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 Exercise [05.15] 
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Post Exercise [05.15]
In the very last paragraph I have made a small change to the file which gives some extra information but does not affect the solution to the exercise. 21/7/2008.

I have made further amendments to incorporate corrections and clarifications after some very useful exchanges with dickdock. 26/7/2008
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Last edited by vasco on 26 Jul 2008, 07:39, edited 3 times in total.

20 Jul 2008, 09:11

Joined: 11 Jul 2008, 05:14
Posts: 9
Post Re: Exercise [05.15]
In the proof I'm struggling with the last section: I can't see how this step works

Quote:
exp(abLogz + 2{\pi}inb) = exp(abw)
\Rightarrow abLogz + 2{\pi}inb = abw


If we can just equate exponentials, is there anything to prove (as you seem to say in the note at the end), since

(z^a)^b = e^{b{\log}z^a} = z^{ab} = e^{abw} \Rightarrow b{\log}z^a = abw Q.E.D. (As it 'appens that's what I wrote in my copy of the book originally, before I made the tragic mistake of visiting this forum.)

In the last step,

Quote:
\Rightarrow Logz^a + 2{\pi}in = aw
\Rightarrow Logz^a = aw


why does the 2{\pi}in term disappear?

(Just as an aside, it's unfortunate you've chosen logz for the principal value and Logz for the multivalue since it appears that the norm is the other way around, as per this excellent site
http://math.fullerton.edu/mathews/c2003/ComplexFunLogarithmMod.html)


22 Jul 2008, 03:37
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Post Re: Exercise [05.15]
Hi dickdock
As far as using Logz to denote the general value and logz for the principal value: I didn't invent it, I got it from my university textbooks. So I guess people just use different conventions in different parts of the world? Not sure.

Any way (using my original convention), since

Logz=logz+2\pi ik,~k=0,\pm1,\pm2,etc

then

Logz+2\pi in,~n=0,\pm1,\pm2,etc
=logz+2\pi ik+2\pi in,~k=0,\pm1,\pm2,etc,~n=0,\pm1,\pm2,etc
=logz+2\pi i(k+n),~k=0,\pm1,\pm2,etc,~n=0,\pm1,\pm2,etc

putting m=k+n gives

=logz+2\pi im,~m=0,\pm1,\pm2,etc

=Logz

So generally Logz+2\pi ik=Logz,~k=0,\pm1,\pm2,etc

Also if e^{z_1}=e^{z_2} then multiplying both sides by e^{-z_2} gives
e^{z_1}\cdot e^{-z_2}=1
\Rightarrow e^{(z_1-z_2)}=1
\Rightarrow z_1-z_2=2k\pi i,~k=0,\pm1,\pm2,etc
\Rightarrow z_1=z_2+2k\pi i,~k=0,\pm1,\pm2,etc
In our cases below and elsewhere, this 2k\pi i can always be absorbed into Logz as shown above.

Quote:
If we can just equate exponentials, is there anything to prove (as you seem to say in the note at the end), since

(z^a)^b = e^{b{\log}z^a} = z^{ab} = e^{abw} \Rightarrow b{\log}z^a = abw Q.E.D. (As it 'appens that's what I wrote in my copy of the book originally, before I made the tragic mistake of visiting this forum.)

This is not correct. We are trying to show that if we choose a certain value for Log z in (3), out of the infinity of values it can have, and we call that value w, then we must choose, out of the infinity of values that Logz^a in (2) can have, the value aw.
You must always remember that (2) and (3) in my original post are multivalued and from looking at (3) and (5) it is clear that they are equal only under certain conditions.
If (2) and (3) are to be equal and Logz for (3) is chosen as w, then

exp\{bLogz^a\}=exp\{abw\}
\Rightarrow bLogz^a=abw
\Rightarrow Logz^a=aw


22 Jul 2008, 16:21

Joined: 11 Jul 2008, 05:14
Posts: 9
Post Re: Exercise [05.15]
University textbooks? That's cheating.

Thanks for the 2{\pi}ki clarification, but putting Logz = {\log}z + 2{\pi}ki and also allowing Logz + 2{\pi}ki = Logz doesn't make much notational sense to me.

Anyway, as you point out above, taking the log of both sides introduces a 2{\pi}ki ambiguity. So when you do

\exp(bLogz^a) = \exp(abw) \Rightarrow bLogz^a = abw

shouldn't there be a 2{\pi}ki term introduced here? Giving

\exp(bLogz^a) = \exp(abw) \Rightarrow bLogz^a = abw + 2{\pi}ki

which appears to affect what follows.


24 Jul 2008, 22:59
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Post Re: Exercise [05.15]
dickdock wrote:
Thanks for the 2{\pi}ki clarification, but putting Logz = {\log}z + 2{\pi}ki and also allowing Logz + 2{\pi}ki = Logz doesn't make much notational sense to me.


I see what you mean about the notation. Maybe the following will help:

Strictly speaking you should put

Logz + 2{\pi}ki = Logz,~k=0,\pm1,\pm2~etc~~~~~~~~(1)

Logz represents an infinity of values whereas logz is only one value. So (1) means that if you

- choose a value for k
- take the infinity of values of Logz and add 2\pi ik to each one then you will have another infinity of values for Logz which will be the same as the original set.
It's better to think of Logz as representing an infinite set of values and then think of (1) as saying that if we take the set represented by Logz and add 2\pi ikto each of them we end up with the same set.
There is probably a better notation in Set Theory but I'm not familiar with it. Anyway I hope the accompanying words help to clarify what I mean.

Quote:
Anyway, as you point out above, taking the log of both sides introduces a 2{\pi}ki ambiguity. So when you do

\exp(bLogz^a) = \exp(abw) \Rightarrow bLogz^a = abw

shouldn't there be a 2{\pi}ki term introduced here? Giving

\exp(bLogz^a) = \exp(abw) \Rightarrow bLogz^a = abw + 2{\pi}ki

which appears to affect what follows.

You are right. I will amend the original post to incorporate all your suggestions.


25 Jul 2008, 06:57

Joined: 22 Sep 2011, 10:17
Posts: 9
Post Re: Exercise [05.15]
Near the end of your solution you utilise: Log(z^a)^b=bLogz^a

is this really legit? I mean Log(z^a)^b=Log[exp{b Logz^a}]=log[exp{b Logz^a}]+2iPik=b Logz^a+2iPik

This means that starting in your proof at the line:

bLogz^a-2iPik=abw

=> bLogz^a+2iPik=abw
=>Log(z^a)^b=abw

It is tempting to pull down the b on the LHS and conclude Log(z^a)=aw, but due to the multivalued nature of the Log it's not clear to me that this is always permissible, after all if it was you could just pull down the a in your (2) and equate straight away with (3)

It seems to pull down a b in this manner is equivalent to choosing out of the infinite values, the k=0 one, and in generally you are only forced to have:

=> bLogz^a+2iPik=abw

or Logz^a=aw-2iPik/b

-------------------------------------

by simply comparing your (2) and (3) one can observe:

bLogz^a-abw=2iPik (after choosing the branch of (3)'s log to correspond to w)

then Logz^a=aw-2iPik/b follows after a bit of algebra.
------------

Ahh! but k must actually be zero, by the same reasoning you used in [5.10]b, if we want to equate (2) and (3) we have something like exp(z1)=exp(z2) you really do need z1=z2, not just z1-z2=2iPik.......in just the same way as it was not valid to equate e=exp(1+2iPi) in [5.10]b.


22 Sep 2011, 11:51
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