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 Exercise 05.15 - proposed solution 
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Joined: 11 Jul 2008, 05:14
Posts: 9
Post Exercise 05.15 - proposed solution
Prove (w^a)^b = w^{ab} (w nonzero) requires choosing logw^a = aw_0 where w_0 = the specific RHS choice of logw.

Dealing with the special cases where a and/or b are zero:
If a is zero then (w^0)^b = e^{blogw^0} = e^{b(0(logw))} = e^{b0} = 1 = w^{0b} and there's nothing to prove.
If b is zero, then (w^a)^0 = 1 = w^{a0} and there's nothing to prove.

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

Let w_0 = logr_0 + i{\theta}_0

We note first off that w^a = e^{alogw} = e^{a(logr_0 + i({\theta}_0 + 2{\pi}k))}

So logw^a = loge^{a(logr_0 + i({\theta}_0 + 2{\pi}k))} = a(logr_0 + i({\theta}_0 + 2{\pi}k))(1 + 2{\pi}mi)
Giving logw^a= a(logr_0 - 2{\pi}m({\theta}_0 + 2{\pi}k) + i({\theta}_0 + 2{\pi}k + 2{\pi}mlogr_0))

To begin, by definition

(w^a)^b = e^{b(logw^a)} = e^{b(a(logr_0 - 2{\pi}m({\theta}_0 + 2{\pi}k) + i({\theta}_0 + 2{\pi}k + 2{\pi}mlogr_0)))} and
w^{ab} = e^{ab(logw)} = e^{ab(logr_0 + i{\theta}_0)}

We require these to be equal, which means the moduli must match and the arguments must differ by a multiple of 2{\pi}, hence:

Equating the moduli:
ab(logr_0 - 2{\pi}m({\theta_0} + 2{\pi}k)) = ablogr_0
-ab2{\pi}m(\theta}_0 + 2{\pi}k) = 0
-m({\theta}_0 + 2{\pi}k) = 0 {Eqn. 1}

Equating the arguments (give or take multiples of 2{\pi}):
ab({\theta}_0 + 2{\pi}k + logr_02{\pi}m) = ab{\theta}_0 + 2{\pi}n
ab(2{\pi}k + logr_02{\pi}m) = 2{\pi}n
ab(k + logr_0m) = n

Since the LHS is a complex number, and n is an integer, this equation can only be true in general if n = 0. So

ab(k + logr_0m) = 0
k + logr_0m = 0
k = -logr_0m

Subbing into { Eqn. 1}:
-m({\theta}_0 - 2{\pi}logr_0m) = 0
And since m is an integer and {\theta}_0 and logr_0 are real, the only solution in general is m = 0. Hence k = 0.

Hence, logw^a = a(logr_0 + i{\theta}_0) = aw_0


16 Jul 2008, 16:46
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Exercise 05.15 - proposed solution
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 w^a = e^{alogw} = e^{a(logr_0 + i({\theta}_0 + 2{\pi}k))}

So logw^a = loge^{a(logr_0 + i({\theta}_0 + 2{\pi}k))} = a(logr_0 + i({\theta}_0 + 2{\pi}k))(1 + 2{\pi}mi)


This should be:
logw^a = loge^{a(logr_0 + i({\theta}_0 + 2{\pi}k))} = a(logr_0 + i({\theta}_0 + 2{\pi}k))+ 2{\pi}mi

Quote:
To begin, by definition

(w^a)^b = e^{b(logw^a)} = e^{b(a(logr_0 - 2{\pi}m({\theta}_0 + 2{\pi}k) + i({\theta}_0 + 2{\pi}k + 2{\pi}mlogr_0)))} and
w^{ab} = e^{ab(logw)} = e^{ab(logr_0 + i{\theta}_0)}

We require these to be equal, which means the moduli must match and the arguments must differ by a multiple of 2{\pi}, hence:

Equating the moduli:
ab(logr_0 - 2{\pi}m({\theta_0} + 2{\pi}k)) = ablogr_0
-ab2{\pi}m(\theta}_0 + 2{\pi}k) = 0
-m({\theta}_0 + 2{\pi}k) = 0 {Eqn. 1}


The modulus of for example e^{ab(logr_0 + i{\theta}_0)}is not what you say above. Don't forget that a and b are complex. For example:
if you have an expression e^{(c+ i{\theta})} c complex then you cannot say that the modulus of the expression is e^c. If for eample c=c_r+ic_i then
e^{c+i{\theta}}=e^{{c_r}+i(c_i+\theta)} and the modulus is e^{c_r}

There may be other errors which are consequences of these. I hope these comments will help you to review your proposed solution.
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
Post Re: Exercise 05.15 - proposed solution
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
Post Re: Exercise 05.15 - proposed solution
Prove (w^a)^b = w^{ab} (w nonzero) requires choosing {\log}w^a = aw_0 where w_0 = the specific RHS choice of {\log}w.

Let w_0 = \log|w| + i{\theta}_0, a = x_a + iy_a, and b = x_b +iy_b.

We note first off that

{\log}w^a = {\log}e^{a{\log}w} = {\log}e^{(x_a + iy_a)(\log|w| + i({\theta}_0 + 2{\pi}k))}
= (x_a + iy_a)(\log|w| + i({\theta}_0 + 2{\pi}k)) + i2{\pi}m
= x_a\log|w| - y_a({\theta}_0 + 2{\pi}k) + i(x_a({\theta}_0 + 2{\pi}k) + y_a\log|w| + 2{\pi}m)

By definition

(w^a)^b = e^{b{\log}w^a} = e^{(x_b + iy_b)(x_a\log|w| - y_a({\theta}_0 + 2{\pi}k) + i(x_a({\theta}_0 + 2{\pi}k) + y_a\log|w| + 2{\pi}m))}
= e^{x_ax_b\log|w| - y_ax_b({\theta}_0 + 2{\pi}k) - x_ay_b({\theta}_0 + 2{\pi}k) - y_ay_b\log|w| - y_b2{\pi}m + i(blah...)}

We require this to be equal to

w^{ab} = e^{ab{\log}w} = e^{(x_ax_b - y_ay_b + i(x_ay_b + y_ax_b))(log|w| + i{\theta}_0)}
= e^{(x_ax_b - y_ay_b)\log|w| - (x_ay_b + y_ax_b){\theta}_0 + i(blah...)}

so if we equate the moduli we get

x_ax_b\log|w| - y_ax_b({\theta}_0 + 2{\pi}k) - x_ay_b({\theta}_0 + 2{\pi}k) - y_ay_b\log|w| - y_b2{\pi}m
= (x_ax_b - y_ay_b)\log|w| - (x_ay_b + y_ax_b){\theta}_0

Cancelling terms reduces this to

-y_ax_b2{\pi}k - x_ay_b2{\pi}k - y_b2{\pi}m = 0

And we're left with

y_bm = -(y_ax_b + x_ay_b)k

Since y_b and (y_ax_b + x_ay_b) 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 {\log}w^a = x_a\log|w| - y_a{\theta}_0 + i(x_a{\theta}_0 + y_a\log|w|) = aw_0.


22 Jul 2008, 01:41
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