The Road to Reality

Trouble with 5.10 and what I am supposed to do.
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Author:  Eyedrop [ 25 May 2008, 23:30 ]
Post subject:  Trouble with 5.10 and what I am supposed to do.

I'm having trouble with problem 5.10

Resolve this paradox; e=e^{1+i2\pi}, so e^{1+i2\pi}=(e^{1+i2\pi})^{1+i2\pi}=e^{(1+i2\pi)(1+i2\pi)}=e^{1-4\pi^2}

What am I supposed to do? Find out where a rule is broken along the way to the end?

As far as I am able to find out on the calculator (e^{1+i2\pi})^{1+i2\pi} is not equal to e^{(1+i2\pi)(1+i2\pi)} so the fault must be there somewhere. Am i supposed to find out how the fault happens?

It seems like in this case the (b^{n})^m=b^{nm} doesn't hold

If I take (e^{1+2i\pi})^{1+2i\pi} and split it into e^{1+i2\pi}*(e^{i2\pi})^{1+i2\pi} then it is till equal to e.

Let's say m=i2Pi and n={1+i2\pi} then (e^{m})^{n} should be equal to e^{mn}

I need some help and pointers to what properties i should use.

:editing and trying to figure out the lateX language

Author:  Shaun Culver [ 26 May 2008, 00:49 ]
Post subject:  Re: Trouble with 5.10 and what I am supposed to do.

It seems that this exercise deserves a "smiley face" - and not a "severely confused face".

The "order of operations" rule seems to be violated: Parentheses > exponents & roots > multiplication & division > addition & subtraction

(e^{1+2 \pi i})^{1+2 \pi i}=(e)^{1+2 \pi i}

Here, I have applied the "parentheses" rule before the "exponents..." rule. This replaces:

(e^{1+2 \pi i})^{1+2 \pi i}=e^{1+4 \pi i-4 \pi^2}

I may be wrong...just making an observation.

Author:  Eyedrop [ 26 May 2008, 14:41 ]
Post subject:  Re: Trouble with 5.10 and what I am supposed to do.

How about cases where you can't contract inside the paranthesis.

Let's say we got (e^{a+bi})^{c+di} Is it possible to contract that without knowing what a, b, c and d is?

Anyone know of any good sources which discuss what being raised in the power of a complex number means?

My best bet would be to treat it as a complex number z; (e^a*e^{bi})^{c+di} and then do the same again and end up with z; (e^{a})^{c+di}*(e^{bi})^{c+di}=

This was obviosly wrong as when I try to substitute a and c for 1 and b and d for 2Pi I end up with e^{(1-4\pi^2)+i(4\pi)

The part with e^{ac}*e^{adi}=e^1e^{i2\pi}=e if a and c is 1 and d is 2pi.
So (e^{bi})^{c+di} should end up with the value 1 which it does if I substitute the numbers. Is there a way to contract that expression more without knowing the value of b, c and d? Seems like the trouble comes when I try to raise the imaginary part in a complex number.

If I divide (e^{bi})^{c+di} into e^{bi}^c*e^{bi}^{di}=e^{cbi}*e^{bi}^{di} Then I get 1*e^{i2\pi}^{i2\pi}.

(e^{i2\pi})^{i2\pi}=1 How do i manually solve this? Or what is (e^{bi})^{di} Can I say that since 2pi is the argument of w=1*e^{2\pii} We need to find the principal arguement first which is 0 and then we get (e^0)^{2\pii}

Sorry for my noobishness in advance.

Author:  Sameed Zahoor [ 27 May 2008, 09:36 ]
Post subject:  Re: Trouble with 5.10 and what I am supposed to do.

The solution is not as simple as it seems.I think I know the reason but it will have to wait till I "formalise" it.

Author:  Shaun Culver [ 27 May 2008, 13:57 ]
Post subject:  Re: Trouble with 5.10 and what I am supposed to do.

What is your opinion on my "order of operations" argument?

Author:  Sameed Zahoor [ 31 May 2008, 11:09 ]
Post subject:  Re: Trouble with 5.10 and what I am supposed to do.

Shaun Culver wrote:
What is your opinion on my "order of operations" argument?

In the real number system


It does not matter which operation you perform first.This is one of the basic laws of exponents.In fact,the operation which signifies raising to a power is defined in such a way so that these rules hold.However,in the case of complex numbers this particular rule does not hold due to the ambiguity in the complex logarithm unless...

Author:  ZZR Puig [ 01 Jun 2008, 18:07 ]
Post subject:  Re: Trouble with 5.10 and what I am supposed to do.

The supposed paradox lies in the mistake of considering the exponential function e^zas a number instead of a as function.

We know that:


We can now consider e^z=f(z), so the previous equation is equivalent to:


And in the general case:


That has nothing to do with the equation:


Which is clearly false.

As Shaun Culver said I think it's fairly easy, although for a different reason. It's all a matter of a notation misunderstanding. If we had choosen a different notation criteria we could have written instead:


If this explanation is found acceptable I can post it in the solutions forum.

Author:  ZZR Puig [ 03 Jun 2008, 23:56 ]
Post subject:  Re: Trouble with 5.10 and what I am supposed to do.

I've been giving a bit more thought to the paradox issue, as the previously proposed solution didn't completely pease me.

The problem was that by definition the function f(x)=e^x is defined so the following equation holds:


Even if it's not that obvious whether \left(f(z)\right)^z=f\left(f(z)z\right) also holds for every z or not.

I now see that the origin of the paradox is a misunderstanding of the initial equation stated.

What e=e^{1+i2\pi} really means is that e=e^{1+i2\pi}=ee^{i2\pi}

The extension or generalization of this equation would be:

Therefore e=e^{1+i2\pi}\neq\left(e^{1+i2\pi}\right)^{1+i2\pi}

Another example of the supposed paradox that is more obviously wrong could be this one:
e^2=e^{2+i2\pi}\Rightarrow e=e^{1+i\pi}=\left(e^{1+i\pi}\right)^{1+i\pi}

Author:  damian [ 06 Jun 2008, 12:35 ]
Post subject:  Re: Trouble with 5.10 and what I am supposed to do.

Hi, I am not totally convinced. Why do you say that: Therefore $e=e^{1+2 \pi i} \ne ({e^{{(1+2 \pi i)}})^{1+2 \pi i}}$ ?.

I think the second inequaliy that you wrote should still be an equality. But I still do not know the resolution of the paradox.

I think that $e=e^{1+2 \pi i} = ({e^{{(1+2 \pi i)}})^{1+2 \pi i}}$ holds, but $e=e^{1+2 \pi i} = ({e^{{(1+2 \pi i)}})^{1+2 \pi i}} \ne e^{(1 + 2 \pi i)(1+2 \pi i)}$, but why?


Author:  damian [ 06 Jun 2008, 13:56 ]
Post subject:  Re: Trouble with 5.10 and what I am supposed to do.

I think I have a solution that pleases me.

$(e^a)^b= e^{ab} only if b is real, but does not hold for any b non real. To prove it I think that you have to take logarithms on both sides and use that
$Ln e^{(ab)}=Re(ab)+i(Im(ab)+2 \pi n)$ and $Ln (e^a)^b=b(Ln e^{Re(a)}+i(Im(a)+2 \pi m))= bRe(a)+ib(Im(a)+2 \pi m)$, with n and m integers and Re and Im denoting the real and imaginary part. Choosing n and m to be zero, which I think that you can do, because all of the logarithmic solutions give rise to the same result upon exponentiacion, both sides are equal if b is real.

I am not 100% sure, though. Does anyone find any mistakes?

Author:  vasco [ 08 Jun 2008, 08:52 ]
Post subject:  Re: Trouble with 5.10 and what I am supposed to do.

I think the solution is much simpler than you all seem to think. Roger Penrose gives a strong hint in section 5.5 of his book where he says:
The best way of avoiding these problems (ie 05.10) appears to be to adopt the rule that the notation w to the power z is used only when a particular choice of log w has been specified

He then goes on to say :
in the special case of e to the power z, the tacit convention is always to take the particular choice log e = 1 ....

so if we take the first statement in ex 05.10 i.e.
e =e to the power (1+2*pi*i) and take logarithms of both sides we obtain

log e = log {e to the power (1+2*pi*i)} = 1+2*pi*i

which contradicts the tacit assumption that log e =1

So the upshot of all this is that the first statement:

e =e to the power (1+2*pi*i) is false based on our choice that log e =1

Author:  ZZR Puig [ 08 Jun 2008, 15:42 ]
Post subject:  Re: Trouble with 5.10 and what I am supposed to do.

One thing is to establish conventions to avoid ambiguities or other possible problems, and another one is that certain expressions are mathematically correct or not.

The origin of the paradox is to consider e=e^{1+i2\pi}=\left(e^{1+i2\pi}\right)^{1+i2\pi} as a correct expression when it's not. In fact, e=e^{1+i2\pi}\neq\left(e^{1+i2\pi}\right)^{1+i2\pi} but instead e=e^{1+i2\pi}=e^{1+i2k\pi}.

In short, e^{1+i2k\pi}\neq{e^{(1+i2\pi)^k} even if both terms of the expression are equal for k=1.

Author:  vasco [ 09 Jun 2008, 16:22 ]
Post subject:  Re: Trouble with 5.10 and what I am supposed to do.

I think my solution is correct and depends totally on removing ambiguities, since w^z is multivalued. Here is a more detailed explanation:

The general power w^z

Let’s define Log w=log |w| + i * arg w

This means that Log w is an infinitely many-valued function of w.

The principal value of Log w, obtained by giving arg w its principal value, will be denoted by log w, since it is identical with the ordinary logarithm when w is real and positive.

The principal value of arg z is defined as that which satisfies –pi<arg z<=pi

If z & w denote any complex numbers then we can define the principal value of the power w^z to be the number uniquely determined by the equation

w^z= exp{z * logw}, see Penrose 5.4 (w not = 0)

where log w is the principal value of Log w.

if we choose other values of Log w we obtain other sets of values of the power which can be called subsidiary values.

Once we have chosen a value for Log w then w^z has a unique value for each z, called its principal value.

The following formula contains ALL possible values of w^z

w^z=exp{z * (log w + 2*k*pi*i)}. Formula 1

The principal value of w^z is obtained by setting k=0.

Let’s look at the case where w=e and z=1 i.e. w^z=e^1

Substituting in formula 1 gives

w^z=exp{(log e+ 2*k*pi*i)}.

We obtain the principal value for w^z by setting k=0

w^z=e as expected.

Now if we look at the case where w=e and z=1+2*pi*i and apply formula 1 it gives us

w^z= exp{(1+2*pi*i)(log e + 2*k*pi*i)}

We want to produce the principal value of w^z again for consistency, so we set k=0, which gives


These 2 values for e^z are distinct principal values when z has the two values 1 and 1+2*pi*i.

It is therefore incorrect to equate them as is done at the beginning of exercise 05.10. This is what leads to the paradox.
In order to be consistent and give w^z a unique value for each z then we must always use the same value of Log w. In the above I chose the principal value of Log e which is 1. This is the usual convention.

Author:  vasco [ 10 Jun 2008, 06:27 ]
Post subject:  Re: Trouble with 5.10 and what I am supposed to do.

Here is an observation which people may find useful.

In my last post I stated that

$w^z=exp\{z (\log w + 2k\pi i)\}.$ Formula 1

gives ALL possible values of $w^z$.

Let’s investigate $(w^z)^p$ where p is a complex number

Using formula 1 above, we can see that

$(w^z)^p=exp\{p(\log w^z+2k\pi i)\}= exp\{p(z\log w+2k\pi i)\}= exp\{pz\log w+2k\pi ip\} $
$=exp\{z(p\log w+2k\pi i\frac{p}{z})\}$
$=exp\{z(\log w^p+2k\pi i\frac{p}{z})\}=(w^p)^z$ only if p=z and/or k=0

So generally $(w^z)^p=(w^p)^z $ for all w and p ONLY if we use principal values

Author:  damian [ 10 Jun 2008, 08:56 ]
Post subject:  Re: Trouble with 5.10 and what I am supposed to do.

Hi Vasco!

Thank you for your explanations! It has been very useful and thanks to it many of the ideas about logarthmics of complex numbers are now very clear in my head.

Nevertheless, even if I cannot find any errors in your explanation I find hard to accept your conclusion that e is not equal to $e^{1+2 \pi i}$. Even if as you say they might be distinct principal values, the value as such is the same, is it not?. On the other hand I do not agree with ZZR Puig either. On the other hand, I do not agree with me either because my explanation does not go to the root of the problem. So I do not know. I think that you might probably be very close to the final solution, but there is still some things missing in this apparently easy but incredible problem.

Could you give a little further explanation of why e and $e^{1+2 \pi i}$ are not the same even if intuitively they seem the same?

I have to say that deep in my heart I still believe Eyedrop is right in acknowledging where the problem lies. Eyedrop, have you investigated it further?

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