(Exer 7.28(a)) Find a power series for $\frac1z$ centred at $z_0=1$. --> I got this.
(Exer 7.28(b)) Find a power series for $Ln(z)$ centred at $z_0=1$. --> I'm not sure how to do this.
For the real $\ln(x)$,
$ \ \ \ \ \ \ \ \ \ \ $ I integrate the power series of a rational function $f$ over some interval $\gamma_x$ whose left point is some $x_0 \in \mathbb R$ and whose right end point is $x$. I could directly the power series for $f(t) = \frac 1 t$ over $\gamma_x = [1,x], x_0=1$. I could also integrate the power series for $f(t) = \frac{1}{1+t}$ over $[0,x],x_0=0$ and then make a change of variables to get from $\ln(1+x)$ to $\ln(x)$. These $\gamma_x$'s and $x_0$'s ensure that I get $\ln(x)$ and not, say, $\ln(x)+7$.
For complex $Ln(z)$,
$ \ \ \ \ \ \ \ \ \ \ $ the antiderivatives of $\frac 1 z$ include:
- the principal $Ln$ branch s.t. $Ln(z) = \ln|z| + iArg(z)$
- any other $\mathscr Ln$ branch s.t. $\mathscr Ln(z) = \ln|z| + i\mathscr Arg(z)$
- $\ln|z|$
- $\int_{\gamma_z}\frac{dw}{w} \forall \gamma_z \subset G$ piecewise smooth paths from $z_0 \in G$ to $z$
-
Questions:
(Q1): If $\exists \gamma_z, z_0$ and $G$ to ensure I have a power series for the principal $Ln$, then what are they?
(Q2): If not, then how else do I get a power series for the principal $Ln$ with the power series of $\frac 1 z$? (Or $\frac{1}{1-z}$, $\frac{1}{1+z}$, etc depending on your preference)
(Q3): Is this even possible to do w/o Taylor or Laurent (discussed starting Ch8)?
It seems you know the Taylor series, centered at $1,$ of $\log x$ on the real line. For $0<x<2$ this is
$$\log x = (x-1)-(x-1)^2/2 + (x-1)^3/3 - (x-1)^4/4 +\cdots.$$
We'd like to show that if $x$ is replaced by $z\in D(1,1),$ we get $\log z,$ the principal branch of the logarithm. To do this, define
$$\tag 1 f(z) = (z-1)-(z-1)^2/2 + (z-1)^3/3 - (z-1)^4/4 +\cdots,\,\, z\in D(1,1).$$
Then $f$ is analytic in $D(1,1),$ and
$$f'(z) = 1-(z-1) + (z-1)^2 - (z-1)^3 +\cdots$$
for $z\in D(1,1).$ This is a geometric series whose sum is $1/(1+(z-1)) = 1/z,$ as you found. You also know $\log'(z) = 1/z$ throughout the domain of $\log z,$ which contains $D(1,1).$ Thus in $D(1,1),$ the derivative of $f(z)-\log z$ is identically $0.$ Because $D(1,1)$ is a region, it follows that $f(z)-\log z$ is constant in $D(1,1).$ To find this constant, plug in $z=1.$ $(1)$ makes it clear that $f(1)=0,$ and we know $\log 1=0.$ Thus $f(z)-\log z$ vanishes in $D(1,1),$ i.e., $f(z) = \log z$ in $D(1,1).$ Thus $(1)$ is the desired power series of $\log z$ centered at $1.$