What is the natural log of -0.9?

Mar 17, 2018

$\ln \left(- 0.9\right) = \ln \left(0.9\right) + i \pi \approx - 0.10536 + i \pi$

Explanation:

Note that the natural logarithm is supposed to be the inverse of the exponential function ${e}^{x}$. So the answer to our question is a solution of:

${e}^{x} = - 0.9$

Note however that ${e}^{x} > 0$ for all real values of $x$.

So there is no real value of $x$ which is a candidate for the natural logarithm.

The exponential function ${e}^{x}$ is applicable to complex numbers, so we can look for complex solutions of ${e}^{x} = - 0.9$.

Note that Euler's identity tells us that:

${e}^{i \pi} + 1 = 0$

So we find:

${e}^{i \pi + \ln 0.9} = {e}^{i \pi} \cdot {e}^{\ln 0.9} = - 1 \cdot 0.9 = - 0.9$

In fact the principal value of the complex natural logarithm of $- 0.9$ is $\ln 0.9 + i \pi$.

"Principal value" ?

Any number of the form $x = \ln 0.9 + \left(2 k + 1\right) \pi i$ (with $k$ an integer) will satisfy ${e}^{x} = - 0.9$.

By convention, the principal value of $\ln \left(r {e}^{i \theta}\right)$ is $\ln r + i \theta$ for $\theta \in \left(- \pi , \pi\right]$.

To find the value of $\ln 0.9$ we can use:

$\ln \left(1 + t\right) = t - {t}^{2} / 2 + {t}^{3} / 3 - {t}^{4} / 4 + \ldots$

So:

$\ln \left(1 - t\right) = t + {t}^{2} / 2 + {t}^{3} / 3 + {t}^{4} / 4 + \ldots$

and:

$\ln \left(0.9\right) = \ln \left(1 - 0.1\right) = - \left(0.1 + \frac{0.01}{2} + \frac{0.001}{3} + \frac{0.0001}{4} + \ldots\right)$

$\approx - 0.10536$

So:

$\ln \left(- 0.9\right) = \ln \left(- 0.9\right) + i \pi \approx - 0.10536 + i \pi$