1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

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ANSWERS 599

lk. i (! + n) n, where n is an integer.

Section 6.1. Complex Integrals: page 197


la. 2 - 3i.

le. 1.

le. J21T' /8 + VZ/2 - 1 + i ( VZ/2 - v'zn /8).



  1. Using (6-8), roo e-•tdt = Lim rT e- '^1 dt = Jim (-le-•T + le-•(Ol) =
    Jo T-oo Jo T - oo z z
    l z + T-oo Lim ( -l e-zTz ). Show that Re( z) > 0 implies this last limit equals zero.

  2. ThL~ follows from (6-8), and the fact that if 1L and v are different iable, then
    f is differentiable, and ft [~U(t))
    2
    ] =f(t)f'(t).


Section 6.2. Contours and Contour Integrals: page 211


la. Ci : z1 (t) = 2eit, 0 ~ t ~ ~· C2: z 2 (t) = - t + i (2-t), 0 ~ t ~ 2.

Sa. The approximation simplifies to -2J2 + 2 :::l - 0.828427.


3b. -~.


5a. -32i.

5b. - 8ni.

7a. O.

7c. -2ni.

7e. i-2.


7g. -4- in.

9a. 2ni.

9b. 0.




    • 1 + 2i 3 ·




1 3. -2e.


  1. exp (1 + i) - 1.
    17. sin (1 + i).

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