# Taylor series expansion of $\log[z]$ about $z=1$ (different branches)

I realize this is not the fastest way of getting a Taylor's series expansion of $f(z)=\log(z)$ about $z=1$. But here goes. I am assuming I am working on the principal branch of the logarithm ($-\pi<\theta<\pi$). I am assuming that $f(1)=\log(1)=0$. That's the branch I am on.

Next, some derivatives:

$f'(z)=1/z$

$f''(z)=-1/z^2$

$f'''(z)=2/z^3$

$f^{(4)}(z)=-6/z^4$

...

$f^{(n)}(z)=(-1)^{n+1}(n-1)!/z^n$

Hence,

$$f(z)=\sum_{n=1}^{\infty}c_n(z-1)^n,$$

where

$$c_n=\frac{f^{(n)}(1)}{n!}=\frac{(-1)^{n+1}(n-1)!}{n!(1)^n}=\frac{(-1)^{n+1}}{n}.$$

Thus:

$$f(z)=(z-1)-\frac{(z-1)^2}{2}+\frac{(z-1)^3}{3}-\frac{(z-1)^4}{4}+\frac{(z-1)^5}{5}-\cdots$$

And checking, $f(1)=0$, which is my assumption above.

Question: Now, suppose I start the problem again, only this time I assume I am a different branch of the logarithm. Let's assume that this time we are working on the branch $\pi<\theta<3\pi$, so that $f(1)=\log(1)=2\pi i$.

If I repeat the calculations above, what would they now look like, and what would be the answer for my series?

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Selecting different branches has an effect of shifting the value of logarithm by an integer multiple of $2\pi$. (Though the exact amount of shift may differ, depending on the point.) –  sos440 Mar 31 '13 at 4:55

The various branches of the logarithm only differ by an added constant multiple of $2\pi i$. So just add $2k\pi i$ to your series (for any integer $k$) to obtain the desired series for any other branch of the logarithm.
OK, I believe this is correct, but this is not the question I am asking. The question is this: If I repeat the calculations above, what would they look like if I were working on the branch $\pi<\theta<3\pi$, and how would the $2\pi i$ makes its appearance in the final answer? –  David Mar 31 '13 at 5:49
@David, like I said, just add $2\pi i$ to your power series. All of the calculations will be the same because an added constant will not change any of the derivatives. If $Log(z)$ is the principal branch, then your branch is just $Log(z) + 2\pi i$. –  Antonio Vargas Mar 31 '13 at 5:50
The derivative of the other branch is still $1/z$, so your calculation is the same, except for the constant term $f(1)$. –  GEdgar Mar 28 at 0:39
What we really have is, $$f(z)=\sum_{n=0}^{\infty}c_n(z-1)^n,$$ and in your case, you originally started with $c_0 = f^{(0)}(1) = f(1) = 0$. This is the only part of the calculation that changes. If you want $f(1) = 2\pi i$, then $c_0 = 2 \pi i$. In other words, as the other answer said, just add the constant $2 \pi i$.
Thus, for you, $$f(z)=2 \pi i + (z-1)-\frac{(z-1)^2}{2}+\frac{(z-1)^3}{3}-\cdots$$