can anyone please help me with this complex analysis question ?

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  • kb
    Lv 7
    7 years ago
    Favorite Answer

    Without using the Residue Theorem:

    Note that the integrand has singularities at z^2 = i

    ==> z = [e^(πi/2 + 2πik)]^(1/2) = ±(1 + i)/√2.

    So, z^2 + i = (z - (1 + i)/√2)(z + (1 + i)/√2).

    ------------

    Now, we apply the Generalized Cauchy Integral Formula, deforming C into two disjoint closed contours

    C₁ and C₂, which enclose z = (1 + i)/√2 and z = -(1 + i)/√2, respectively (and separately).

    (i) ∫c₁ e^(iz) dz/(z^2 + i)^2

    = ∫c₁ e^(iz) dz/[(z - (1 + i)/√2)(z + (1 + i)/√2)]^2

    = ∫c₁ [e^(iz)/(z + (1 + i)/√2)^2] dz/(z - (1 + i)/√2)^2

    = 2πi * (d/dz) [e^(iz)/(z + (1 + i)/√2)^2] {at z = (1 + i)/√2}, by Gen. Cauchy

    = (πi/4) e^((-1 + i)/√2) [(1 + i)√2 + 2], after some work.

    (ii) ∫c₂ e^(iz) dz/(z^2 + i)^2

    = ∫c₂ [e^(iz)/(z - (1 + i)/√2)^2] dz/(z + (1 + i)/√2)^2

    = 2πi * (d/dz) [e^(iz)/(z - (1 + i)/√2)^2] {at z = -(1 + i)/√2}, by Gen. Cauchy

    = (πi/4) e^((1 - i)/√2) [(-1-i)√2 + 2].

    -----------

    Therefore, ∫c e^(iz) dz/(z^2 + i)^2

    = (πi/4) e^((-1 + i)/√2) [(1 + i)√2 + 2] + (πi/4) e^((1 - i)/√2) [(-1-i)√2 + 2].

    I hope this helps!

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