I dont know how to make the Laurent series expansion of this functions $$\frac{1}{(z+1)(z-3)(z+4)} ,3<|z+1|<4;$$
$$\frac{(z+2)^2}{(z-i)^2} , 2<|z+i|<\infty$$
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1What are your first steps to solve and what is the difficulty for you? Then we are able to give you a hint or just one step towards the solution – Applied mathematician Jan 07 '13 at 19:37
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I'm assuming you know how to expand stuff like $1/(z+1)$. Can you think of a partial fraction decomposition of the above? – Alex R. Jan 07 '13 at 19:38
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the problem is that i havent even idea how to solve this – molniya Jan 07 '13 at 19:40
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Here is a relates problem. – Mhenni Benghorbal Jan 07 '13 at 20:29
1 Answers
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For these types of questions it is important to know the Laurent expansion of $(z-a)^{-1}$ for both $|z|\leq a$ and $|z| \geq a$. I hope you can easily convince yourself that $$\tag{1} \frac{1}{z-a} = -\frac{1}{a} \sum_{k=0}^\infty \left(\frac{z}{a}\right)^n, \qquad |z| \leq a $$ and $$\tag{2} \frac{1}{z-a} = \frac{1}{z} \sum_{k=0}^\infty \left(\frac{a}{z}\right)^n, \qquad |z| \geq a.$$
I will show you how it works for the first question $$\frac{1}{(z+1)(z-3)(z+4)} ,3<|z+1|<4;$$ and I believe you can the do it yourself for the second. The Laurent series should be expanded in a ring centered at $z_0=-1$. Thus, we introduce $w=z+1$. We then have the problem $$\frac{1}{w(w-4)(w+3)} ,3<|w|<4;$$
Next we perform a partial fraction expansion and obtain $$\frac{1}{w(w-4)(w+3)}= -\frac{1}{12 w}+\frac{1}{21 (w+3)}+\frac{1}{28 (w-4)} .$$ For each of the terms you should now think whether you have to apply (1) or (2).
Fabian
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