A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka

(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 $z$



(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)?

  • 3
    $\begingroup$ The origin is a branch point, so there are no power series centered at the origin. Laurent or Taylor. This is because a converging power series is holomorphic in an annulus, but the existence of a branch cut means that $\log$ is not. The usual thing is to get one for $\log(1+z)$ by the process you described: integerate the series of $1/(1+z)$. You can, of course, center a series anywhere else. A Tayloir series centered at $z_0$ will then have radius of convergence $R=|z_0|$ when you arrange the branch cut not to disturb that. $\endgroup$ – Jyrki Lahtonen Aug 5 '18 at 13:25
  • 3
    $\begingroup$ It is unclear what you question is. Please explain clearly what you are trying to answer and what you have tried so far. $\endgroup$ – Somos Aug 5 '18 at 13:25
  • $\begingroup$ @Somos I put my questions at the end...? $\endgroup$ – BCLC Aug 5 '18 at 14:04
  • $\begingroup$ @JyrkiLahtonen Thanks! I checked again, and it was power series around $z_0=1$. For $\mathbb R$, I know how to do it. The thing is, how do I know that I'm getting the same answer in $\mathbb C$ without Taylor or Laurent series? There's a possibility this could be an error in the textbook, which introduces Taylor and Laurent in the next chapter. But if it's possible to do without Taylor or Laurent, then how do I know integrating and making change of variables will give me a power series for $Ln(z)$ and not any of the other antiderivatives? I mean, what's your $\gamma_z$ and $z_0$? $\endgroup$ – BCLC Aug 5 '18 at 14:10
  • 1
    $\begingroup$ I assume you know something about power series, like the fact that they define analytic functions in their discs of convergence, and can be term wise differentiated or integrated in the usual ways? $\endgroup$ – zhw. Aug 10 '18 at 20:48

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 $1.$

  • $\begingroup$ Thanks zhw.! I'm analysing now. ... ok done. Brilliant precision and translation from $\mathbb R$. Exactly what I was looking for. $\endgroup$ – BCLC Aug 12 '18 at 9:45

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.