Singular part of $f(z) = \frac{z}{(2 + \log z)^{2}}$

Question

I'm asked to find the "singular part" of $f(z) := \frac{z}{(2 + \log z)^{2}}$ (I think this means the principal part of the Laurent series -- this question is from an older qualifying exam at my university). I know that $f$ has a pole at $z = e^{-2}$ of order $2$, and that this is the only pole. I've tried to integrate $f$ over a circle about $e^{-2}$ to at least get $a_{-1}$ in Laurent. This doesn't seem to get me anywhere. The limit calculation of residues is horrendous for this function, and I keep making errors. Regardless, as this is a previous qual question, I presume the solution isn't too overbearing. There must be a more efficient way to calculate the "singular part" of $f$.

I am also asked to calculate the radius of convergence of the power series of $f - (\text{singular part})$.

I've found a relevant calculation here: Residue of $f(z) = \frac{z}{1-\cos(z)}$ at $z=0$. I was hoping to do the same with this problem, but I'm not familiar with big-O notation. If my calculations are right, the power series for $h(z) := 2 + \log z$ is

$$(z-e^{-2})\sum_{n=1}^{\infty} \frac{(-1)^{n+1}e^{-2n}}{n} (z-e^{-2})^{n-1}.$$

How would I go about using the technique in the link (if applicable)?

Answer 1

We can write $f(z)=\frac a {z-e^{-2}} +\frac b {(z-e^{-2})^{2}}+g(z)$ where $g$ is analytic in a neighborhood of $e^{-1}$. The singular part is just $\frac a {z-e^{-2}} +\frac b {(z-e^{-2})^{2}}$. All you have to do is to find $a$ and $b$. For this consider $f(z)(z-e^{-2})^{2}$. This is an analytic function and you have to find the constant term and the coefficient of $z-e^{-2}$ in its power seriess. Can you take it from here?

Answer 2

Just some ideas for the computation of $a$ and $b$ -

Put $p:=e^{-2}$ and $h(z)=(2+\log(z))^2.$ Then the first three derivatives of $h$ are

\begin{align*} h'(z) &=\frac{2(2+\log(z))}{z}\\ h''(z) &=-2\frac{1+\log(z)}{z^2}\\ h'''(z) &=2\frac{z+2z\log(z)}{z^4} \end{align*}

Since $h$ has a zero of order $2$ at $p$, we can write $h(z)=(z-p)^2\varphi(z)$ in a neighborhood of $p$ where $\varphi$ is analytic and $\varphi(p)\neq 0.$ More specifically,

\begin{align*} \varphi(z) &= \sum_{n=2}^{\infty}\frac{h^{(n)}(p)}{n!}(z-p)^{n-2}\\ \varphi'(z) &=\sum_{n=3}^{\infty}\frac{h^{(n)}(p)}{n!}(n-2)(z-p)^{n-3} \end{align*}

so we see $\varphi(p)=\frac{h''(p)}{2}=e^4$ and $\varphi'(p)=\frac{h'''(p)}{3!}=-e^6.$ Since $a$ is the residue of $f$ at $p$, and $f$ has a double pole at $p$, the residue is computed

\begin{align*} a &=\lim_{z\to p}\frac{d}{dz}\left[(z-p)^2f(z)\right]\\ &=\lim_{z\to p}\frac{d}{dz}\left[(z-p)^2\frac{z}{(z-p)^2\varphi(z)}\right]\\ &=\lim_{z\to p}\frac{d}{dz}\left[\frac{z}{\varphi(z)}\right]\\ &=\lim_{z\to p}\frac{\varphi(z)-z\varphi'(z)}{\varphi(z)^2}\\ &=\frac{\varphi(p)-p\varphi'(p)}{\varphi(p)^2}\\ &=2e^{-4}. \end{align*}

Computing $b$ is easier, by writing

$$f(z)=\frac{a}{z-p}+\frac{b}{(z-p)^2}+g(z)$$

where $g$ is analytic in a neighborhood of $p$, we find \begin{align*} b &=\lim_{z\to p}\left(a(z-p)+b+g(z)(z-p)^2\right)\\ &=\lim_{z\to p}(z-p)^2f(z)\\ &=\lim_{z\to p}\frac{z}{\varphi(z)}\\ &=\frac{p}{\varphi(p)}\\ &=e^{-6}. \end{align*}