $(i^4)^i=e^-2pi-8k pi=(i^i)^4$.
â gimusi
Aug 24 at 10:29
add a comment |Â
1
You mean $i=e^fracipi2+2kpi i$ etc.
â Lord Shark the Unknown
Aug 24 at 8:36
@LordSharktheUnknown Opssss yes of course! I fix that. Thanks
â gimusi
Aug 24 at 8:39
But, if I do $i=e^fracipi2+2 k pi i$ then $i^4=e^2pi i-8k pi i$ and $(i^4)^i=e^-2pi+8k pi$...but is not true, because $(i^i)^4 != (i^4)^i$
â Perico Cruel
Aug 24 at 9:24
What I mean is: Is that true: $((e^(i*(2/pi) + 2*k*pi*i)) ^i)^4 = ((e^(i*(2/pi) + 2*k*pi*i)) ^4)^i$ ?
â Perico Cruel
Aug 24 at 9:30
@PericoCruel If we consider the multivalued $i^4=e^ileft(fracpi2+2kpiright)$ then $(i^4)^i=e^-2pi-8k pi=(i^i)^4$.
â gimusi
Aug 24 at 10:29
1
1
You mean $i=e^fracipi2+2kpi i$ etc.
â Lord Shark the Unknown
Aug 24 at 8:36
You mean $i=e^fracipi2+2kpi i$ etc.
â Lord Shark the Unknown
Aug 24 at 8:36
@LordSharktheUnknown Opssss yes of course! I fix that. Thanks
â gimusi
Aug 24 at 8:39
@LordSharktheUnknown Opssss yes of course! I fix that. Thanks
â gimusi
Aug 24 at 8:39
But, if I do $i=e^fracipi2+2 k pi i$ then $i^4=e^2pi i-8k pi i$ and $(i^4)^i=e^-2pi+8k pi$...but is not true, because $(i^i)^4 != (i^4)^i$
â Perico Cruel
Aug 24 at 9:24
But, if I do $i=e^fracipi2+2 k pi i$ then $i^4=e^2pi i-8k pi i$ and $(i^4)^i=e^-2pi+8k pi$...but is not true, because $(i^i)^4 != (i^4)^i$
â Perico Cruel
Aug 24 at 9:24
What I mean is: Is that true: $((e^(i*(2/pi) + 2*k*pi*i)) ^i)^4 = ((e^(i*(2/pi) + 2*k*pi*i)) ^4)^i$ ?
â Perico Cruel
Aug 24 at 9:30
What I mean is: Is that true: $((e^(i*(2/pi) + 2*k*pi*i)) ^i)^4 = ((e^(i*(2/pi) + 2*k*pi*i)) ^4)^i$ ?
â Perico Cruel
Aug 24 at 9:30
@PericoCruel If we consider the multivalued $i^4=e^ileft(fracpi2+2kpiright)$ then $(i^4)^i=e^-2pi-8k pi=(i^i)^4$.
â gimusi
Aug 24 at 10:29
@PericoCruel If we consider the multivalued $i^4=e^ileft(fracpi2+2kpiright)$ then $(i^4)^i=e^-2pi-8k pi=(i^i)^4$.
â gimusi
Aug 24 at 10:29
add a comment |Â
up vote
2
down vote
The complex exponential does not satisfy $a^bc=(a^b)^c$.
Of course the question is how you define the complex exponential in the first place. In gimusi's answer, it is a multi-valued function. Personally, I prefer to define it as single-valued (which means it is not defined or discontinuous somewhere).
Some information is on Wikipedia at https://en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers
More application-oriented textbooks (such as Wunsch, Complex Variables with Applications or Brown-Churchill, Complex Variables and Applications) define $a^b$ as multi-valued. More theory-oriented textbooks (Lang, Complex Analysis or Stein-Shakarchi, Complex Analysis) tend to make $a^b$ single-valued, but only defined on a subset of the complex numbers, and dependent on a choice of a branch of the logarithm.
add a comment |Â
up vote
2
down vote
The complex exponential does not satisfy $a^bc=(a^b)^c$.
Of course the question is how you define the complex exponential in the first place. In gimusi's answer, it is a multi-valued function. Personally, I prefer to define it as single-valued (which means it is not defined or discontinuous somewhere).
Some information is on Wikipedia at https://en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers
More application-oriented textbooks (such as Wunsch, Complex Variables with Applications or Brown-Churchill, Complex Variables and Applications) define $a^b$ as multi-valued. More theory-oriented textbooks (Lang, Complex Analysis or Stein-Shakarchi, Complex Analysis) tend to make $a^b$ single-valued, but only defined on a subset of the complex numbers, and dependent on a choice of a branch of the logarithm.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The complex exponential does not satisfy $a^bc=(a^b)^c$.
Of course the question is how you define the complex exponential in the first place. In gimusi's answer, it is a multi-valued function. Personally, I prefer to define it as single-valued (which means it is not defined or discontinuous somewhere).
Some information is on Wikipedia at https://en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers
More application-oriented textbooks (such as Wunsch, Complex Variables with Applications or Brown-Churchill, Complex Variables and Applications) define $a^b$ as multi-valued. More theory-oriented textbooks (Lang, Complex Analysis or Stein-Shakarchi, Complex Analysis) tend to make $a^b$ single-valued, but only defined on a subset of the complex numbers, and dependent on a choice of a branch of the logarithm.
The complex exponential does not satisfy $a^bc=(a^b)^c$.
Of course the question is how you define the complex exponential in the first place. In gimusi's answer, it is a multi-valued function. Personally, I prefer to define it as single-valued (which means it is not defined or discontinuous somewhere).
Some information is on Wikipedia at https://en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers
More application-oriented textbooks (such as Wunsch, Complex Variables with Applications or Brown-Churchill, Complex Variables and Applications) define $a^b$ as multi-valued. More theory-oriented textbooks (Lang, Complex Analysis or Stein-Shakarchi, Complex Analysis) tend to make $a^b$ single-valued, but only defined on a subset of the complex numbers, and dependent on a choice of a branch of the logarithm.
answered Aug 24 at 8:43
Kusma
2,697214
2,697214
add a comment |Â
add a comment |Â
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2
It's dangerous to write $a^b$ unless either $a$ is a positive real or $b$ is an integer.
â Lord Shark the Unknown
Aug 24 at 8:21