Why is $2pi i neq 0?$ [duplicate]

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up vote
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down vote

favorite

9

This question already has an answer here:

• 
  What is wrong with this fake proof $e^i = 1$?
  
  1 answer
  
  

We know that $e^pi i = -1$ because of de Moivre's formula. ($e^pi i = cos pi + isin pi = -1).$

Suppose we square both sides and get $e^2pi i = 1$(which you also get from de Moivre's formula), then shouldn't $2pi i=0$? What am I missing here?

complex-analysis exponential-function fake-proofs

share|cite|improve this question

edited Dec 11 '16 at 21:10

6005

34.4k749122

asked May 21 '13 at 1:15

Sidd Singal

2,53731630

marked as duplicate by 23rd, Brandon Carter, Micah, Amzoti, Asaf Karagila May 21 '13 at 8:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  19
  

  
  
  
  
  
  

  
  The exponential map on the complex plane is not injective. You're right to say that both $e^2pi i=1$ and $e^0=1$, but without injectivity, we cannot conclude that $2pi i=0$.
  – Jared
  May 21 '13 at 1:16
  
  

  
  
  
  


• 
  
  
  

  
  21
  

  
  
  
  
  
  

  
  It's the same as why $-1neq 1$, despite the fact that $(-1)^2=1^2$.
  – Glen O
  May 21 '13 at 1:17
  

  
  
  
  


• 
  
  
  

  
  14
  

  
  
  
  
  
  

  
  I'm 70 inches tall and so is Joe Smith. Since we're both the same height, I must be Joe Smith, right? No. Just because $f(x) = f(y)$ doesn't mean that $x=y$.
  – MJD
  May 21 '13 at 1:19
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Thanks everyone. I just did not realize that logarithm functions were not injective in the complex realm.
  – Sidd Singal
  May 21 '13 at 1:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  sin(0) = sin(2pi) so 0 = 2pi, how does that work?
  – imranfat
  May 21 '13 at 3:24
  

  
  
  
  

 | 
show 2 more comments

up vote
14
down vote

favorite

9

This question already has an answer here:

• 
  What is wrong with this fake proof $e^i = 1$?
  
  1 answer
  
  

We know that $e^pi i = -1$ because of de Moivre's formula. ($e^pi i = cos pi + isin pi = -1).$

Suppose we square both sides and get $e^2pi i = 1$(which you also get from de Moivre's formula), then shouldn't $2pi i=0$? What am I missing here?

complex-analysis exponential-function fake-proofs

share|cite|improve this question

edited Dec 11 '16 at 21:10

6005

34.4k749122

asked May 21 '13 at 1:15

Sidd Singal

2,53731630

marked as duplicate by 23rd, Brandon Carter, Micah, Amzoti, Asaf Karagila May 21 '13 at 8:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  19
  

  
  
  
  
  
  

  
  The exponential map on the complex plane is not injective. You're right to say that both $e^2pi i=1$ and $e^0=1$, but without injectivity, we cannot conclude that $2pi i=0$.
  – Jared
  May 21 '13 at 1:16
  
  

  
  
  
  


• 
  
  
  

  
  21
  

  
  
  
  
  
  

  
  It's the same as why $-1neq 1$, despite the fact that $(-1)^2=1^2$.
  – Glen O
  May 21 '13 at 1:17
  

  
  
  
  


• 
  
  
  

  
  14
  

  
  
  
  
  
  

  
  I'm 70 inches tall and so is Joe Smith. Since we're both the same height, I must be Joe Smith, right? No. Just because $f(x) = f(y)$ doesn't mean that $x=y$.
  – MJD
  May 21 '13 at 1:19
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Thanks everyone. I just did not realize that logarithm functions were not injective in the complex realm.
  – Sidd Singal
  May 21 '13 at 1:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  sin(0) = sin(2pi) so 0 = 2pi, how does that work?
  – imranfat
  May 21 '13 at 3:24
  

  
  
  
  

 | 
show 2 more comments

up vote
14
down vote

favorite

9

up vote
14
down vote

favorite

9

9

This question already has an answer here:

• 
  What is wrong with this fake proof $e^i = 1$?
  
  1 answer
  
  

We know that $e^pi i = -1$ because of de Moivre's formula. ($e^pi i = cos pi + isin pi = -1).$

Suppose we square both sides and get $e^2pi i = 1$(which you also get from de Moivre's formula), then shouldn't $2pi i=0$? What am I missing here?

complex-analysis exponential-function fake-proofs

share|cite|improve this question

edited Dec 11 '16 at 21:10

6005

34.4k749122

asked May 21 '13 at 1:15

Sidd Singal

2,53731630

This question already has an answer here:

• 
  What is wrong with this fake proof $e^i = 1$?
  
  1 answer
  
  

We know that $e^pi i = -1$ because of de Moivre's formula. ($e^pi i = cos pi + isin pi = -1).$

Suppose we square both sides and get $e^2pi i = 1$(which you also get from de Moivre's formula), then shouldn't $2pi i=0$? What am I missing here?

This question already has an answer here:

• 
  What is wrong with this fake proof $e^i = 1$?
  
  1 answer
  
  

complex-analysis exponential-function fake-proofs

share|cite|improve this question

edited Dec 11 '16 at 21:10

6005

34.4k749122

asked May 21 '13 at 1:15

Sidd Singal

2,53731630

share|cite|improve this question

share|cite|improve this question

edited Dec 11 '16 at 21:10

6005

34.4k749122

edited Dec 11 '16 at 21:10

6005

34.4k749122

edited Dec 11 '16 at 21:10

6005

34.4k749122

34.4k749122

asked May 21 '13 at 1:15

Sidd Singal

2,53731630

asked May 21 '13 at 1:15

Sidd Singal

2,53731630

asked May 21 '13 at 1:15

Sidd Singal

2,53731630

2,53731630

marked as duplicate by 23rd, Brandon Carter, Micah, Amzoti, Asaf Karagila May 21 '13 at 8:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by 23rd, Brandon Carter, Micah, Amzoti, Asaf Karagila May 21 '13 at 8:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  19
  

  
  
  
  
  
  

  
  The exponential map on the complex plane is not injective. You're right to say that both $e^2pi i=1$ and $e^0=1$, but without injectivity, we cannot conclude that $2pi i=0$.
  – Jared
  May 21 '13 at 1:16
  
  

  
  
  
  


• 
  
  
  

  
  21
  

  
  
  
  
  
  

  
  It's the same as why $-1neq 1$, despite the fact that $(-1)^2=1^2$.
  – Glen O
  May 21 '13 at 1:17
  

  
  
  
  


• 
  
  
  

  
  14
  

  
  
  
  
  
  

  
  I'm 70 inches tall and so is Joe Smith. Since we're both the same height, I must be Joe Smith, right? No. Just because $f(x) = f(y)$ doesn't mean that $x=y$.
  – MJD
  May 21 '13 at 1:19
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Thanks everyone. I just did not realize that logarithm functions were not injective in the complex realm.
  – Sidd Singal
  May 21 '13 at 1:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  sin(0) = sin(2pi) so 0 = 2pi, how does that work?
  – imranfat
  May 21 '13 at 3:24
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  19
  

  
  
  
  
  
  

  
  The exponential map on the complex plane is not injective. You're right to say that both $e^2pi i=1$ and $e^0=1$, but without injectivity, we cannot conclude that $2pi i=0$.
  – Jared
  May 21 '13 at 1:16
  
  

  
  
  
  


• 
  
  
  

  
  21
  

  
  
  
  
  
  

  
  It's the same as why $-1neq 1$, despite the fact that $(-1)^2=1^2$.
  – Glen O
  May 21 '13 at 1:17
  

  
  
  
  


• 
  
  
  

  
  14
  

  
  
  
  
  
  

  
  I'm 70 inches tall and so is Joe Smith. Since we're both the same height, I must be Joe Smith, right? No. Just because $f(x) = f(y)$ doesn't mean that $x=y$.
  – MJD
  May 21 '13 at 1:19
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Thanks everyone. I just did not realize that logarithm functions were not injective in the complex realm.
  – Sidd Singal
  May 21 '13 at 1:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  sin(0) = sin(2pi) so 0 = 2pi, how does that work?
  – imranfat
  May 21 '13 at 3:24
  

  
  
  
  

19

19

The exponential map on the complex plane is not injective. You're right to say that both $e^2pi i=1$ and $e^0=1$, but without injectivity, we cannot conclude that $2pi i=0$.
– Jared
May 21 '13 at 1:16

The exponential map on the complex plane is not injective. You're right to say that both $e^2pi i=1$ and $e^0=1$, but without injectivity, we cannot conclude that $2pi i=0$.
– Jared
May 21 '13 at 1:16

21

21

It's the same as why $-1neq 1$, despite the fact that $(-1)^2=1^2$.
– Glen O
May 21 '13 at 1:17

It's the same as why $-1neq 1$, despite the fact that $(-1)^2=1^2$.
– Glen O
May 21 '13 at 1:17

14

14

I'm 70 inches tall and so is Joe Smith. Since we're both the same height, I must be Joe Smith, right? No. Just because $f(x) = f(y)$ doesn't mean that $x=y$.
– MJD
May 21 '13 at 1:19

I'm 70 inches tall and so is Joe Smith. Since we're both the same height, I must be Joe Smith, right? No. Just because $f(x) = f(y)$ doesn't mean that $x=y$.
– MJD
May 21 '13 at 1:19

3

3

Thanks everyone. I just did not realize that logarithm functions were not injective in the complex realm.
– Sidd Singal
May 21 '13 at 1:27

Thanks everyone. I just did not realize that logarithm functions were not injective in the complex realm.
– Sidd Singal
May 21 '13 at 1:27

1

1

sin(0) = sin(2pi) so 0 = 2pi, how does that work?
– imranfat
May 21 '13 at 3:24

sin(0) = sin(2pi) so 0 = 2pi, how does that work?
– imranfat
May 21 '13 at 3:24

 | 
show 2 more comments

4 Answers
4

active

oldest

votes

up vote
32
down vote



accepted

You have shown that $e^2pi i = e^0$. This does not imply $2pi i = 0$, because $e^z$ is not injective.
You have to give up your intuition about real functions when you move them to the complex plane, because they change a lot. $e^z$ is actually periodic for complex $z$.

share|cite|improve this answer

answered May 21 '13 at 1:17

6005

34.4k749122

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think the thing about real functions may be a red herring. As Glen O pointed out in comments, there are plenty of noninjective functions. For example, $(-1)^2 = 1^2$, and yet $-1ne 1$. Or even $3cdot 0 = 4cdot 0$, and yet $3ne 4$.
  – MJD
  May 21 '13 at 1:20
  
  

  
  
  
  


• 
  
  
  

  
  20
  

  
  
  
  
  
  

  
  Sure, but I don't think that's where the OP is confused. I think they understand that $x^2$ and $0$ are noninjective functions, but they assume without thinking about it that $e^x$ is injective, because in their experience (with real $e^x$) it is.
  – 6005
  May 21 '13 at 1:24
  
  

  
  
  
  

add a comment | 

up vote
11
down vote

The $log$ function is multi-valued on $mathbbC^*$ (you can however choose a "branch" of it; see Wikipedia). At any rate, just because
$$e^2pi i=e^0$$
does not imply that $2pi i=0$.

share|cite|improve this answer

edited May 21 '13 at 1:22

answered May 21 '13 at 1:17

Zev Chonoles

108k16217408

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I did not figure such a big change occurs when moving over to the complex plane and figured it would be as intuitive as with real numbers. Thank you (and other answerers/commenters).
  – Sidd Singal
  May 21 '13 at 1:24
  

  
  
  
  

add a comment | 

up vote
11
down vote

It's like saying that, because $sinpi = sin0$, that $pi = 0$. Not all functions have perfect inverses; sin being one of them. In the complex numbers, $e^z$ doesn't either.

You're implicitly going: $e^2pi i = e^0 implies lne^2pi i = lne^0 implies 2 pi i = 0$. The error here is that $forall x in mathbbC lne^x = x$ is not true! $ln$ isn't even a function, just like the naive version of $arcsin(x)$. You have to make a choice of range, which is usually $Im(x) in (-pi, pi]$.

share|cite|improve this answer

answered May 21 '13 at 1:24

Henry Swanson

9,54912151

add a comment | 

up vote
3
down vote

Actually, the following is true:

$ e^2npi i = 1  forall  n in mathbbZ $

As a special case of $n = 0$ this gives what you know: $e^0 = 1$, but as mentioned, this is only a special case of the general formula given above.

You could as well assume that $2pi = 0$ because of

$sin 2pi = sin 0 land cos 2pi = cos 0$.

As others already mentioned, this stuff just isn't injective, so conclusions relying on the injectivity can be wrong.

share|cite|improve this answer

answered May 21 '13 at 7:59

Alfe

736417

add a comment | 

4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
32
down vote



accepted

You have shown that $e^2pi i = e^0$. This does not imply $2pi i = 0$, because $e^z$ is not injective.
You have to give up your intuition about real functions when you move them to the complex plane, because they change a lot. $e^z$ is actually periodic for complex $z$.

share|cite|improve this answer

answered May 21 '13 at 1:17

6005

34.4k749122

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think the thing about real functions may be a red herring. As Glen O pointed out in comments, there are plenty of noninjective functions. For example, $(-1)^2 = 1^2$, and yet $-1ne 1$. Or even $3cdot 0 = 4cdot 0$, and yet $3ne 4$.
  – MJD
  May 21 '13 at 1:20
  
  

  
  
  
  


• 
  
  
  

  
  20
  

  
  
  
  
  
  

  
  Sure, but I don't think that's where the OP is confused. I think they understand that $x^2$ and $0$ are noninjective functions, but they assume without thinking about it that $e^x$ is injective, because in their experience (with real $e^x$) it is.
  – 6005
  May 21 '13 at 1:24
  
  

  
  
  
  

add a comment | 

up vote
32
down vote



accepted

You have shown that $e^2pi i = e^0$. This does not imply $2pi i = 0$, because $e^z$ is not injective.
You have to give up your intuition about real functions when you move them to the complex plane, because they change a lot. $e^z$ is actually periodic for complex $z$.

share|cite|improve this answer

answered May 21 '13 at 1:17

6005

34.4k749122

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think the thing about real functions may be a red herring. As Glen O pointed out in comments, there are plenty of noninjective functions. For example, $(-1)^2 = 1^2$, and yet $-1ne 1$. Or even $3cdot 0 = 4cdot 0$, and yet $3ne 4$.
  – MJD
  May 21 '13 at 1:20
  
  

  
  
  
  


• 
  
  
  

  
  20
  

  
  
  
  
  
  

  
  Sure, but I don't think that's where the OP is confused. I think they understand that $x^2$ and $0$ are noninjective functions, but they assume without thinking about it that $e^x$ is injective, because in their experience (with real $e^x$) it is.
  – 6005
  May 21 '13 at 1:24
  
  

  
  
  
  

add a comment | 

up vote
32
down vote



accepted

up vote
32
down vote



accepted

You have shown that $e^2pi i = e^0$. This does not imply $2pi i = 0$, because $e^z$ is not injective.
You have to give up your intuition about real functions when you move them to the complex plane, because they change a lot. $e^z$ is actually periodic for complex $z$.

share|cite|improve this answer

answered May 21 '13 at 1:17

6005

34.4k749122

You have shown that $e^2pi i = e^0$. This does not imply $2pi i = 0$, because $e^z$ is not injective.
You have to give up your intuition about real functions when you move them to the complex plane, because they change a lot. $e^z$ is actually periodic for complex $z$.

share|cite|improve this answer

answered May 21 '13 at 1:17

6005

34.4k749122

share|cite|improve this answer

share|cite|improve this answer

answered May 21 '13 at 1:17

6005

34.4k749122

answered May 21 '13 at 1:17

6005

34.4k749122

answered May 21 '13 at 1:17

6005

34.4k749122

34.4k749122

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think the thing about real functions may be a red herring. As Glen O pointed out in comments, there are plenty of noninjective functions. For example, $(-1)^2 = 1^2$, and yet $-1ne 1$. Or even $3cdot 0 = 4cdot 0$, and yet $3ne 4$.
  – MJD
  May 21 '13 at 1:20
  
  

  
  
  
  


• 
  
  
  

  
  20
  

  
  
  
  
  
  

  
  Sure, but I don't think that's where the OP is confused. I think they understand that $x^2$ and $0$ are noninjective functions, but they assume without thinking about it that $e^x$ is injective, because in their experience (with real $e^x$) it is.
  – 6005
  May 21 '13 at 1:24
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think the thing about real functions may be a red herring. As Glen O pointed out in comments, there are plenty of noninjective functions. For example, $(-1)^2 = 1^2$, and yet $-1ne 1$. Or even $3cdot 0 = 4cdot 0$, and yet $3ne 4$.
  – MJD
  May 21 '13 at 1:20
  
  

  
  
  
  


• 
  
  
  

  
  20
  

  
  
  
  
  
  

  
  Sure, but I don't think that's where the OP is confused. I think they understand that $x^2$ and $0$ are noninjective functions, but they assume without thinking about it that $e^x$ is injective, because in their experience (with real $e^x$) it is.
  – 6005
  May 21 '13 at 1:24
  
  

  
  
  
  

1

1

I think the thing about real functions may be a red herring. As Glen O pointed out in comments, there are plenty of noninjective functions. For example, $(-1)^2 = 1^2$, and yet $-1ne 1$. Or even $3cdot 0 = 4cdot 0$, and yet $3ne 4$.
– MJD
May 21 '13 at 1:20

I think the thing about real functions may be a red herring. As Glen O pointed out in comments, there are plenty of noninjective functions. For example, $(-1)^2 = 1^2$, and yet $-1ne 1$. Or even $3cdot 0 = 4cdot 0$, and yet $3ne 4$.
– MJD
May 21 '13 at 1:20

20

20

Sure, but I don't think that's where the OP is confused. I think they understand that $x^2$ and $0$ are noninjective functions, but they assume without thinking about it that $e^x$ is injective, because in their experience (with real $e^x$) it is.
– 6005
May 21 '13 at 1:24

Sure, but I don't think that's where the OP is confused. I think they understand that $x^2$ and $0$ are noninjective functions, but they assume without thinking about it that $e^x$ is injective, because in their experience (with real $e^x$) it is.
– 6005
May 21 '13 at 1:24

add a comment | 

up vote
11
down vote

The $log$ function is multi-valued on $mathbbC^*$ (you can however choose a "branch" of it; see Wikipedia). At any rate, just because
$$e^2pi i=e^0$$
does not imply that $2pi i=0$.

share|cite|improve this answer

edited May 21 '13 at 1:22

answered May 21 '13 at 1:17

Zev Chonoles

108k16217408

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I did not figure such a big change occurs when moving over to the complex plane and figured it would be as intuitive as with real numbers. Thank you (and other answerers/commenters).
  – Sidd Singal
  May 21 '13 at 1:24
  

  
  
  
  

add a comment | 

up vote
11
down vote

The $log$ function is multi-valued on $mathbbC^*$ (you can however choose a "branch" of it; see Wikipedia). At any rate, just because
$$e^2pi i=e^0$$
does not imply that $2pi i=0$.

share|cite|improve this answer

edited May 21 '13 at 1:22

answered May 21 '13 at 1:17

Zev Chonoles

108k16217408

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I did not figure such a big change occurs when moving over to the complex plane and figured it would be as intuitive as with real numbers. Thank you (and other answerers/commenters).
  – Sidd Singal
  May 21 '13 at 1:24
  

  
  
  
  

add a comment | 

up vote
11
down vote

up vote
11
down vote

The $log$ function is multi-valued on $mathbbC^*$ (you can however choose a "branch" of it; see Wikipedia). At any rate, just because
$$e^2pi i=e^0$$
does not imply that $2pi i=0$.

share|cite|improve this answer

edited May 21 '13 at 1:22

answered May 21 '13 at 1:17

Zev Chonoles

108k16217408

The $log$ function is multi-valued on $mathbbC^*$ (you can however choose a "branch" of it; see Wikipedia). At any rate, just because
$$e^2pi i=e^0$$
does not imply that $2pi i=0$.

share|cite|improve this answer

edited May 21 '13 at 1:22

answered May 21 '13 at 1:17

Zev Chonoles

108k16217408

share|cite|improve this answer

share|cite|improve this answer

edited May 21 '13 at 1:22

edited May 21 '13 at 1:22

edited May 21 '13 at 1:22

answered May 21 '13 at 1:17

Zev Chonoles

108k16217408

answered May 21 '13 at 1:17

Zev Chonoles

108k16217408

answered May 21 '13 at 1:17

Zev Chonoles

108k16217408

108k16217408

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I did not figure such a big change occurs when moving over to the complex plane and figured it would be as intuitive as with real numbers. Thank you (and other answerers/commenters).
  – Sidd Singal
  May 21 '13 at 1:24
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I did not figure such a big change occurs when moving over to the complex plane and figured it would be as intuitive as with real numbers. Thank you (and other answerers/commenters).
  – Sidd Singal
  May 21 '13 at 1:24
  

  
  
  
  

I did not figure such a big change occurs when moving over to the complex plane and figured it would be as intuitive as with real numbers. Thank you (and other answerers/commenters).
– Sidd Singal
May 21 '13 at 1:24

I did not figure such a big change occurs when moving over to the complex plane and figured it would be as intuitive as with real numbers. Thank you (and other answerers/commenters).
– Sidd Singal
May 21 '13 at 1:24

add a comment | 

up vote
11
down vote

It's like saying that, because $sinpi = sin0$, that $pi = 0$. Not all functions have perfect inverses; sin being one of them. In the complex numbers, $e^z$ doesn't either.

You're implicitly going: $e^2pi i = e^0 implies lne^2pi i = lne^0 implies 2 pi i = 0$. The error here is that $forall x in mathbbC lne^x = x$ is not true! $ln$ isn't even a function, just like the naive version of $arcsin(x)$. You have to make a choice of range, which is usually $Im(x) in (-pi, pi]$.

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answered May 21 '13 at 1:24

Henry Swanson

9,54912151

add a comment | 

up vote
11
down vote

It's like saying that, because $sinpi = sin0$, that $pi = 0$. Not all functions have perfect inverses; sin being one of them. In the complex numbers, $e^z$ doesn't either.

You're implicitly going: $e^2pi i = e^0 implies lne^2pi i = lne^0 implies 2 pi i = 0$. The error here is that $forall x in mathbbC lne^x = x$ is not true! $ln$ isn't even a function, just like the naive version of $arcsin(x)$. You have to make a choice of range, which is usually $Im(x) in (-pi, pi]$.

share|cite|improve this answer

answered May 21 '13 at 1:24

Henry Swanson

9,54912151

add a comment | 

up vote
11
down vote

up vote
11
down vote

It's like saying that, because $sinpi = sin0$, that $pi = 0$. Not all functions have perfect inverses; sin being one of them. In the complex numbers, $e^z$ doesn't either.

You're implicitly going: $e^2pi i = e^0 implies lne^2pi i = lne^0 implies 2 pi i = 0$. The error here is that $forall x in mathbbC lne^x = x$ is not true! $ln$ isn't even a function, just like the naive version of $arcsin(x)$. You have to make a choice of range, which is usually $Im(x) in (-pi, pi]$.

share|cite|improve this answer

answered May 21 '13 at 1:24

Henry Swanson

9,54912151

It's like saying that, because $sinpi = sin0$, that $pi = 0$. Not all functions have perfect inverses; sin being one of them. In the complex numbers, $e^z$ doesn't either.

You're implicitly going: $e^2pi i = e^0 implies lne^2pi i = lne^0 implies 2 pi i = 0$. The error here is that $forall x in mathbbC lne^x = x$ is not true! $ln$ isn't even a function, just like the naive version of $arcsin(x)$. You have to make a choice of range, which is usually $Im(x) in (-pi, pi]$.

share|cite|improve this answer

answered May 21 '13 at 1:24

Henry Swanson

9,54912151

share|cite|improve this answer

share|cite|improve this answer

answered May 21 '13 at 1:24

Henry Swanson

9,54912151

answered May 21 '13 at 1:24

Henry Swanson

9,54912151

answered May 21 '13 at 1:24

Henry Swanson

9,54912151

9,54912151

add a comment | 

add a comment | 

up vote
3
down vote

Actually, the following is true:

$ e^2npi i = 1  forall  n in mathbbZ $

As a special case of $n = 0$ this gives what you know: $e^0 = 1$, but as mentioned, this is only a special case of the general formula given above.

You could as well assume that $2pi = 0$ because of

$sin 2pi = sin 0 land cos 2pi = cos 0$.

As others already mentioned, this stuff just isn't injective, so conclusions relying on the injectivity can be wrong.

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answered May 21 '13 at 7:59

Alfe

736417

add a comment | 

up vote
3
down vote

Actually, the following is true:

$ e^2npi i = 1  forall  n in mathbbZ $

As a special case of $n = 0$ this gives what you know: $e^0 = 1$, but as mentioned, this is only a special case of the general formula given above.

You could as well assume that $2pi = 0$ because of

$sin 2pi = sin 0 land cos 2pi = cos 0$.

As others already mentioned, this stuff just isn't injective, so conclusions relying on the injectivity can be wrong.

share|cite|improve this answer

answered May 21 '13 at 7:59

Alfe

736417

add a comment | 

up vote
3
down vote

up vote
3
down vote

Actually, the following is true:

$ e^2npi i = 1  forall  n in mathbbZ $

As a special case of $n = 0$ this gives what you know: $e^0 = 1$, but as mentioned, this is only a special case of the general formula given above.

You could as well assume that $2pi = 0$ because of

$sin 2pi = sin 0 land cos 2pi = cos 0$.

As others already mentioned, this stuff just isn't injective, so conclusions relying on the injectivity can be wrong.

share|cite|improve this answer

answered May 21 '13 at 7:59

Alfe

736417

Actually, the following is true:

$ e^2npi i = 1  forall  n in mathbbZ $

As a special case of $n = 0$ this gives what you know: $e^0 = 1$, but as mentioned, this is only a special case of the general formula given above.

You could as well assume that $2pi = 0$ because of

$sin 2pi = sin 0 land cos 2pi = cos 0$.

As others already mentioned, this stuff just isn't injective, so conclusions relying on the injectivity can be wrong.

share|cite|improve this answer

answered May 21 '13 at 7:59

Alfe

736417

share|cite|improve this answer

share|cite|improve this answer

answered May 21 '13 at 7:59

Alfe

736417

answered May 21 '13 at 7:59

Alfe

736417

answered May 21 '13 at 7:59

Alfe

736417

736417

add a comment | 

add a comment |