calculate-z-1-z-z-1-z-2-dz-with-is-the-circle-z-C-z-3-2-

Question Number 37298 by math khazana by abdo last updated on 11/Jun/18

c a l c u l a t e \int_{γ} \frac{z + 1}{z (z - 1) (z + 2)} d z w i t h γ i s t h e

c i r c l e γ = {z \in C / ∣ z ∣ = \frac{3}{2}}

Commented by prof Abdo imad last updated on 15/Jun/18

l e t φ (z) = \frac{z + 1}{z (z - 1) (z + 2)} t b e p o l e s o f φ a r e

0, 1, - 2 (- 2 i s o u t o f t h e c i r c l e γ)

\int_{γ} φ (z) d z = 2 i π {R e s (φ, 0) + R e s (φ, 1)}

R e s (φ, 0) = \frac{1}{- 2} = - \frac{1}{2}

R e s (φ, 1) = \frac{2}{3} \Rightarrow

\int_{γ} φ (z) d z = 2 i π {- \frac{1}{2} + \frac{2}{3}} = 2 i π . \frac{1}{6} = \frac{i π}{3} .