Can anyone help me out with finding the Laurent series of $f(z)=\dfrac{1}{(z-1)^2}+\dfrac{1}{z-2}$ in $\{z \in \Bbb C: 2<|z-4|<3\}$?
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3The Laurent series in which annulus? – Daniel Fischer Jan 21 '14 at 20:59
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Sorry I forgot to say, I edited my post. – Roos Jansen Jan 21 '14 at 21:07
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A related technique. – Mhenni Benghorbal Jan 21 '14 at 21:19
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Just an addition to Mhenni's answer: You can obtain the power series for $\frac{1}{(1-w)^2}$ by taking the derivative of the series for $\frac{1}{(1-w)}$. (Be careful with the signs!) – Braindead Jan 22 '14 at 00:27
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A related technique. Here is how you advance.
$$ f(z)=\dfrac{1}{(z-1)^2}+\dfrac{1}{z-2} =\dfrac{1}{((z-4)+3)^2}+\dfrac{1}{(z-4)+2}$$
$$ = \dfrac{1}{9\left(\frac{(z-4)}{3}+1\right)^2}+\dfrac{1}{(z-4)(1+\frac{2}{z-4})} $$
$$ = \dfrac{1}{9\left(w+1\right)^2}+\dfrac{1}{(z-4)(1+t)}, $$
where
$$ w= \frac{(z-4)}{3}\,\quad t= \frac{2}{z-4}. $$
Now, recalling the geometric series, we have, the power series of $g(w)=\frac{1}{(1+w)(1+w)}$ will converge for $|w|<1$ which implies $|z-4| <3 $. On the other hand, the power series $h(t)=\frac{1}{1+t}$ will converge for $|t|<1$ which gives $ |z-4|>2 $.
Mhenni Benghorbal
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