I'm having problem understanding how to get to this solution. log(-1+0i)=ipi Thank you for your help!
Follow Math Help Forum on Facebook and Google+
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!
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: $\log(z)=\log(|z|)+{\bf{i}}\arg(z)$ $\log(|-1|)=\log(1)+\bf{i}\pi$.