Many series representations allow for domain extension. That is, the series for $\sin(x)$ allows you to find the value of $\sin(i)$ or similar. However, when I take the series for $\ln(x)$, it doesn't seem to actually map to the complex domain, and I am not sure why.
The series I have is: $$ \ln(x) = \sum\limits_{n=1}^{\infty} (-1)^{n+1}\frac{1}{n}(x - 1)^n $$ If I expand it out to four terms, I get: $$ \ln(x) = (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} + \frac{(x - 1)^4}{4} $$ Now, I realize that, with only four terms, it is not going to be incredibly accurate, but nonetheless it should be in the ballpark. Additionally, this should converge if $0 < x < 2$. However, I'm not for certain whether or not the complex $i$ would converge, nor actually how to tell on complex numbers. So that could be the problem, but then I don't know how to figure out which complex values this converges for.
If I substitute $i$ in I get $\frac{2}{3} + \frac{8}{3}i$, while the actual value is $\frac{i \pi}{2}$. In fact, if I extend it to a full ten terms of the expansion, I get $-3.45 + 4.37i$ (rounded for simplicity), which is even further away.
Anyway, I'm sure that my lack of knowledge of complex analysis is biting me, but I thought I'd see if there was a simple answer, and if there was a way to figure the complex radius of convergence of the Taylor series for $\ln(x)$ about $1$.
Taylor series of holomorphic functions converge up to the nearest singularity. As exists an holomorphic extension of $\sin:\Bbb R \longrightarrow\Bbb R$ the Taylor series of $\sin$ converges in $\Bbb C$. But in the case of the logarithm...