# Can I find the natural log of a negative number?

Nov 5, 2015

Yes, if $x < 0$ then the principal value of $\ln \left(x\right)$ is $\ln \left(- x\right) + i \pi$

#### Explanation:

The Real valued function ${e}^{x} : \mathbb{R} \to \left(0 , \infty\right)$ is one to one, with inverse function $\ln \left(x\right) : \left(0 , \infty\right) \to \mathbb{R}$.

We can extend the definition of ${e}^{x}$ to the Complex valued function ${e}^{z} : \mathbb{C} \to \mathbb{C} \setminus \setminus \left\{0\right\}$, but this is a many to one function, so it has no inverse function, unless we do something to limit the domain of ${e}^{z}$ or the range of $\ln z$.

For example, if we limit the domain of ${e}^{z}$ to the set $\left\{a + i b \in \mathbb{C} : - \pi < b \le \pi\right\}$, then it is a one to one function with inverse function:

$\ln \left(z\right) : \mathbb{C} \setminus \setminus \left\{0\right\} \to \left\{a + i b \in \mathbb{C} : - \pi < b \le \pi\right\}$

If $x < 0$, then ${e}^{\ln \left(- x\right) + i \pi} = {e}^{\ln} \left(- x\right) {e}^{i \pi} = - x \cdot - 1 = x$