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I am trying to express a function with Taylor and Laurent series. I've been reading my textbook and also various online resources, but I still can't follow any of the example problems. Here's what I understand so far.

I have a function of a complex-valued number z, and its denominator is $0$ at $z= 1$ and $2$.

$$f(z)=\frac{5-z}{z^2-3z+2}$$

From this, I think I should have three separate series. One for $|z|<1$, one for $1<|z|<2$, and one for $|z|>2$. I think I managed to get the first one by deriving the Taylor series for $f$, but I know that this only has a radius of convergence of $1$.

How can I proceed to derive the other two series? One resource I read stated that when $z$ is greater than the radius of convergence, I can use the fact that $1/z<R$ to work with a geometric series, but I am confused on the algebraic manipulation.

Any feedback/constructive criticism is appreciated.

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2 Answers 2

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Note that$$\frac{5-z}{z^2-3z+2}=\frac3{z-2}-\frac4{z-1}$$and so, if $1<\lvert z\rvert<2$, then\begin{align}f(z)&=-\frac3{2-z}+\frac4{1-z}\\&=-\frac{\frac32}{1-\frac z2}+\frac4{1-z}\\&=-\frac32\sum_{n=0}^\infty\left(\frac z2\right)^n-4\sum_{n=-\infty}^{-1}z^n\text{ (since $1<\lvert z\rvert<2$)}\\&=-3\sum_{n=0}^\infty\frac{z^n}{2^{n+1}}-4\sum_{n=-\infty}^{-1}z^n.\end{align}Can you deal with the case $\lvert z\rvert>2$ now?

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  • $\begingroup$ I understand that after partial fraction decomposition, you chose to divide the first term's numerator and denominator by 2 since the annulus goes up to z<2. As for the second term, could you explain what's going on? You said "since 1<|z|<2", but I have a hard time relating that back to the Laurent series you wrote. Thank you. $\endgroup$
    – Skipher
    Commented Feb 18, 2019 at 18:28
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    $\begingroup$ Here, I am just using the fact that$$\frac1{1-z}=\begin{cases}\sum_{n=0}^\infty z^n&\text{ if }\lvert z\rvert<1\\-\sum_{n=-\infty}^{-1}z^n&\text{ if }\lvert z\rvert>1.\end{cases}$$ $\endgroup$ Commented Feb 18, 2019 at 18:34
  • $\begingroup$ That makes sense, thank you! $\endgroup$
    – Skipher
    Commented Feb 19, 2019 at 3:18
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You'll have a partial fraction expansion in terms of $1/(z-1)$ and $1/(z-2)$ so you'll need Laurent expansions for each. Use that $$\frac1{z-a}=-\frac1 a\frac1{(1-z/a)}=\sum_{n=0}^\infty-\frac{z^n}{a^{n+1}}$$ if $|z|<|a|$ and $$\frac1{z-a}=\frac1 z\frac1{(1-a/z)}=\sum_{n=1}^\infty a^nz^{-n-1}$$ if $|z|>|a|$.

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  • $\begingroup$ Just to clarify, do I combine both summations to express the function, or do I define it as piecewise depending on |z|? $\endgroup$
    – Skipher
    Commented Feb 18, 2019 at 18:32
  • $\begingroup$ As you say, "I think I should have three separate series..." @Skipher $\endgroup$ Commented Feb 19, 2019 at 3:20

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