When a and b are real numbers, ln a^b = b ln a, but if not, you can’t garantee. e^ix = cos x + isen x, so e^ix = e^i(x+2pi). You can see that exponential is not one-to-one in the complex field, that’s why ln behaves so strangely.

MÃquei

e^ipi = -1
ln -1 = i(pi + 2kpi), k integer
If you let k = 1 and 3 at the same time
2 = 6

BradenThe ln(i^2) one is actually kind of interesting. How would you address it?

Alexis Beaujet6ln(i) = 2ln(i)

ln(i^6)=ln(i^2)

exp(ln(i^6))=exp(ln(i^2))

i^6 = i^2

iÂ²*iÂ²*iÂ² = iÂ²

-1 * -1 * -1 = -1

-1 *1=-1

1=1

MÃqueiWhen a and b are real numbers, ln a^b = b ln a, but if not, you can’t garantee. e^ix = cos x + isen x, so e^ix = e^i(x+2pi). You can see that exponential is not one-to-one in the complex field, that’s why ln behaves so strangely.

MÃqueie^ipi = -1

ln -1 = i(pi + 2kpi), k integer

If you let k = 1 and 3 at the same time

2 = 6

BradenThanks, Miquei!