find-1-C-e-z-2-z-2-4z-3-dz-C-z-2-5-2-cosx-x-2-2x-2-dx-

Question Number 146977 by tabata last updated on 16/Jul/21

f i n d

(1) \int_{C} \frac{e^{z^{2}}}{z^{2} + 4 z + 3} d z, C :∣ z - 2 ∣= 5

(2) \int_{- \infty}^{\infty} \frac{c o s x}{x^{2} + 2 x + 2} d x

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

1) Φ = \int_{C} \frac{e^{z^{2}}}{z^{2} + 4 z + 3} dz withC \to∣ z - 2 ∣= 5

Δ^{^{'}} = 4 - 3 = 1 \Rightarrow z_{1} = - 2 + 1 = - 1 and z_{2} = - 2 - 1 = - 3

f (z) = \frac{e^{z^{2}}}{(z - z_{1}) (z - z_{2})} residus \Rightarrow \int_{C} f (z) dz = 2 i π {Res (f, z_{1}) + Res (f, z_{2})}

= 2 i π {\frac{e^{z_{1}^{2}}}{z_{1} - z_{2}} + \frac{e^{z_{2}^{2}}}{z_{2} - z_{1}}} = 2 i π {\frac{e}{2} + \frac{e^{9}}{- 2}} = i π {e - e^{9}}

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

Ψ = \int_{- \infty}^{+ \infty} \frac{cosx}{x^{2} + 2 x + 2} dx = Re (\int_{- \infty}^{+ \infty} \frac{e^{ix}}{x^{2} + 2 x + 2} dx)

let φ (z) = \frac{e^{iz}}{z^{2} + 2 x + 2} poles of φ ?

Δ^{^{'}} = 1 - 2 = - 1 \Rightarrow z_{1} = - 1 + i and z_{2} = - 1 - i \Rightarrow φ (z) = \frac{e^{iz}}{(z - z_{1}) (z - z_{2})}

residus \Rightarrow \int_{R} φ (z) dz = 2 i π Res (φ / z_{1}) = 2 i π . \frac{e^{{iz}_{1}}}{z_{1} - z_{2}}

= 2 i π . \frac{e^{i (- 1 + i)}}{2 i} = π e^{- 1} . e^{- i} = \frac{π}{e} (\cos (1) - isin (1)) \Rightarrow

Ψ = \frac{π}{e} \cos (1) (1 = 1 radian)