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Exercise [5.10]e
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Author:  Azrael84 [ 23 Sep 2011, 16:08 ]
Post subject:  Exercise [5.10]e

This is my attempt at a solution essentially it boils down to:

If we want (e^a)^b=e^{ab} (1) to hold then after taking Log(e)=log(e)=1 on the RHS we are forced into Log(e^a)=alog(e)=a on the LHS as mentioned by Penrose in [5.15].

So in order for (1) to hold we are assuming the branch Log(e^(1+i2k.pi))=(1+i2k.pi). However the original assertion of the paradox was that e=e^{(1+i2k.pi)} and taking logarithms each side: Log(e)=Log(e^{1+i2k.pi}) . Now because we took Log(e)=log(e)=1 earlier we should do again for consistency forcing us to conclude Log(e^{1+i2k.pi}) =1 contrary to our earlier conclusion that Log(e^(1+i2k.pi))=(1+i2k.pi) that we had to assume for the validity of (1).

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Author:  Azrael84 [ 23 Sep 2011, 17:24 ]
Post subject:  Re: Exercise [5.10]e

Another solution based on a different method:

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