Summation of series

APPLICATIONS OF COMPLEX VARIABLES
y
Y
R
X
O
x
Figure 25.5 A contour for locating the zeros of a polynomial that occur in
the first quadrant of the Argand diagram.
(iii) Y O: z = iy and so arg h = tan−1 y/(y 4 + 1), which starts at O(R −3 ) and finishes at
0 as y goes from large R to 0. It never reaches π/2 because y 4 + 1 = 0 has no real
positive root. Thus ∆Y O [arg h] = 0.
Hence for the complete contour ∆C [arg h] = 0 + 2π + 0 + O(R −3 ), and, if R is allowed
to tend to infinity, we deduce from (25.17) that h(z) has one zero in the first quadrant.
Furthermore, since the roots occur in conjugate pairs, a second root must lie in the fourth
quadrant, and the other pair must lie in the second and third quadrants.
To show that the zeros lie within the given annulus in the z-plane we apply Rouché’s
theorem, as follows.
(i) With C as |z| = 3/2, f = z 4 , g = z + 1. Now |f| = 81/16 on C and |g| ≤ 1 + |z| <
5/2 < 81/16. Thus, since z 4 = 0 has four roots inside |z| = 3/2, so also does
z 4 + z + 1 = 0.
(ii) With C as |z| = 2/3, f = 1, g = z 4 + z. Now f = 1 on C and |g| ≤ |z 4 | + |z| =
16/81 + 2/3 = 70/81 < 1. Thus, since f = 0 has no roots inside |z| = 2/3, neither
does 1 + z + z 4 = 0.
Hence the four zeros of h(z) = z 4 + z + 1 occur one in each quadrant and all lie between
the circles |z| = 2/3 and |z| = 3/2. A further technique useful for locating the zeros of functions is explained in
exercise 25.8.
25.4 Summation of series
We now turn to an application of contour integration which at first sight might
seem to lie in an unrelated area of mathematics, namely the summation of infinite
series. Sometimes a real infinite series with index n, say, can be summed with the
help of a suitable complex function that has poles on the real axis at the various
positions z = n with the corresponding residues at those poles equal to the values
of the terms of the series. A worked example provides the best explanation of
how the technique is applied; other examples will be found in the exercises.
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25.4 SUMMATION OF SERIES
By considering
0
C
π cot πz
dz,
(a + z)2
where a is not an integer and C is a circle of large radius, evaluate
∞
n=−∞
1
.
(a + n)2
The integrand has (i) simple poles at z = integer n, for −∞ < n < ∞, due to the factor
cot πz and (ii) a double pole at z = −a.
(i) To find the residue of cot πz, put z = n + ξ for small ξ:
cot πz =
cos(nπ + ξπ)
cos nπ
1
≈
=
.
sin(nπ + ξπ)
(cos nπ)ξπ
ξπ
The residue of the integrand at z = n is thus π(a + n)−2 π −1 .
(ii) Putting z = −a + ξ for small ξ and determining the coefficient of ξ −1 gives§
π
π cot πz
= 2 cot(−aπ + ξπ)
(a + z)2
ξ
π
d
= 2 cot(−aπ) + ξ
+ ··· ,
(cot πz)
ξ
dz
z=−a
so that the residue at the double pole z = −a is given by
π[−π cosec 2 πz]z=−a = −π 2 cosec 2 πa.
Collecting together these results to express the residue theorem gives
N
0
π cot πz
1
2
2
I=
dz
=
2πi
−
π
cosec
πa
,
2
(a + n)2
C (a + z)
n=−N
(25.23)
where N equals the integer part of R. But as the radius R of C tends to ∞, cot πz → ∓i
(depending on whether Im z is greater or less than zero, respectively). Thus
dz
I<k
,
(a + z)2
which tends to 0 as R → ∞. Thus I → 0 as R (and hence N) → ∞, and (25.23) establishes
the result
∞
π2
1
=
.
2
2
(a
+
n)
sin
πa
n=−∞
Series with alternating signs in the terms, i.e. (−1)n , can also be attempted
in this way but using cosec πz rather than cot πz, since the former has residue
(−1)n π −1 at z = n (see exercise 25.11).
§
This again illustrates one of the techniques for determining residues.
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