Is a logarithm with base 1 defined in the field of complex numbers? I have not found any information about this. In real numbers, this is uncertain because $ \ln(1) = 0 $ and $ \log_a(b)= \frac {\ln(b)} {\ln(a)}$.
And it's dividing by zero. But in the field of complex numbers:
$ \ln(1)=2πik $
$ \log_1(z)= \frac {\log_e(z)} {2πik} $
One reasonable definition comes from the observation that $(-1)^2=1$. Thus, one can assume $\log_1 z= \frac12\log_{-1} z$.
This way, the multi-valued logarithm will be
$$\operatorname{Log_1}z=\frac12\operatorname{Log_{-1}}z=\frac{\ln z+i\pi k}{2\ln(-1)}=\frac{\ln z+i\pi k}{2i\pi}=\frac{\ln z}{2i\pi}+k, k\in \mathbb{Z}$$
The simplest expression for a branch will be $\log_1 z=\frac{\ln z}{2i\pi}$, but it should be noted that this is not an established convention.
All definitions of $\log_ax$ and $a^x$ should satisfy $a^{\log_ax}=x$. At least, I'm not sure I've ever seen it otherwise.
So you need definitions that satisfy $1^{\log_1x}=x$. Let's assume $1^x$ works in the usual way. It returns a single number, $1$, regardless of what $x$ is. If $x$ is a number, it would have to work this way to satisfy $1^{\log_1x}=x$. (If you want to work with some different, unusual definition of $1^x$, you'd need to start by offering that definition in precise terms.)
So where does this leave you? $1^{\log_1x}=1$, and what could "$\log_1x$" be to satisfy that? It could be any number at all. But only when $x=1$ will $1^{\log_1x}=x$ be satisfied.
So $\log_11$ should be viewed as an indeterminate expression, like $\frac00$ or $1^{\infty}$. And $\log_1x$ for $x\neq1$ is more severely undefined, similar to $\frac{z}0$ when $z\neq0$.
A logarithm to the base $+1$ seems difficult to define because of that issue with division by zero on one logarithmic branch.
We can consider a limited definition for base $-1$ and argument of $\pm1$. To wit,
$\log_{-1}(+1)\equiv0\bmod2\tag{even}$
$\log_{-1}(-1)\equiv1\bmod2\tag{odd}$
If we take any branch of $\ln(+1)$ and divide by any branch of $\ln(-1)$, the result is a rational fraction that invariably reduces to residue $0\bmod2$. Dividing any branch of $\ln(-1)$ by any branch of $\ln(-1)$ similarly gives $1\bmod2$. Addition of the $\bmod2$ residues matches up with multiplication of the $\pm1$ arguments, so this discrete logarithm indeed has all the properties we normally expect of a logarithm.
We may define similar logarithms using other roots of unity as bases, for example a base that is a primitive cube root of unity can define logarithms with different residues $\bmod3$.
Recall that one definition of exponentiation is
$$a^b = \exp(b \log a)$$
Where $\log a$ denotes the (possibly non-real) number $z$ such that $e^z = a$.
Recall that by Euler's Formula, $e^{x+iy} = e^x(\cos(y) + i\sin(y))$. If we let $x = 0$ and $y$ be an integer multiple of $2\pi$, then:
$$e^{i2\pi k} = e^0(\cos(2\pi k) + i\sin(2\pi k)) = 1, k\in\mathbb{Z}$$
And thus, we can define:
$$\log 1 = 2\pi ik$$
The standard definition of $\log 1$ is $0$, which is what you get from $k = 0$. But we can just use a nonstandard branch of the complex log function. Anyhow, by the change-of-base formula:
$$\log_1 z = \frac{\log z}{\log 1}$$
If $z = re^{i\theta}$, then:
$$\log_1 z = \frac{\log r + i\theta}{2\pi i k} = \frac{\theta - i\log r}{2\pi k}, k \ne 0$$
So yes, it works, provided that you don't mind getting an imaginary logarithm when $z$ is real (and $r \ne 1$).
