using-residue-theorem-evaluate-z-3-zsecz-z-1-2-dz-

Question Number 147035 by rs4089 last updated on 17/Jul/21

u s i n g r e s i d u e t h e o r e m

e v a l u a t e \int_{∣ z ∣= 3} \frac{z s e c z}{{(z - 1)}^{2}} d z

Answered by mathmax by abdo last updated on 17/Jul/21

Ψ = \int_{∣ z ∣= 3} \frac{z}{cosz {(z - 1)}^{2}} dz \Rightarrow Ψ = 2 i π Res (f, 1)

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

= \lim_{z \to 1} {\frac{z}{cosz}}^{(1)} = \lim_{z \to 1} \frac{cosz + zsinz}{\cos^{2} z} = \frac{\cos (1) + \sin (1)}{\cos^{2} (1)}

\Rightarrow Ψ = 2 i π \times \frac{\cos (1) + \sin (1)}{\cos^{2} (1)}

Answered by Olaf_Thorendsen last updated on 17/Jul/21

Ω = \int_{∣ z ∣= 3} \frac{z \sec z}{{(z - 1)}^{2}} d z

Ω = 2 i π {Res}_{1} f

Ω = 2 i π \times \frac{1}{(2 - 1)!} . \lim_{z \to 1} \frac{\partial^{2 - 1}}{\partial z^{2 - 1}} {(z - 1)}^{2} f (z)

Ω = 2 i π . \lim_{z \to 1} \frac{\partial}{\partial z} (z \sec z)

Ω = 2 i π . \lim_{z \to 1} (\frac{\cos z + z \sin z}{\cos^{2} z})

Ω = 2 i π (\frac{\cos (1) + \sin (1)}{\cos^{2} (1)})