What is the Taylor series expansion of $\frac{z}{\ln{z+1}}+z $ centered on $z=1+i$?

Question

I am experimenting with the behavior of collage level calculus, mixed with complex numbers. In particular Taylor series polynomials with a complex variable & coefficients. I found this question in a book, and have no idea how to begin. Is there a process for adapting the way we do Taylor series over the reals into $ \mathbb{C} $ ?

Answer 1

FYI,

I would suggest the following substitution:

$z=e^u$ then $f(z=e^u)=\dfrac{e^u}{u+1}$

We know that $e^x=\sum\limits_{n=0}^\infty \dfrac{x^n}{n!}$ so we get:

$f(u)=\sum\limits_{n=0}^\infty \dfrac{u^n}{n!}\frac{1}{1+u}$

Substitute back $u=lnz$

$f(e^{\ln z})=\sum\limits_{n=0}^\infty \dfrac{{(\ln z})^n}{n!}\dfrac{1}{1+\ln z}\tag1$

At $z=1+i=\sqrt2 e^{\frac{i\pi}{4}}$

$f(1+i)=\sum\limits_{n=0}^\infty \dfrac{{(\ln 2+\frac{i\pi}{2}})^n}{2^nn!}\dfrac{2}{2+\ln 2+\frac{i\pi}{2}}\tag2$

Answer 2

Some things are fairly easy to work with, but dividing by a logarithm is trouble. We have $\ln z = -1$ at $z=e^{-1}$, so the power series centered at $1+i$ for that quotient will have radius of convergence $\sqrt{1+(1-e^{-1})^2}=\sqrt{2-2e^{-1}+e^{-2}}$. I find it very unlikely that there will be a reasonable closed form for the coefficients of the power series, given that.

Have you transcribed the question accurately? Was it, perhaps, to find some number of terms instead?

Answer 3

As @jmerry answered, I do not hink that it would be easy to find the general form of the coefficients of the Taylor series.

But, for sure, $$\frac{z}{\log (z)+1}+z=a+\frac{a}{\log (a)+1}+ \left(\frac{\log (a)}{(\log (a)+1)^2}+1\right)(z-a)+\frac{ (1-\log (a))}{2 a (\log (a)+1)^3}(z-a)^2 +\frac{ \left(\log ^2(a)+2 \log (a)-5\right)}{6 a^2 (\log (a)+1)^4}(z-a)^3+O\left((z-a)^4\right)$$ Just make $a=1+i$ and simplify a little since $\frac 1{2a}=\frac{1-i}4$ , and $\frac 1{6a^2}=-\frac{i}{12}$ and the problem is, for sure, with $\log(1+i)$ as you know.