Struggling to compute a power series for a complex value function I am struggling to compute the power series expansion of $$f(z) = \frac{1}{2z+5}$$ about $z=0$, where $f$ is a complex function. I tried comparing it to the geometric series as follows,$$ f(z) = \frac{1}{2z+5} = \frac{1}{1-\omega} = \sum_{n=0}^\infty\omega^n $$ which gives us $$ 2z+ 5 = 1 - \omega \implies \omega = -2z - 4 \implies f(z) = \sum_{n-0}^\infty(-2z + 4)^n = \sum_{n=0}^\infty(-2)^n(z+2)^n$$ which is clearly not even an expansion about $z=0$. This is the wrong answer as well according to my textbook which stated that the answer was $$\sum_{n=0}^\infty\frac{2^n}{5^{n+1}}(-1)^nz^n$$
could someone please explain where I have gone wrong.
 A: You've gone wrong because
$$
\frac{1}{1-\omega}=\sum_{n\geq 0}\omega^n
$$
assumes that ${|\omega|<1}$. In your case, this would mean ${|2z+4|<1}$, i.e. ${|z+2|<\frac{1}{2}}$. You have calculated the series about ${z=-2}$ with radius of convergence ${\frac{1}{2}}$.
Instead, consider
$$
\frac{1}{2z+5} = \frac{1}{5}\left(\frac{1}{\frac{2}{5}z+1}\right) = \frac{1}{5}\left(\frac{1}{1-\left(\frac{-2}{5}z\right)}\right)
$$
can you take it from here?
A: To get the expansion about $z=0$, we need to ensure $\omega$ doesn't contain an additive term and is just some scaling of $z$. By considering $5f(z)=\frac{1}{2z/5+1}=\frac{1}{1-\omega}$ and solving for $\omega$, we get $\omega =-\frac{2}{5}z$ and thus $5f(z)=\sum_{n=0}^{\infty}\frac{2^n}{5^n}(-1)^nz^n$, which gives the required answer.
A: $$\begin{array}{lcl}
\dfrac{1}{2 z + 5} & = & \dfrac{1}{5} \dfrac{1}{1 + \dfrac{2 z}{5}} \\[3mm]
& = & \displaystyle \dfrac{1}{5} \sum_{n = 0}^{+\infty} (-1)^n \left(\dfrac{2 z}{5}\right)^n \\[3mm]
& = & \displaystyle \sum_{n = 0}^{+\infty} \dfrac{2^n}{5^{n + 1}} (-1)^n z^n \\[3mm]
\end{array}$$
with the condition :
$$\left|\dfrac{2 z}{5}\right| < 1$$
