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Contradiction in Euler’s Identity? $2ipi = 0$ [duplicate]

How can zero divided by $i$ be $2pi$?Separating equation into real and imaginary parts and eliminating integralFor which complex $a,,b,,c$ does $(a^b)^c=a^{bc}$ hold?Why is it valid to multiply both sides of an equation by its complex conjugate?Euler Identity Derivation-Is this allowed?Use Euler’s formula to write an exponential function in the form $a + ib$Proving a complex identitySomething strange regarding Euler's IdentityHow has the definition of the complex logarithm evolved since Cotes (1682-1716)?Good treatment of Euler’s formulaNatural Logarithm contradiction?

0

$begingroup$

This question already has an answer here:

• 
  How can zero divided by $i$ be $2pi$?
  
  3 answers
  
  

Many of us know Euler’s famous identity:

$e^{ipi} + 1 = 0$.

But if we add -1 (subtract 1) to both sides we get:

$e^{ipi} = -1$

then natural log of both sides and:

$ipi = ln(-1)$

Next we multiply both sides by 2:

$2ipi = 2ln(-1)$

Which by basic logarithm rules is equal to:

$2ipi = ln({-1}^2) = ln(1) = 0$

so

$2ipi = 0$

which can’t be true as either 2, i or $pi$ would have to equal 0. My best guess is that logarithms just plainly aren’t defined for negative values, I just assumed this was just true logarithms bounded under the real numbers, though I might be wrong. If someone could help it would be great.

Thanks.

complex-numbers eulers-constant

share|cite|improve this question

edited Mar 13 at 6:47

Blue

49.1k870156

asked Mar 13 at 6:38

L. McDonaldL. McDonald

8010

$endgroup$

marked as duplicate by Lord Shark the Unknown, Matthew Towers, J. W. Tanner, Leucippus, Eevee Trainer Mar 14 at 2:58

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Natural logarithm is for positive numbers, not complex numbers. Note that $e^{0}=e^{2pi i}$ but $0 neq 2pi i$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 13 at 6:42
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, and as said below, it is defined, but differently. Thanks.
  $endgroup$
  – L. McDonald
  Mar 13 at 6:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What i mean is that the multi-valued logarithm in the complex plane is not called natural logarithm and properties of natural logarithm don't hold for it.
  $endgroup$
  – Kavi Rama Murthy
  Mar 13 at 6:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, I missed that in your comment.
  $endgroup$
  – L. McDonald
  Mar 13 at 6:47
  

  
  
  
  

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  How can zero divided by $i$ be $2pi$?
  
  3 answers
  
  

Many of us know Euler’s famous identity:

$e^{ipi} + 1 = 0$.

But if we add -1 (subtract 1) to both sides we get:

$e^{ipi} = -1$

then natural log of both sides and:

$ipi = ln(-1)$

Next we multiply both sides by 2:

$2ipi = 2ln(-1)$

Which by basic logarithm rules is equal to:

$2ipi = ln({-1}^2) = ln(1) = 0$

so

$2ipi = 0$

which can’t be true as either 2, i or $pi$ would have to equal 0. My best guess is that logarithms just plainly aren’t defined for negative values, I just assumed this was just true logarithms bounded under the real numbers, though I might be wrong. If someone could help it would be great.

Thanks.

complex-numbers eulers-constant

share|cite|improve this question

edited Mar 13 at 6:47

Blue

49.1k870156

asked Mar 13 at 6:38

L. McDonaldL. McDonald

8010

$endgroup$

marked as duplicate by Lord Shark the Unknown, Matthew Towers, J. W. Tanner, Leucippus, Eevee Trainer Mar 14 at 2:58

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Natural logarithm is for positive numbers, not complex numbers. Note that $e^{0}=e^{2pi i}$ but $0 neq 2pi i$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 13 at 6:42
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, and as said below, it is defined, but differently. Thanks.
  $endgroup$
  – L. McDonald
  Mar 13 at 6:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What i mean is that the multi-valued logarithm in the complex plane is not called natural logarithm and properties of natural logarithm don't hold for it.
  $endgroup$
  – Kavi Rama Murthy
  Mar 13 at 6:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, I missed that in your comment.
  $endgroup$
  – L. McDonald
  Mar 13 at 6:47
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  How can zero divided by $i$ be $2pi$?
  
  3 answers
  
  

Many of us know Euler’s famous identity:

$e^{ipi} + 1 = 0$.

But if we add -1 (subtract 1) to both sides we get:

$e^{ipi} = -1$

then natural log of both sides and:

$ipi = ln(-1)$

Next we multiply both sides by 2:

$2ipi = 2ln(-1)$

Which by basic logarithm rules is equal to:

$2ipi = ln({-1}^2) = ln(1) = 0$

so

$2ipi = 0$

which can’t be true as either 2, i or $pi$ would have to equal 0. My best guess is that logarithms just plainly aren’t defined for negative values, I just assumed this was just true logarithms bounded under the real numbers, though I might be wrong. If someone could help it would be great.

Thanks.

complex-numbers eulers-constant

share|cite|improve this question

edited Mar 13 at 6:47

Blue

49.1k870156

asked Mar 13 at 6:38

L. McDonaldL. McDonald

8010

$endgroup$

share|cite|improve this question

edited Mar 13 at 6:47

Blue

49.1k870156

asked Mar 13 at 6:38

L. McDonaldL. McDonald

8010

share|cite|improve this question

edited Mar 13 at 6:47

Blue

49.1k870156

asked Mar 13 at 6:38

L. McDonaldL. McDonald

8010

edited Mar 13 at 6:47

Blue

49.1k870156

edited Mar 13 at 6:47

Blue

49.1k870156

asked Mar 13 at 6:38

L. McDonaldL. McDonald

8010

asked Mar 13 at 6:38

L. McDonaldL. McDonald

8010

1 Answer
1

active

oldest

votes

4

$begingroup$

The (complex) logarithm is a multivalued function. The solution to $$ e^{z} = w$$ is given by $$ z= log w + 2pi i n$$ where $n$ is an arbitrary integer, $n=dots, -2,-1,0,1,2,dots$.

In your case, you will get
$$ 2pi i = 0 + 2pi in$$ which is clearly no contradiction (as $n=1$ makes this an equality).

share|cite|improve this answer

edited Mar 13 at 6:43

answered Mar 13 at 6:42

FabianFabian

20k3774

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you please edit this so it gives a definitively answer... thanks
  $endgroup$
  – L. McDonald
  Mar 13 at 6:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. That is better.
  $endgroup$
  – L. McDonald
  Mar 13 at 6:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @L.McDonald If you feel it is a definitive answer then mark it with a $color{green}checkmark$
  $endgroup$
  – Chase Ryan Taylor
  Mar 13 at 7:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know, but it was late at night for me, and I didn’t have time for the 10 minute break.
  $endgroup$
  – L. McDonald
  Mar 14 at 4:49
  

  
  
  
  

add a comment | 

4

$begingroup$

The (complex) logarithm is a multivalued function. The solution to $$ e^{z} = w$$ is given by $$ z= log w + 2pi i n$$ where $n$ is an arbitrary integer, $n=dots, -2,-1,0,1,2,dots$.

In your case, you will get
$$ 2pi i = 0 + 2pi in$$ which is clearly no contradiction (as $n=1$ makes this an equality).

share|cite|improve this answer

edited Mar 13 at 6:43

answered Mar 13 at 6:42

FabianFabian

20k3774

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you please edit this so it gives a definitively answer... thanks
  $endgroup$
  – L. McDonald
  Mar 13 at 6:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. That is better.
  $endgroup$
  – L. McDonald
  Mar 13 at 6:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @L.McDonald If you feel it is a definitive answer then mark it with a $color{green}checkmark$
  $endgroup$
  – Chase Ryan Taylor
  Mar 13 at 7:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know, but it was late at night for me, and I didn’t have time for the 10 minute break.
  $endgroup$
  – L. McDonald
  Mar 14 at 4:49
  

  
  
  
  

add a comment | 

4

$begingroup$

The (complex) logarithm is a multivalued function. The solution to $$ e^{z} = w$$ is given by $$ z= log w + 2pi i n$$ where $n$ is an arbitrary integer, $n=dots, -2,-1,0,1,2,dots$.

In your case, you will get
$$ 2pi i = 0 + 2pi in$$ which is clearly no contradiction (as $n=1$ makes this an equality).

share|cite|improve this answer

edited Mar 13 at 6:43

answered Mar 13 at 6:42

FabianFabian

20k3774

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you please edit this so it gives a definitively answer... thanks
  $endgroup$
  – L. McDonald
  Mar 13 at 6:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. That is better.
  $endgroup$
  – L. McDonald
  Mar 13 at 6:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @L.McDonald If you feel it is a definitive answer then mark it with a $color{green}checkmark$
  $endgroup$
  – Chase Ryan Taylor
  Mar 13 at 7:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know, but it was late at night for me, and I didn’t have time for the 10 minute break.
  $endgroup$
  – L. McDonald
  Mar 14 at 4:49
  

  
  
  
  

add a comment | 

$begingroup$

The (complex) logarithm is a multivalued function. The solution to $$ e^{z} = w$$ is given by $$ z= log w + 2pi i n$$ where $n$ is an arbitrary integer, $n=dots, -2,-1,0,1,2,dots$.

In your case, you will get
$$ 2pi i = 0 + 2pi in$$ which is clearly no contradiction (as $n=1$ makes this an equality).

share|cite|improve this answer

edited Mar 13 at 6:43

answered Mar 13 at 6:42

FabianFabian

20k3774

$endgroup$

share|cite|improve this answer

edited Mar 13 at 6:43

answered Mar 13 at 6:42

FabianFabian

20k3774

answered Mar 13 at 6:42

FabianFabian

20k3774

answered Mar 13 at 6:42

FabianFabian

20k3774