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

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)?

## 1 Answer

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?

• It seems that finding the constant term and the coefficient of $z-e^{-2}$ is the same as what I've described above. I know that there exists a $g(z), a, b$ as you have, but finding $a, b$ is the problem. I must be missing something! – Fred May 2 at 23:54
• If $h$ is analytic in a neighborhood of $c$ then the power series for $k$ starts with $h(c)+h'(c)(z-c)$. So all you to do is find $h(c)$ and $h'(c)$ where $c=e^{-2}$ and $h(z)=(z-e^{-2})^{2} f(z)$. You can find these suing L'Hopital's Rule. – Kavi Rama Murthy May 3 at 0:00
• Yes. As mentioned, I have tried L'Hospital's rule to find these values, but the computations are a little ridiculous! However, your input is still valid and I'll attempt a solution with this method. – Fred May 3 at 0:22