Exercise 05.15 - proposed solution

dickdock · 16 Jul 2008, 16:46

Prove $(w^a)^b = w^{ab}$ (w nonzero) requires choosing logw^a = aw_0

where

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