find the Laurent series of the function z/(z^2+z-2) in the region |z|>2
Thanks for your help
Thanks for your help
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Note that z/(z^2 + z - 2)
= z/((z + 2)(z - 1))
= (1/3) [2/(z + 2) + 1/(z - 1)], via partial fractions
= (1/3) * (1/z) [2/(1 + 2/z) + 1/(1 - 1/z)], by factoring
= (1/3) * (1/z) [Σ(n = 0 to ∞) 2(-2/z)^n + Σ(n = 0 to ∞) (1/z)^n], by geometric series
= (1/3) * Σ(n = 0 to ∞) [((-1)^n 2^(n+1) + 1)/z^(n+1)].
This converges when both |-2/z| < 1 and |1/z| < 1 ==> |z| > 2.
I hope this helps!
= z/((z + 2)(z - 1))
= (1/3) [2/(z + 2) + 1/(z - 1)], via partial fractions
= (1/3) * (1/z) [2/(1 + 2/z) + 1/(1 - 1/z)], by factoring
= (1/3) * (1/z) [Σ(n = 0 to ∞) 2(-2/z)^n + Σ(n = 0 to ∞) (1/z)^n], by geometric series
= (1/3) * Σ(n = 0 to ∞) [((-1)^n 2^(n+1) + 1)/z^(n+1)].
This converges when both |-2/z| < 1 and |1/z| < 1 ==> |z| > 2.
I hope this helps!