# Complex Analysis Taylor Series

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.

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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

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.

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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?

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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 $$\frac{\left( -1/2-1/2\, i \right) \left( 1/2+1/2\,i \right)^k} {k+1}.$$

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