1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

282 CHAPTER 7 • TAYLOR AND LAURENT SERIES

• Corollary 7.6 If f has a pole of order k at the point a, then g (z) = 7h> has a

removable singularity at a. If we define g (a)= 0, then g (z) has a zero of order
k at a. •

• Corollary 7. 7 lf f and g have poles of orders m and n, respectively, at
  the point a, then their product h (z) = f (z) g (z) has a pole of order m + n

at a. •

t Corollary 7 .8 Let f and g be analytic with zeros of orders m and n, respec-

tively, at a. Then their quotient h (z) = £f# has the following behavior.

i. If m > n, then h has a removable singularity at a. If we define h (a) = 0,

then h has a zero of order m -n at a.

ii. If m < n, then h has a pole of order n -m at a.

iii. If m = n, then h has a removable singularity at a and can be defined so that

his analytic at a by h (a) = lim h (z). •

z-o

•EXAMPLE 7.12 Locate the zeros and poles of h(z) = "";z and determine

their order.

Solution In Section 5.4 we saw that the zeros of f (z) = sin z occur at the

points mr, where n is an integer. Because f^1 (mr) = cosmr =f 0, the zeros off

are simple. Similarly, the function g(z) = zcosz has simple zeros at the points
0 and (n + ~) 11", where n is an integer. From the information given, we find that

h(z) =~behaves as follows:

i. h has simple zeros at mr, where n = ±1, ±2, ... ;

ii. h has simple poles at ( n + D ?r, where n is an integer; and

iii. h is analytic at 0 if we define h (O) = lim h (z) = 1.

z~o

• EXAMPLE 7. 13 Locate the poles of g (z) = sz•+ 216 z•+s and specify their
  order.

Solution The roots of the quadratic equat ion 5z^2 + 26z + 5 = 0 occur at

the points - 5 and - i. If we replace z with z^2 in this equation, the function

f (z) = 5z^4 + 26 z^2 + 5 has simple zeros at the points ±iv's and± Jg. Corollary

7.5 implies t hat g has simple poles at ±iv'5 and ± Js.