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21.1 Power Series 723

then f ( j) (ζ) = 0for j = 0,1,··· ,m − 1, while f (m) (ζ) = 0. Therefore, the order of a zero is the
smallest integer such that the derivative of that order is nonzero at ζ.
In the proof of Theorem 21.4, we actually showed that, in some disk about an isolated zero
of f of order m, we can write
m
f (z) = (z − z) g(z)
where g(z) = 0 on this disk. This fact is important in its own right.

EXAMPLE 21.5
2
Let f (z) = z sin(z). Then f has an isolated zero at 0. Compute
2

f (z) = 2z sin(z) + z cos(z),
2

f (z) = 2sin(z) + 4z cos(z) − z sin(z),
and
2

f (z) = 6cos(z) − 6z sin(z) − z cos(z).

Now f (0) = f (0) = f (0) = 0, while f (0) = 0. Therefore, f has a zero of order 3 at 0.

3
We will write f (z) = z g(z) where g(z) is differentiable and nonzero is on some disk about
0. Use the Maclaurin expansion to write
1 1
3
5
sin(z) = z + z + z + ··· ,
3! 5!
so
1 1
2 3 5 7
f (z) = z sin(z) = z + z + z + ···
3! 5!
1 1

2
4
= z 3 1 + z + z + ···
3! 5!
3
= z g(z),
where g(z) = 0 on a disk about 0.
m
One immediate ramification of being able to write f (z) = (z − ζ) g(z) with g(z) = 0in
some disk about ζ is that, under certain conditions, the orders of the zeros of products add and
orders of zeros of quotients subtract (reminiscent of a logarithm). To be specific, suppose h has
a zero of order m at ζ, and k has a zero of order n at ζ. Then
1. h(z)k(z) has a zero of order m + n at ζ.
2. If n < m, then h(z)/k(z) has a zero of order m − n at ζ.

m
n
To see why statement (1) is true, write h(z) = (z − ζ) α(z) and k(z) = (z − ζ) β(z), where
α(z) and β(z) are nonzero in some open disk D about ζ. Then
h(z)k(z) = (z − ζ) m+n α(z)β(z)
and α(z)β(z) = 0in D,so h(z)k(z) has a zero of order m + n at ζ. Statement (2) is proved
similarly.
These facts will be important when we consider the order of poles as singularities of
functions.

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October 14, 2010 15:35 THM/NEIL Page-723 27410_21_ch21_p715-728