is there an $i$ of logarithms with negative bases and/or negative arguments?

Question

As you probably know, $\log_{a}(b)$ is undefined if $a<0$ and/or if $b<0$. Can a constant such as $i$ be used but for logarithms with negative bases and/or negative arguments? Can it even be defined like $i$ so that it is a unit such that every other number of this theoretical set of logarithms with negative bases and/or negative arguments can be expressed as a multiple of it? (Just like $i$ and the set of imaginary numbers).

Answer 1

The logarithm can be extended to the complex numbers by defining the complex logarithm by $$\operatorname{Log} z = \ln |z| + i \operatorname{arg} z$$ where $\ln |z|$ is the real logarithm of the modulus of $z$ and $\operatorname{arg} z$ is the angle of $z$ in the complex plane for some branch cut.

Answer 2

I will only consider here the case $a=e$ (aka the natural log). For any nonzero, complex base, you could change to the natural log with a complex argument.

The answer is, yes, you can define a log function for a general complex argument, but it gets a bit complicated. In the complex plane, the log function requires a branch cut to distinguish the exact branch. The reason for this is the exponential is actually periodic in the complex plane, so it has infinite candidates for an inverse.

I won't go into the details too much, but a complex log can be defined as follows. $$ \ln(z) = \ln(re^{i\theta}) = \ln(r) + i\theta.$$ As $re^{i\theta} = re^{i\theta + 2\pi i}$, you have a level of freedom in choosing $\theta$. It is standard practice to take the principle branch as $-\pi < \theta < \pi$. If you wanted $\ln(-1)$, you might take the branch $0 < \theta < 2\pi$ to get $\ln(-1) = i\pi$.

Answer 3

Euler's formula opened to door to defining logarithms for arbitrary complex numbers.

If $e$ is the base of the natural logarithm, $i^2=-1$ and $H$ is the measure of a half-angle, then

$$e^{ix}=\cos(Hx/)+i \sin(Hx/)$$

We get Euler's Identity

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

from, this we derive

$$e^{i2z}=1$$

when $z$ is an integer.

and derive

$$ln(-1)=i+i2z=i(2z+1)$$

and

$$ln(i)=\frac{i(2z+1)}{2}$$

From here in out, we use the usual rules of logarithms to use arbitrary complex arguments and bases.