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 Exercise [5.10]e 
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Joined: 02 Oct 2010, 15:29
Posts: 2
Post Exercise [5.10]e
The problem says
(e^(1+2*pi*i))^(1+2*pi*i)=e^(1+4*pi*i-4*pi^2)
To get this equation, the identity
(a^m)^n=a^(m*n)
has to be used. This identity is true for all real m and n, but it doesn't hold in general when m and n are complex. Assuming this identity holds where it really doesn't is the source of the "paradox".
See this http://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities.

I know these equations are difficult to read, but I couldn't get LaTex to work. The LaTex equations I tried to put in this post are below. I would greatly appreciate it if someone could look at them and tell me why they won't show up.
(e^{1+2{\pi}i})^{1+2{\pi}i}=e^{1+4{\pi}i-4\pi^2}
(a^{m})^n=a^{mn}


Last edited by badassmofo on 03 Oct 2010, 16:11, edited 1 time in total.

02 Oct 2010, 21:58
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Exercise [05.10]e
Some of the information in your link is confusing. I will expand on this in a later post.


Last edited by vasco on 03 Oct 2010, 09:18, edited 2 times in total.

03 Oct 2010, 07:14
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Post Re: Exercise [05.10]e
The function w=a^z where a is real and w and z are complex, is a multi-function.
That is, for any value of z there is an infinite number of values of w=a^z.
So if we change the meaning of = to mean "the set of" then we can say w=a^z.
We can see this kind of behaviour without going to complex functions:
y=x^(1/2) where x and y are real, is a multifunction. If we choose x=4 then y=+2 or -2, so y has two values for every x and we can't say +2=-2, but we can say y=x^(1/2) if we understand = to mean "the set of". For every value of x there are two values of y and for every value of y there is one value of x.
Here's an example of a similar paradox:
2=4^(1/2)=4^(1/2)=(4 x 1)^(1/2)=4^(1/2) x 1^(1/2)=2 x -1=-2
In the case w=a^z where w, a, z are all complex also results in w being a multivalued function and as long as we realise this and use = to mean "the set of" then we are OK again.
It's true that if we try and do ordinary arithmetic with them, then we will run into difficulty. We should use set theory or define them on the Riemann sphere.
The apparently strange results shown in your link:
http://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities
are due entirely to this multivaluedness.

Don't forget to have a look at the other solutions to this exercise.
The exercises in the forum default to date/time order. You need to sort them by subject to see the previous solutions or look at the sorted solutions https://sites.google.com/site/vascoprat/rtr-solutions/chapter-5

Note: The Latex problem is something we know about and doesn't look like being solved any time soon. The best solution is to produce a pdf file from your Latex and attach the pdf file to your post.


03 Oct 2010, 08:51
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