There are $p$ solutions to $\sqrt[\frac{p}q]1$, if $\frac{p}q$ is a fraction in lowest terms. I have found on this website that an irrational exponent has infinite roots.
But what about $\sqrt[a+bi]1$? How can you find out how many solutions there are to that based on $a$ and $b$?
On
First note that $$\sqrt[a+ib]1=1^{\frac{1}{a+ib}}=1^{\frac{a-ib}{a^2+b^2}}=\left(\sqrt[a^2+b^2]1\right)^{a-ib}=1^{a-ib}=1^a\cdot 1^{-ib}=1^{-ib}$$
Now you can write $1=e^{Log \;1}$ where $Log\; 1$ is the complex logarithm of $1$, that is a multivalued function with values $Log\; 1=0,\pm 2\pi,\pm 4\pi,...$.
So if we take the principal value $Log\; 1=0$ we have: $$ 1^{-ib}=e^{0\cdot (-ib)}=e^0=1 $$ but for the other values of $Log \;1$ we have infinitely many other values of the exponential and of the starting root.
Let's say we have the following equation: $$x^{a+bi}=1$$ Then, we can say $x=e^{\ln x}$, so we multiply the exponents: $$e^{\ln x(a+bi)}=1$$ Take the $\ln$ of both sides: $$\ln x(a+bi)=2\pi ni \text{ for some } n \in \Bbb{Z}$$ Divide both sides by $a+bi$: $$\ln x=\frac{2\pi ni}{a+bi} \text{ for some } n \in \Bbb{Z}$$ Simplify the right side: $$\ln x=\frac{2\pi nai}{a^2+b^2}+\frac{2\pi nb}{a^2+b^2} \text{ for some } n \in \Bbb{Z}$$ Take both sides to the power of $e$: $$x=e^{\frac{2\pi nai}{a^2+b^2}+\frac{2\pi nb}{a^2+b^2}} \text{ for some } n \in \Bbb{Z}$$
Thus, if $b \neq 0$, there are an infinite number of solutions because each solution differ by a real factor from the $\frac{2\pi nb}{a^2+b^2}$ part of the exponent.
However, if $b=0$ and $a+bi \in \Bbb{R}$, then that part of the exponent is always $0$ and we have just: $$x=e^{\frac{2\pi nai}{a^2+b^2}}=e^{\frac{2\pi nai}{a^2+0^2}}=e^{\frac{2\pi ni}{a}} \text{ for some } n \in \Bbb{Z}$$ Thus, when $n=ta$ for $t \in \Bbb{Z}$ we get the same solution as $n=0$ since $e^{2\pi i}=1$. However, this is only possible if $a$ is rational, so rational exponents have the same number of solutions as their numerator while irrational exponents have an infinite number of solutions.