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I need to find the expansion of Laurent series to $\frac{1}{( z-1)( z+i)}$, specifying the singularities and the ring in which it is valid. I tried to use the example of the function $\frac{1}{( z-1)( z+2)}$ as a base, but I couldn't develop it as a series of powers for $z = -i$. I couldn't find a substitution in the sum of $n = 0$ to infinity. Someone can give a light on it? I know I have to do as a geometric progression.

PCeltide
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Mateus L M
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1 Answers1

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Note that$$\frac1{(z-1)(z+i)}=\frac12\left(\frac{-1+i}{1-z}+\frac{1+i}{1-iz}\right).$$Now, in order to see what happens on an annulus centered at $1$, use the fact that$$\frac{1+i}{1-iz}=\frac{1+i}{1-i-i(z-1)}.$$And, in order to see what happens on an annulus centered at $-i$, use the fact that$$\frac{-1+i}{1-z}=\frac{-1+i}{1+i-(z+i)}.$$

José Carlos Santos
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  • Thank you very much, I found the sum for z = 1 and the interval, through the substitution in the geometric progression, but this second part I had not found because I was making the substitution for 1 / (z-1) + (i-1). – Mateus L M Apr 07 '21 at 16:30