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Let $\displaystyle\sum_{n=-\infty}^{\infty}a_nz^n$ be the Laurent series expansion of $\displaystyle f(z)=\frac{1}{2z^2-13z+15}$ in the annulus $\displaystyle\frac{3}{2}<|z|<5$. Then $\dfrac{a_1}{a_2}=5.$

But I don't get $\dfrac{a_1}{a_2}=5$. I solved like this: $$\frac{1}{2z^2-13z+15}=\frac{1}{15}(1+\frac{2z^2-13z}{15})^-1 = \frac{1}{15}{(1-\frac{2z^2}{15}+\frac{13z}{15}+\frac{4z^4+169z^2-52z^3}{225}+...)}.$$ I get $a_1=\dfrac{13}{15^2}$ and $a_2=\dfrac{-2}{15^2}+\frac{169}{15^3}=\dfrac{139}{15^3}$. Then $\dfrac{a_1}{a_2}=\dfrac{195}{139}=1.4$ Where I am wrong.

Somebody help me.

  • 0
    Write down a factors of this polynomial and use partial fraction. – 2017-02-01
  • 0
    See [here](https://math.stackexchange.com/questions/2285139/laurent-series-expansion-of-a-given-function) for your answer – 2017-05-18

0 Answers 0