Question

What is the natural log of `-0.9`?

Answer 1

`ln (-0.9) = ln (0.9) + ipi ~~ -0.10536 + ipi`

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^(ipi) + 1 = 0`

So we find:

`e^(ipi + ln 0.9) = e^(ipi) * e^(ln 0.9) = -1 * 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 + (2k+1)pii` (with `k` an integer) will satisfy `e^x = -0.9`.

By convention, the principal value of `ln (r e^(i theta))` is `ln r + i theta` for `theta in (-pi, pi]`.

To find the value of `ln 0.9` we can use:

`ln (1+t) = t-t^2/2+t^3/3-t^4/4+...`

So:

`ln (1-t) = t+t^2/2+t^3/3+t^4/4+...`

and:

`ln (0.9) = ln (1-0.1) = -(0.1+0.01/2+0.001/3+0.0001/4+...)`

`~~ -0.10536`

So:

`ln (-0.9) = ln (-0.9) + ipi ~~ -0.10536+ipi`