calculate-z-5-2-z-2-z-2-9-z-i-2-dz-

Question Number 146547 by mathmax by abdo last updated on 13/Jul/21

calculate \int_{∣ z ∣= 5} \frac{2 - z^{2}}{(z^{2} + 9) {(z - i)}^{2}} dz

Answered by Olaf_Thorendsen last updated on 14/Jul/21

f (z) = \frac{2 - z^{2}}{(z^{2} + 9) {(z - i)}^{2}}

f (z) = \frac{2 - z^{2}}{(z - 3 i) (z + 3 i) {(z - i)}^{2}}

Ω = \int_{∣ z ∣= 5} f (z) d z

{Res}_{3 i} f = \lim_{z \to 3 i} (z - 3 i) f (z) = \frac{11 i}{24}

{Res}_{- 3 i} f = \lim_{z \to - 3 i} (z + 3 i) f (z) = - \frac{11 i}{96}

{Res}_{i} f = \frac{1}{(2 - 1)!} \lim_{z \to i} \frac{\partial^{2 - 1}}{\partial z^{2 - 1}} {(z - i)}^{2} f (z)

{Res}_{i} f = \lim_{z \to i} \frac{\partial}{\partial z} (\frac{2 - z^{2}}{z^{2} + 9})

{Res}_{i} f = \lim_{z \to i} (\frac{- 2 z (z^{2} + 9) - (2 - z^{2}) 2 z}{{(z^{2} + 9)}^{2}})

{Res}_{i} f = (\frac{- 2 i \times 8 - 3 \times 2 i}{64}) = - \frac{11 i}{32}

Ω = \sum_{k} {Res}_{z_{k}} f = \frac{11 i}{24} - \frac{11 i}{96} - \frac{11 i}{32} = 0

Commented by mathmax by abdo last updated on 14/Jul/21

thank you sir .

Answered by qaz last updated on 14/Jul/21

poles all inside circle ∣ z ∣= 5,

sum of function residues is 0 . i think we dont need to evalue .

Answered by mathmax by abdo last updated on 14/Jul/21

f (z) = \frac{2 - z^{2}}{(z^{2} + 9) {(2 - i)}^{2}} \Rightarrow f (z) = \frac{2 - z^{2}}{(z - 3 i) (z + 3 i) {(z - i)}^{2}}

the poles of f are 3 i, - 3 i andi (all interior in the circle ∣ z ∣= 5)

residus theorem \Rightarrow \int_{∣ z ∣= 5} f (z) dz = 2 i π {Res (f, 3 i) + Res (f, - 3 i) + Res (f, i)}

Res (f, 3 i) = \frac{2 - {(3 i)}^{2}}{6 i {(2 i)}^{2}} = - \frac{11}{12 i}

Res (f, - 3 i) = \frac{2 - {(- 3 i)}^{2}}{(- 6 i) {(- 4 i)}^{2}} = - \frac{11}{6 . 16 i} = - \frac{11}{96 i}

Res (f, i) = \lim_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} f (z)}^{(1)}

= \lim_{z \to i} {\frac{2 - z^{2}}{z^{2} + 9}}^{(1)} = \lim_{z \to i} \frac{- 2 z (z^{2} + 9) - 2 z (2 - z^{2})}{{(z^{2} + 9)}^{2}}

= \lim_{z \to i} \frac{- 18 z - 4 z}{{(z^{2} + 9)}^{2}} = \frac{- 22 i}{8^{2}} = - \frac{22 i}{64} = - \frac{11 i}{32} \Rightarrow

\int_{∣ z ∣= 5} f (z) dz = 2 i π {- \frac{11}{12 i} - \frac{11}{96 i} + \frac{11}{32 i}} = \dots .

Commented by mathmax by abdo last updated on 14/Jul/21

sorry Res (f, 3 i) = - \frac{11}{24 i} \dots!