1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

8.1 • THE RESIDUE THEOREM 299

where

. z^2 + 3z + 2

A = Res[zf(z),O] = hm = - 2,

• 
  
  
  
  • 0 z - 1
  
  
  
  

B = Res[f,O] =Lim d zZ +^3 z +^2
•-O dz z - 1

= Lim (2z + 3)(z - 1) - ~z

2

+ 3z + 2) = _

5

and

z-o (z - 1) '

. z^2 + 3z + 2

C = Res[!, 1) = lim 2 = 6.

z - 1 Z

Thus,

z^2 + 3z+ 2 - 2 5 6

----z2 (z - 1) -- --z 2 - +z --z - 1 ·

-------~EXERCISES FOR SECTION 8 .1

1. Find Res[f,O) for

(a) f(z)=z-^1 expz.

(b) f(z)=z-^3 cosh4z.

(c) f (z) = cscz.

(d) f(z)=z2+ 4z+5.

z^2 + z

(e) f (z) = cotz.

(f) f (z) = z -^3 cos z.

(g) J(z) = z-^1 sinz.

(h) f(z) = z2+~z+ 5.

z

(i) f (z) = exp (1 + ~).

(j) f (z) = z^4 sin(~).

(k) J (z) = z-^1 csc z.

(1) f (z) = z-^2 csc z.

(m) f (z) = exp(4z)-1.

sm^2 z

(n) f (z) = z-^1 csc^2 z.