$a_{−3}$ in the Laurent expansion of $f(z)$ in the region $0 < |z| < 1$. $f(z)=\frac{1}{2z^3}-\frac{1}{z^3+i}$

Question

I am looking to find the $a_{−3}$ in the Laurent expansion of $f(z)$ in the region $0 < |z| < 1$.

$$f(z)=\frac{1}{2z^3}-\frac{1}{z^3+i}$$

I can't get my head around this Laurent Series:

$$\frac{1}{2z^3}=\frac{1}{2(z^3-\frac{1}{2})+1}=\frac{1}{2(z^3-\frac{1}{2})(1+\frac{1}{2(z^3-\frac{1}{2})})}$$

$$\frac{1}{2(z^3-\frac{1}{2})}(\sum_{n=0}^{\infty}(-1)^n(\frac{1}{2(z^3-\frac{1}{2})})^n$$

I really don't think I am doing this correctly and I am not making much progress. All help is appreciated.

Answer 1

Let$$g(z)=\frac12-\frac{z^3}{z^3+i}.$$Then $g$ is analytic and $g(0)=\frac12$. So, the Taylor series of $g$ centered at $0$ is like$$\frac12+b_1z+b_2z^2+b_3z^3+\cdots$$(actually, $b_1=b_2=0$, but that's not important) and therefore$$f(z)=\frac{g(z)}{z^3}=\frac12z^{-3}+b_1z^{-2}+b_2z^{-1}+b_3+\cdots$$So, $a_{-3}=\frac12$.