1. Sep 22, 2011

### LAHLH

Hi,

Can anyone help me resolve and understand this paradox:

$$e=e^{1+2i\pi}$$

and so

$$e=\left[e^{1+2\pi i}\right]^{1+2\pi i}=e^{1+4\pi i-4\pi^2}=e^{1-4\pi^2}$$

which is obviously fallacious. This paradox is from Roger Penrose Road to Reality and is currently hurting my head. I keep thinking I'm pretty close, but I don't feel I've fully grasped the resolution yet, could someone spell it out?

thanks

2. Sep 22, 2011

### pwsnafu

First $e^{i\pi} = -1$. This is called Euler's formula.
Secondly, and more importantly, for arbitrary complex numbers $(a^{b})^c \neq a^{bc}$. There are times when you get equality, and times when you don't.

3. Sep 22, 2011

### LAHLH

Yeah, I know these things but still lack a full understanding I feel.

I am a physicist btw not amateur reading the book (all the more shameful that I can't quite get my head around this given I've taken courses in complex analysis and all that :s).

My ideas so far:

$$w^z=e^{zLn(w)}=e^{z(lnw+k2\pi i)}$$ again I use ln as principal branch and capitalised Ln as multivalued logarithm.

So $$e^{z}=e^{zLne}=e^{z(lne+k2\pi i)}=e^{z(1+k2\pi i)}$$

One can think of e^z as the multivalued function E(z) which maps a given z to a multitude of values, the value we get depends on our definition of Ln(e) (that is to say, which branch we take).

Considering E(1) this could equal e^1 (k=0) or it could equal $$e^{1+2\pi i}$$ (k=1) just depending on which branch we take. But it is not necessarily true that $$e^1=E(1)= e^{1+2\pi i}$$ implies that $$e=e^{1+2\pi i}$$. Anymore than it is true that $$5=sqrt(25)=-5$$ implies that 5=-5. Although I think in this example because of Euler, one can say $$e=e^{1+2\pi i}$$ without issue.

Although I've read other accounts actually critiquing this first line of the fallacy too, but I can't see a problem. (see e.g. http://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities)

4. Sep 22, 2011

### LAHLH

The article explains this part of the fallacy as: "It is that e is a real number whereas the result of e1+2πin is a complex number better represented as e+0i." But I can't see what the difference is between a real x and a complex number x+0i??

But then yes, somehow applying the formula $$(a^b)^c=a^{bc}$$ affects which principal value is chosen in a non-consistent way...this is what I'm struggling to see...

5. Sep 22, 2011

### jackmell

I think the paradox emerges from branching in Complex Analysis. When the sum of the arguments is greater than 2pi, then it's crossing over any single-valued branch of the $\arg$ multifunction. That is:

$$(e^a)^b=e^{ab}$$

if

$$|Im(a)|+|Im(b)|<2\pi$$

Or if just the principal branch is used, then each argument has to be less than $\pi$

So that we can write:

$$\left(e^{1+(\pi+0.01)i}\right)^{1+(\pi+0.01)i}\neq e^{[1+(\pi+0.01)i]^2}$$

but:

$$\left(e^{1+(\pi-0.01)i}\right)^{1+(\pi-0.01)i}= e^{[1+(\pi-0.01)i]^2}$$

Last edited: Sep 22, 2011
6. Sep 22, 2011

### LAHLH

Interesting jackmell, seems to be along the correct lines.

Also in the Penrose book, he states that $$(w^a)^b=w^{ab}$$ where once we have made a choice for Logw on RHS we must (for LHS) make the choice Logw^a=aLogw

We have here $$(e^{1+2\pi i})^{1+2\pi i} =e^{(1+2\pi i)(1+2\pi i)}$$ on the RHS the choice of Loge=1, then we must make the choice for LHS $$Log(e^{1+2\pi i})=(1+2\pi)Loge=(1+2\pi)$$

However in the very first equation we had $$e=e^{1+2\pi i}$$ which taking logs on each side assumes that $$Log(e^{1+2\pi i})=Log(e)=1$$, where Log(e)=1 and not 1+i2pi so as to be consistent with how we chose it on RHS of first equation originally.

I don't know if this argument is true or not, but it seems like it should involve something like this, if anyone can spell it out I'd be grateful.

Last edited: Sep 22, 2011
7. Sep 22, 2011

### LAHLH

I've attached my attempt at resolution. I don't know if it's correct so if any one could comment that would great, or if anyone has a simple solution..

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8. Sep 22, 2011

### Robert1986

The problem is that you are trying to think of complex numbers as though they are real. That is the so-called paradox.

9. Sep 22, 2011

### deluks917

(ea)b = eab does not hold in the complex plane in general.

10. Sep 23, 2011

### icystrike

I'm pretty sure it is covered in complex analysis

11. Sep 23, 2011

### LAHLH

Yes of course, I know $$(w^b)^c=w^{bc}$$ doesn't hold in the complex plane. It was mentioned on post #2 of this thread, and it even says it next to the problem in the Penrose book, and yes of course I have studied it. BUT the problem is a little more than just saying that this law doesn't hold for complex numbers in general: because it does hold for complex numbers under certain assumptions....namely once you make a choice for Logw=k on the RHS, if you choose Log(w^b)=bk on LHS the law $$(w^b)^c=w^{bc}$$ holds!

The question really is where does the inconsistency in branch cut choices arise.

My belief at this point is that you are making two different choices of branch of the function $$Log(e^{1+2\pi i})$$ (one in formula #1 and a different one in assuming the identity $$(w^b)^c=w^{bc}$$ and that is why you get into trouble .....