Law of exponent for complex numbers

Question

Why does the following law

$$(z_1z_2)^w=z_1^wz_2^w$$

sometimes fail, such as when $z_1=z_2=-1$ and $w=-i$?

And what are the conditions I need for $z_1$ and $z_2$ so that the above identity hold for all complex values of w?

Answer 1

The if we write $z_i = r_ie^{i\theta_i}$ and $w = x + iy$ then we have $$(z_1z_2)^w = e^{\ln(r_1r_2)x}e^{i\ln(r_1r_2)y}e^{i(\theta_1+\theta_2) x}e^{-y(\theta_1 + \theta_2)}$$

The same formal identity holds for $z_1^wz_2^w.$ The reason the two calculations come out different is that when calculating $(z_1z_2)^w,$ you reduce $\theta_1+\theta_2$ mod $2\pi$ after you multiply $z_1$ and $z_2$, but before you exponentiate. Thus the problematic factor is $e^{-y(\theta_1+\theta_2)}$ which will be a factor of $e^{2\pi y}$ different between the two calculations if such a reduction takes place. Therefore the two calculations come out the same whenever $\theta_1 + \theta_2 < 2\pi$.

That's not the case in your example, since for $z_1 = z_2 = -1,$ $\theta_1 = \theta_2 = \pi$ so you reduce $\theta_1 + \theta_2 = 2\pi$ to $0$ in one of the calculations. Thus the answers are different by a factor of $e^{2\pi}.$

Answer 2

Note first that when $w=0$, then clearly the equality holds. So let $w \neq 0$. Then $$(z_1z_2)^w = z_1^wz_2^w \iff \exp(w \ln(z_1z_2)) = \exp(w \ln(z_1))\exp(w \ln(z_2)) \iff \exp(w \ln(z_1z_2)) = \exp(w(\ln(z_1)+\ln(z_2))) \iff w \ln(z_1z_2) = w(\ln(z_1)+\ln(z_2)) \iff \ln(z_1z_2) = \ln(z_1)+\ln(z_2)$$ Hence we found the required condition on $z_1$ and $z_2$ such that the equality holds, namely $$\ln(z_1z_2) = \ln(z_1)+\ln(z_2).$$