Originally Posted by

**adkinsjr** The argument of a logarithmic function can never be negative.

It can be, but the solution would be in Complex. That's what I am wondering.

Originally Posted by

**e^(i*pi)** There are no real solutions but there are in $\displaystyle \mathbb{C}$

$\displaystyle log(-x) = log(x) + i\pi$

Thanks. So, if I follow this form, let's say I am calculating log(-4):

Therefore,

log(-4) = log4 + i/pi

log(-4) (approximately)= 0.602 + i/pi

Is this the way to calculate it? What would the value of i/pi be here? My calculator states that it would be approximately 1.36i, but I don't understand how it managed to multiply (sqrt) -1 by 3.14