1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

ANSWERS 601

11. 
    

    lnv'2-t+i(lnv'2+~-1).

    
    


12. 
    

    Log (1 + i) - Log (2) + Log (2 + i) =-~In J2 +In J5 + i (~+Arc tan 4).

    
    

15. Parametrize C with z (t) = z1 + (zz -z 1) t, 0 ~ t ~ 1. Then we see that

fc ldz = Jd z' (t) dt = f 0

1

(z2 - zi) dt = (z2 - zi) ti:~~= z2 - z1.

l 7a. J5( cos(0.46) + i sin(0.46)) - 3.

19. We know that an antiderivative of the function f g' +gf' is Jg by the product
    rule. Since f g' and gf' are analytic (explain why!), Theorem 6.9 gives us
    fc!f(z)g'(z)+g(z)f'(z)]dz = f(z)g(z)I;~;:. The conclusion follows
    from this.

Section 6.5. Integral Representations for Analytic Functions: p age

239

1. 47ri.


2. -i~.
   5. -i~.


3. 27ri.

(^9) · (n2,,; -1)! ·

1 1. i - ii.

13a. i7r sinh 1.

13b. i7r sinh 1.

15 a. 71'.

15b. -71'.

17. o.


18. Let f (z) = (z^2 - lf, which is analytic everywhere. By Cauchy's integral
    < lOrm ul as, •n n ( z ) - 1 J (n) ( ) I [ n! f f({) d'"] Th J ·
    2 "n! z - z•n! 2ii1 Jc ({-•)"+^1 .,. e cone us1on
    follows from this. Show the details.

Section 6.6. The Theorems of Morera and Liouville and Some Appli-

cations: page 247

la. (z+ 1 +i)(z+ l -i)(z-1 +i) (z - 1 - i).

l e. (z + i) (z -i) (z + 2 - i) (z - 2 -i).