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Exercise [05.15]
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Supporter Joined: 07 Jun 2008, 08:21
Posts: 235
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
In the proof I'm struggling with the last section: I can't see how this step works

Quote:  If we can just equate exponentials, is there anything to prove (as you seem to say in the note at the end), since 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:  why does the 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 Supporter Joined: 07 Jun 2008, 08:21
Posts: 235
Hi dickdock
As far as using to denote the general value and 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 then   putting gives  So generally Also if then multiplying both sides by gives    In our cases below and elsewhere, this can always be absorbed into 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 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 in (3), out of the infinity of values it can have, and we call that value , then we must choose, out of the infinity of values that in (2) can have, the value .
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 for (3) is chosen as , then   22 Jul 2008, 16:21 Joined: 11 Jul 2008, 05:14
Posts: 9
University textbooks? That's cheating.

Thanks for the clarification, but putting and also allowing doesn't make much notational sense to me.

Anyway, as you point out above, taking the log of both sides introduces a ambiguity. So when you do shouldn't there be a term introduced here? Giving which appears to affect what follows.

24 Jul 2008, 22:59 Supporter Joined: 07 Jun 2008, 08:21
Posts: 235
dickdock wrote:
Thanks for the clarification, but putting and also allowing 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  represents an infinity of values whereas is only one value. So (1) means that if you

- choose a value for - take the infinity of values of and add to each one then you will have another infinity of values for which will be the same as the original set.
It's better to think of as representing an infinite set of values and then think of (1) as saying that if we take the set represented by and add to 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 ambiguity. So when you do shouldn't there be a term introduced here? Giving 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
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 Page 1 of 1 [ 6 posts ] 