Is there a constant $k$ which $k^{4n+1}=k^{4n+2}=1, k^{4n+3}=k^{4n}=-1$?

Question

We can be sure, that $$\log(1+i)=\frac{\pi i}{4}+\frac{\log(2)}{2}$$ then if we take $$\log(1+k)=\frac{\pi}{4}+\frac{\log(2)}{2}$$ so sign-alternating is $++--$. Is there a constant $k$ which $k^{4n+1}=k^{4n+2}=1, k^{4n+3}=k^{4n}=-1$?

Answer 1

Notice that

$$k^{4n+1} = k^{4n+2}$$

just means either $k = 0$ or $k = 1$...