Computing the Laurent series of $\frac{1}{z^2}$ around the point $z = 3$

Question

I am trying to compute the Laurent series of $\frac{1}{z^2}$ around the point $z = 3$. At first glance this seems straightforward, but my normal strategy does not work, since I cannot split the fraction. Normally, what I would do is write $$ \frac{1}{z^2} = \frac{A}{z} + \frac{B}{z}, $$ and then compute $$ \frac{1}{z} = \frac{1}{3+z-3} = \frac{1}{3} \cdot \frac{1}{1+\frac{z-3}{3}} = \frac{1}{3} \cdot \sum_{n=0}^\infty \left( -\frac{z-3}{3} \right)^n = \sum_{n=0}^n \frac{(-1)^n}{3^{n+1}} (z-3)^n. $$ And the multiply by the constant $A$ or $B$ and add both sums. However, this ain't working, because if we split the sum like that, we would need $(A+B)z = 1$. How do we proceed in this case?

Answer 1

Actually, since $3$ is not a singularity of $\frac1{z^2}$, your Laurent series is just a power series. Anyway, note that\begin{align}\frac1z&=\frac1{3-(3-z)}\\&=\frac13\cdot\frac1{1+\frac{z-3}3}\\&=\sum_{n=0}^\infty\frac{(-1)^n}{3^{n+1}}(z-3)^n.\end{align}Therefore\begin{align}\frac1{z^2}&=\left(-\frac1z\right)'\\&=-\sum_{n=1}^\infty\frac{(-1)^nn}{3^{n+1}}(z-3)^{n-1}\\&=\sum_{n=0}^\infty(-1)^n\frac{n+1}{3^{n+2}}(z-3)^n.\end{align}

Answer 2

Hint consider series for $1/z$ then differentiate it.

Answer 3

The function $\frac {1}{z^2}$ is analytic about $z=3$

Thus the Taylor Series formula $$ f(z)= f(3) + f'(3)(z-3) + \frac {f''(3)}{2} (z-3)^2+... $$ represents the infinite series about $z=3$