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732 CHAPTER 22 Singularities and the Residue Theorem

If f (z) is a quotient of functions, it is natural to look for poles at places where the denom-
inator vanishes, that is, where the denominator has a zero. With some care, this strategy is
effective.

THEOREM 22.2 Poles of Quotients (1)
Let f (z) = h(z)/g(z) where h and g are differentiable in some open disk about z 0 . Suppose that
h(z 0 ) = 0but g(z) has a zero of order m at z 0 . Then f has a pole of order m at z 0 .

EXAMPLE 22.5
Let
2
z
1 + e + 4z 3
f (z) = 6 .
sin (z)
Then f has a pole of order 6 at 0 because the numerator is differentiable and nonzero at z = 0,
while the denominator is differentiable and has a zero of order 6 at 0.

EXAMPLE 22.6

Let
1
f (z) = .
cos (z)
3
Then f has a pole of order 3 at each zero of cos(z), which are the numbers z = (2n + 1)π/2for
integer n.

Theorem 22.2 does not apply if the numerator also vanishes at z 0 . The example f (z) =
5
sin(z)/z is instructive. The numerator has a zero of order 1 at 0, the denominator has a zero of
order 5 at 0, but the quotient has a pole of order 4 at 0. It appears that the orders of the zeros
of the numerator and denominator subtract to give the order of the pole. That is, zeros appear to
cancel (recall the observations about addition and subtraction of orders of zeros in quotients at
the end of Chapter 21). This is indeed the case.

THEOREM 22.3 Poles of Quotients (2)

Let f (z) = h(z)/g(z), and suppose h and g are differentiable in some open disk about z 0 .Let h
have a zero of order k at z 0 , and let g have a zero of order m at z 0 with m > k. Then f has a pole
of order m − k at z 0 .

If m = k in Theorem 22.3, f has a removable singularity at z 0 (recall Example 22.1). If
m < k, then f does not have a pole at z 0 .

EXAMPLE 22.7
Let
(z − 3π/2) 4
f (z) = .
7
cos (z)

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October 14, 2010 15:37 THM/NEIL Page-732 27410_22_ch22_p729-750