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I'm studying for my upcoming complex analysis qualifying exam by working through problems in past exams. For this problem, I'd like to know (1) if my answer is correct and complete (i.e. whether I've made any errors/omissions), and (2) if there are any better ways of doing this. I'm specifically curious about whether or not there are other areas of convergence that I need to address. Thanks!

Question:

Expand the function $$ f(z) = \frac{1+2z^2}{z^2+z^4} $$ into power series of $z$ in all areas of convergence.

Attempted Solution:

We begin by noting that this function has poles at $z=0$ and $z=\pm i$. Then we proceed by recalling the geometric series $$ \sum_{k=0}^{\infty} z^k = \frac{1}{1-z}, $$ which converges for $|z|<1$. Using this form, we can rewrite $f(z)$ as follows

\begin{align*} f(z) &= \frac{1+2z^2}{z^2+z^4} \\ &= \frac{1}{z^2+z^4} + \frac{2z^2}{z^2+z^4} \\ &= \frac{1}{z^2(1-(-z^2))} + \frac{2z^2}{z^2(1-(-z^2))} \\ &= \frac{1}{z^2}\left(\frac{1}{1-(-z^2)}\right) + 2\left(\frac{1}{1-(-z^2)}\right)\\ &= \left(\frac{1}{z^2}+2\right)\left(\frac{1}{1-(-z^2)}\right)\\ &= \left(\frac{1}{z^2}+2\right) \sum_{k=0}^{\infty} (-z^2)^k \\ &= \sum_{k=0}^{\infty} \left(z^{-2}+2\right)(-1)^k z^{2k} \end{align*}

for $0<|z|<1$, since there is a pole at $z=0$.

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  • $\begingroup$ From the way it is stated, I guess this is one of two parts to the solution. You also need a different series valid in $1<|z|$. $\endgroup$
    – GEdgar
    Commented Jul 5, 2022 at 19:48
  • $\begingroup$ @GEdgar I wondered about that. How would I go about this though? $\endgroup$
    – Serafina
    Commented Jul 5, 2022 at 19:51
  • $\begingroup$ If $|z|>1$, what is the geometric series with sum $1/(z^2+z^4)$ ? $\endgroup$
    – GEdgar
    Commented Jul 5, 2022 at 19:54
  • $\begingroup$ @GEdgar I'm sorry, I'm unclear what you're asking. I thought that with $|z|>1$, the series would simply diverge? $\endgroup$
    – Serafina
    Commented Jul 5, 2022 at 20:49
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    $\begingroup$ @Serafina: Note your series expansion is not correct, since it contains terms with a negative exponent of $z$. So, it is a Laurent series expansion, not a power series expansion. As we have a (double) pole at $z=0$, which is a singularity, a power series expansion at $z=0$ is not feasible. $\endgroup$ Commented Jul 6, 2022 at 7:12

2 Answers 2

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Since we are looking for a power series we are looking for a representation \begin{align*} f(z)=\sum_{k=0}^\infty a_k\left(z-z_0\right)^k \end{align*} evaluated at center $z_0$.

We start with a partial fraction decomposition \begin{align*} \color{blue}{f(z)}&=\frac{1+2z^2}{z^2\left(1+z^2\right)}\\ &\,\,\color{blue}{=\frac{1}{z^2}+\frac{i}{2}\,\frac{1}{z+i}-\frac{i}{2}\,\frac{1}{z-i}}\tag{1} \end{align*}

We have poles at $0$ and $\pm i$. So we can expand each of the terms at $z_0\in\mathbb{C}\setminus\{0,\pm i\}$ in a power series. We obtain

\begin{align*} \frac{1}{z^2}&=\frac{1}{\left(z-z_0+z_0\right)^2}=\frac{1}{z_0^2}\,\frac{1}{\left(1+\frac{z-z_0}{z_0}\right)^2}\\ &=\frac{1}{z_0^2}\sum_{k=0}^\infty (-1)^k(k+1)\left(\frac{z-z_0}{z_0}\right)^k\qquad\qquad |z-z_0|<|z_0|\tag{2.1} \\ \frac{i}{2}\,\frac{1}{z+i}&=\frac{i}{2}\,\frac{1}{(z-z_0)+(z_0+i)}\\ &=\frac{i}{2}\,\frac{1}{z_0+i}\,\frac{1}{1+\frac{z-z_0}{z_0+i}}\\ &=\frac{i}{2}\,\frac{1}{z_0+i}\sum_{k=0}^\infty(-1)^k\left(\frac{z-z_0}{z_0+i}\right)^k\qquad\qquad |z-z_0|<|z_0+i|\tag{2.2} \\ \frac{i}{2}\,\frac{1}{z-i}&=\frac{i}{2}\,\frac{1}{(z-z_0)+(z_0-i)}\\ &=\frac{i}{2}\,\frac{1}{z_0-i}\,\frac{1}{1+\frac{z-z_0}{z_0-i}}\\ &=\frac{i}{2}\,\frac{1}{z_0-i}\sum_{k=0}^\infty(-1)^k\left(\frac{z-z_0}{z_0-i}\right)^k\qquad\qquad |z-z_0|<|z_0-i|\tag{2.3} \end{align*}

Putting (1) and (2.1) to (2.3) together we conclude $f(z)$ admits the power series representation \begin{align*} \color{blue}{f(z)}&=\frac{1+2z^2}{z^2\left(1+z^2\right)}\\ &\,\,\color{blue}{=\sum_{k=0}^\infty(-1)^k\left(\frac{k+1}{z_0^{k+2}}+\frac{i}{2}\frac{1}{(z_0+i)^{k+1}} -\frac{i}{2}\frac{1}{(z_0-i)^{k+1}}\right)\left(z-z_0\right)^k}\\ \end{align*} evaluated at $z_0\in\mathbb{C}\setminus\{0,\pm i\}$ and convergent in \begin{align*} \{z\in\mathbb{C}: |z-z_0|<|z_0| \text{ and } |z-z_0|<|z_0+i| \text{ and } |z-z_0|<|z_0-i|\} \end{align*}

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  • $\begingroup$ Wow, it seems I had a lot of misunderstanding around this one. It took me a bit to walk through and understand your solution above, but it's making sense now. Thank you. So the way that I did the problem would only apply if the question was asking for a Laurent series, is that correct? $\endgroup$
    – Serafina
    Commented Jul 8, 2022 at 2:27
  • $\begingroup$ I think it's a bit confusing to understand the distinction between what is or is not a power series. In some places, a Laurent series is said to be a power series with negative exponents and a Taylor series is said to be a power series with no negative exponents. This makes it seem as though a Laurent series in indeed considered a power series? $\endgroup$
    – Serafina
    Commented Jul 8, 2022 at 2:42
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    $\begingroup$ @Serafina: It's the binomial series expansion. We have the coefficient $\binom{-2}{k}=\binom{2+k-1}{k}(-1)^k=\binom{k+1}{1}(-1)^k=(k+1)(-1)^k$. $\endgroup$ Commented Jul 23, 2022 at 15:01
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    $\begingroup$ @Serafina: Yes, thanks. Good review, typo corrected. The region of convergence is the intersection of the single regions, so that it is admissible to combine them all. $\endgroup$ Commented Jul 24, 2022 at 5:28
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    $\begingroup$ Awesome, makes sense... thank you again! $\endgroup$
    – Serafina
    Commented Jul 25, 2022 at 14:31
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Yes, that's perfect. Note that you can even simplifie more your expression: $$\sum_{k=0}^{\infty} \left(z^{-2}+2\right)(-1)^k z^{2k}=\sum_{k=0}^{\infty}(-1)^kz^{2(k-1)}+2\sum_{k=0}^{\infty}(-1)^kz^{2k}=\\\sum_{k=-1}^{\infty}(-1)^{k+1}z^{2k}+2\sum_{k=0}^{\infty}(-1)^kz^{2k}\\=\frac{1}{z^2}+\sum_{k=0}^{\infty}(-1)^{k+1}z^{2k}+2\sum_{k=0}^{\infty}(-1)^kz^{2k}=\frac{1}{z^2}-\sum_{k=0}^{\infty}(-1)^kz^{2k}+2\sum_{k=0}^{\infty}(-1)^kz^{2k}\\=\frac{1}{z^2}+\sum_{k=0}^{\infty}(-1)^kx^{2k}=\frac{1}{z^2}+1-z^2+z^4+O(z^6)$$ Note that the residue of $f$ in $z=0$ since $a_{-1}=0$

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  • $\begingroup$ Thanks! I didn't see how it could be further simplified. That's helpful! $\endgroup$
    – Serafina
    Commented Jul 5, 2022 at 19:20
  • $\begingroup$ One quick question... In your further simplification, did the 2 multiplied by the second summation term get lost somewhere by mistake or am I missing something? $\endgroup$
    – Serafina
    Commented Jul 5, 2022 at 19:34
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    $\begingroup$ i will edit it to clarify :) $\endgroup$ Commented Jul 5, 2022 at 19:35
  • $\begingroup$ Oh, I see it now. Oops. Thanks again! $\endgroup$
    – Serafina
    Commented Jul 5, 2022 at 19:40

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