calculate-dz-z-3-8-in-those-cases-1-z-C-z-1-2-z-C-z-3-

Question Number 37303 by math khazana by abdo last updated on 11/Jun/18

c a l c u l a t e \int_{γ} \frac{d z}{z^{3} + 8} i n t h o s e c a s e s

1) γ = {z \in C / ∣ z ∣ = 1}

2) γ = {z \in C / ∣ z ∣ = 3}

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

1) l e t φ (z) = \frac{1}{z^{3} + 8}

φ (z) = \frac{1}{(z + 2) (z^{2} - 2 z + 4)} p o l e s o f φ ?

z^{2} - 2 z + 4 = 0

Δ^{^{'}} = 1 - 4 = - 3 = {(i \sqrt{3})}^{2}

z_{1} = 1 + i \sqrt{3} a n d z_{2} = 1 - i \sqrt{3} . b u t

∣ - 2 ∣= 2 > 1 a l s o ∣ z_{1} ∣= 2 > 1, ∣ z_{2} ∣= 2 > 1 \Rightarrow

\int_{γ} φ (z) d z = 0

2) a l l p o l e s o f φ a r e i n t h e i n t e r i o r l f γ s o

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

φ (z) = \frac{1}{(z + 2) (z - z_{1}) (z - z_{2})}

R e s (φ, - 2) = \frac{1}{4 + 4 + 4} = \frac{1}{12}

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

= \frac{1}{(1 + i \sqrt{3} + 2) (2 i \sqrt{3})} = \frac{1}{2 i \sqrt{3} (3 + i \sqrt{3})}

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

= \frac{1}{(1 - i \sqrt{3} + 2) (- 2 i \sqrt{3})} = \frac{1}{- 2 i \sqrt{3} (3 - i \sqrt{3})}

\int_{γ} φ (z) d z = 2 i π {\frac{1}{12} + \frac{1}{2 i \sqrt{3} (3 + i \sqrt{3})} - \frac{1}{2 i \sqrt{3} (3 - i \sqrt{3})}}

= 2 i π {\frac{1}{12} + \frac{1}{2 i \sqrt{3}} (\frac{3 - i \sqrt{3} - 3 - i \sqrt{3}}{12})}

= 2 i π {\frac{1}{12} - \frac{1}{12}} = 0