Finding $f(z)=\log(\frac{(z+1)^2}{(z^2+4)})$ Laurent series

Question

Determine the Laurent series of the function $f(z)=\log(\frac{(z+1)^2}{(z^2+4)})$ on the set $A=\{z|\:2<|z|\}$.

I thought of using the following expansion: $\log(1+z)=\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}z^n }{n},|z|<1$

Using the logarithm properties I tried to arrange the expression so that I could apply the formula: $\log(\frac{(z+1)^2}{(z^2+4)})=\log z^3+2\log(1+\frac{1}{z})-\log(1+\frac{4}{z^2})$

However I do not know what should I do regarding $\log(z^3)$.

The solution provided in the book is: $\sum_\limits{-\infty}^{-1}\frac{1-(-1)^{n}4^{-n} }{n}z^{2n}-\sum_\limits{-\infty}^{0}\frac{2}{2n-1}z^{2n-1}$

Question:

How should I solve the question?

Answer 1

Check your work. There should be no $\log z$ terms involved.

Note that the fraction is homogeneous [which also implies that you could avoid $\log z$ terms completely], so we do the following: $$ f (z) = \log \left(\frac {(1 + z^{-1})^2} { 1 + 4z^{-2}}\right) = 2\log (1 + z^{-1}) - \log (1+4z^{-2}), $$ since $|z| > 2$, $|z^{-1}| < 1/2 < 1, |4z^{-2}|< 1$, then $$ f(z) = 2 \sum_1^{+\infty} \frac {(-1)^{n-1}}n z^{-n} - \sum_1^{+\infty} \frac {(-1)^{n-1}4^n }n z^{-2n} $$ or $$ f(z) = 2 \sum_{-\infty}^1 \frac {(-1)^n}n z^n - \sum_{-\infty}^1 \frac {(-1)^n4^{-n}}n z^{2n}. $$ Now collect the odd-indexed terms and even-indexed terms respectively, then you would reach the answer provided.

Answer 2

Since$$\log(z+1)=z-\frac{z^2}2+\frac{z^3}3-\frac{z^4}4+\cdots,$$you know that$$\log\left((z+1)^2\right)=2\log(z+1)=2z-z^2+\frac{2z^3}3-\frac{2z^4}4+\cdots,$$and that$$\log(4z^2+1)=4z^2-\frac{4^2z^4}2+\frac{4^3z^6}3-\frac{4^4z^8}4+\cdots$$and so$$\log\left(\frac{(z+1)^2}{4z^2+1}\right)=2z-5z^2+\frac23z^3+\frac{15}2z^4+\cdots$$Therefore, if $\lvert z\rvert>2$, $\left\lvert\frac1z\right\rvert<\frac12$ and\begin{align}\log\left(\frac{(z+1)^2}{z^2+4}\right)&=\log\left(\frac{(\frac1z+1)^2}{4\frac1{z^2}+1}\right)\\&=\frac2z-\frac5{z^2}+\frac2{3z^3}+\frac{15}{2z^4}+\cdots\end{align}