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.
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 $\ln(-1)$ and divide 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 ligarithm 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 $\bmod3$.
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 $z\neq0$.
Recall that $y = \log_b x$ means that $b^y = x$. If we allow $b$ to be a complex number, expressed as $b = re^{i\theta}$, then:
$$r^ye^{i\theta y} = x$$
Let $b = 1$, so $r = 1$, and $\theta = 2\pi k$, where $k\in\mathbb{Z}$.
$$e^{i2\pi k y} = x$$ $$\cos(2\pi k y) + i\sin(2\pi k y) = x$$
If $|x| = 1$, then we can choose $k \ne 0$ and define $y = \frac{\arg(x)}{2\pi k}$, thus satisfying $1^y = x$ from a certain point of view.
But if $|x| \ne 1$, then there's no way to satisfy the equation, since no matter which choice of $k$ you make, $|\cos(\theta) + i\sin(\theta)| = 1$.
So, while it's possible to define a “base-1 logarithm” on the unit circle, it's not possible to define one on $\mathbb{C}$ as a whole.
