What is the value of $\ln(-1)$ ?

In Mathematics, if you find the answer in the real world, find it in the complex world.

$\begin{aligned} \ln (-1) & = \ln (i^2) = 2 \ln i & \small \blue{\text{where }i = \sqrt{-1} \text{ denotes the imaginary unit.}} \\ & = 2 \ln \left(e^{\frac \pi 2 i}\right) & \small \blue{\text{By Euler's formula: }e^{\theta i}= \cos \theta + i\sin \theta} \\ & = 2 \times \frac \pi 2 i = \boxed {\pi i} \end{aligned}$

References:

• Imaginary unit
• Euler's formula

If $\ln(-1)=y \Rightarrow -1=e^y$ . Since by Euler's formula $e^{it}=\cos(t)+i\sin(t)$ we find $t=(π + 2kπ)$ , for any $k \in \Z$ . So there is more than one solution, because $e^y$ is periodic (mod $2πi$ ). The case $k=0$ is called the principal solution and gives $y=πi$ .

We know $e^{ix}=\cos x + i \sin x$ and we also know when $x = \pi$ then $e^{i\pi}=-1 \newline e^{i\pi}=e^{\ln(-1)} \newline \boxed{\ln(-1) = \pi i}$

$ln(-1) = (2n + 1) πi$ for any integer n

For this question only n=0 is a given answer but all other solutions are also good enough unless we want to keep ln() as an one-one function and not a one-many relation

$ln(x+iy)$ = $\frac{ln(x^2+y^2)}{2}$ $+ i*arg(x+iy)$ is one possible definition of the natural logarithm which is one-one in the complex plane