1. ## Complex function question

I'm having problem understanding how to get to this solution.

log(-1+0i)=ipi

2. ## Re: Complex function question

There's no solution to this expression. It's an identity.

$\text{ by Euler's formula.}$

$-1 +0i= \cos(\pi) + i \sin(\pi) = e^{i \pi}$

$\text{taking logs on both sides}$

$\log_e(-1+0i) = i \pi$

Thank you!

4. ## Re: Complex function question

Originally Posted by mathfn
I'm having problem understanding how to get to this solution.
log(-1+0i)=ipi
I would expect a student to use the properties of logarithm: $\log(z)=\log(|z|)+{\bf{i}}\arg(z)$
$\log(|-1|)=\log(1)+\bf{i}\pi$.