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Exercise 05.15 - proposed solution
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Joined: 11 Jul 2008, 05:14
Posts: 9
Prove (w nonzero) requires choosing where the specific RHS choice of .

Dealing with the special cases where a and/or b are zero:
If a is zero then and there's nothing to prove.
If b is zero, then and there's nothing to prove.

So from now on w, a and b are all assumed to be nonzero.

Let We note first off that So Giving To begin, by definition and We require these to be equal, which means the moduli must match and the arguments must differ by a multiple of , hence:

Equating the moduli:   {Eqn. 1}

Equating the arguments (give or take multiples of ):   Since the LHS is a complex number, and n is an integer, this equation can only be true in general if n = 0. So   Subbing into { Eqn. 1}: And since m is an integer and and are real, the only solution in general is m = 0. Hence k = 0.

Hence, 16 Jul 2008, 16:46 Supporter Joined: 07 Jun 2008, 08:21
Posts: 235
Hi
I have looked through your proposed solution. In my opinion there are a few errors in your maths as follows:

Quote:
We note first off that So This should be: Quote:
To begin, by definition and We require these to be equal, which means the moduli must match and the arguments must differ by a multiple of , hence:

Equating the moduli:   {Eqn. 1}

The modulus of for example is not what you say above. Don't forget that a and b are complex. For example:
if you have an expression  complex then you cannot say that the modulus of the expression is . If for eample then and the modulus is You may of course disagree with my comments. Please post your comments. I have in the meantime posted my own solution under "solutions". That's arrogance for you!!

20 Jul 2008, 11:12 Joined: 11 Jul 2008, 05:14
Posts: 9
Hi vasco, you're totally right again, thanks for pointing out the mistakes.

After another bout of re-education, I think I've managed to salvage the proof (I'll post it separately after this) by further symbol-crunching but it leaves it even more tedious than before.

Something mildly interesting mathematically is that this time I've only bothered to equate the moduli to get a result. Which I think makes more sense complex-multiplication-wise since both parts of a and b get intertwined with both parts of logw.

Anyway, thanks for the corrections, much appreciated. Now to examine your proof pixel by pixel so I can pick holes, I mean applaud it magnanimously.

22 Jul 2008, 01:24 Joined: 11 Jul 2008, 05:14
Posts: 9
Prove (w nonzero) requires choosing where the specific RHS choice of .

Let , , and .

We note first off that   By definition  We require this to be equal to  so if we equate the moduli we get  Cancelling terms reduces this to And we're left with Since and are arbitrary real numbers and k and m are integers, the only way for this to be true in general is for k = m = 0.

Hence .

22 Jul 2008, 01:41 Page 1 of 1 [ 4 posts ] 