Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So the problem states:

"Say f(z) := log z is the principal branch of the logarithm (the primitive of 1/z on the region C(-infinity,0]). Show that the Taylor series of f(z) about $z_0 = -1 + i$ takes the form $$\log z = \sum_{n=0}^{\infty} a_n(z-(-1+i))^n $$ with

$$a_0 = \log \sqrt{2} + i\frac{3\pi}{4}\,\,\,\text{and}\,\,\,a_n = (-1)^{n+1}\frac{e^{-3\pi in/4}}{n2^n/2}$$

Determine the radius of convergence of this series. Explain why the series does not represent f(z) in its entire disk of convergence."

My main concern here is how do I show $\log(-1+i) = \log \sqrt{2} + i\frac{3\pi}{4} $ and determine the radius of convergence aswell.

share|improve this question
    
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Show", "Determine", etc.) to be rude when asking for help; please consider rewriting your post. –  Micah Feb 26 '13 at 4:57
    
In short, show us you have put some work into the problem yourself, if you expect other people to put some work into it for you. –  Gerry Myerson Feb 26 '13 at 5:00
    
Thanks for advice. –  Mett Feb 26 '13 at 5:25

3 Answers 3

The function $g(z):={1\over z}$ is analytic in $\dot{\mathbb C}:={\mathbb C}\setminus\{0\}$, but has no primitive defined in all of $\dot{\mathbb C}$. The function $g$ however has primitives in suitable subdomains $\Omega\subset\dot{\mathbb C}$, the most famous one being the principal value $${\rm Log}(z):=\log|z|+ i\>{\rm Arg}(z)\ ,$$ which is defined on $\Omega:={\mathbb C}\setminus\{≤0\ {\rm real\ axis}\}$. In particular $${\rm Log}(-1+i)={1\over2}\log2+{3\pi\over4}\>i\ .$$ Standing at the point $p:=-1+i\in\dot{\mathbb C}$ we see that the function $g$ is analytic in a disk $D$ of radius $\sqrt{2}$ around $p$. Therefore $g$ has primitives which are analytic in $D$, whence have a power series development with center $p$ and convergence radius $\sqrt{2}$. These primitives are equal up to an additive constant, and one of them has the value ${1\over2}\log2+{3\pi\over4}\>i$ at $p$. Since ${\rm Log}$ is a primitive of $g$ in the neighborhood of $p$ having exactly this value at $p$ the corresponding series represents ${\rm Log}$ in any domain $\Omega'\subset D\cap\Omega$ containing the point $p$. The points of $\Omega$ lying below the negative real axis do not belong to such an $\Omega'$. Therefore the obtained series does not represent ${\rm Log}$ there.

share|improve this answer
    
The cut may be taken along the positive ray instead of the negative ray too. –  user64494 Sep 16 '13 at 20:11

The first part of your question is really asking how to find the real and imaginary parts of the logarithm. So just write it out! $$ e^{x+ i y} = -1 + i $$ Now use Euler's identity to get simultaneous equations for $x$ and $y$: $$ e^x \cos y = -1 ~,~ e^x \sin y = 1 $$ Hopefully you can solve these (the solution for $y$ is only unique because we choose a particular branch of the logarithm).

As for the second part, what do you know about the radius of convergence of the Taylor series for a holomorphic function?

share|improve this answer

The Maple command $$convert(ln(z), FPS, z = -1+I) $$ produces $$ \ln \left( -1+i \right) +\sum _{k=0}^{\infty } \left( -\frac { \left( 1/2+1/2\,i \right)^k}{2\,k+2}-\frac {i \left( 1/2+1/2\,i \right)^k}{2\,k+2} \right) \left( z+1-i \right) ^{k+1} .$$ Next, $$normal(-(1/2+1/2*I)^k/(2*k+2)-I*(1/2+1/2*I)^k/(2*k+2)) $$ outputs $$\frac{\left( -1/2-1/2\, i \right) \left( 1/2+1/2\,i \right)^k} {k+1}. $$

share|improve this answer
    
PS. $$ evalc(ln(-1+I))$$ outputs $1/2\,\ln \left( 2 \right) +3/4\,i\pi $. –  user64494 Sep 16 '13 at 19:39

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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