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```25.10 HINTS AND ANSWERS
25.23
t = −i and then approaches the origin in the fourth quadrant in a curve that
is ultimately antiparallel to the positive real axis. The other contour, C1 , is the
mirror image of this in the real axis; it is conﬁned to the upper half-plane, passes
through t = i and is antiparallel to the real t-axis at both of its extremities. The
contribution to Jν (z) from the curve Ck is 12 Hν(k) , the function Hν(k) being known
as a Hankel function.
Using the method of steepest descents, establish the leading term in an asymptotic expansion for Hν(1) for z real, large and positive. Deduce, without detailed
calculation, the corresponding result for Hν(2) . Hence establish the asymptotic
form of Jν (z) for the same range of z.
Use the method of steepest descents to ﬁnd an asymptotic approximation, valid
for z large, real and positive, to the function deﬁned by
Fν (z) =
exp(−iz sin t + iνt) dt,
C
where ν is real and non-negative and C is a contour that starts at t = −π + i∞
and ends at t = −i∞.
25.1
25.3
25.5
25.7
25.9
25.11
25.13
25.15
DBED. Eliminate other currents from the equations to obtain IR = ω0 CV0 [ (ω02 −
ω 2 − 2iωω0 )/(ω02 + ω 2 ) ], where ω02 = (LC)−1 ; |IR | = ω0 CV0 ; the phase of IR is
tan−1 [ (−2ωω0 )/(ω02 − ω 2 ) ].
Set c coth u1 = −d, c coth u2 = +d, |c cosech u| = a and note that the capacitance
is proportional to (u2 − u1 )−1 .
ξ = constant, ellipses x2 (a+1)−2 +y 2 (a−1)−2 = c2 /(4a2 ); η = constant, hyperbolae
x2 (cos α)−2 − y 2 (sin α)−2 = c2 . The curves are the cuts −c ≤ x ≤ c, y = 0 and
|x| ≥ c, y = 0. The curves for η = 2π are the same as those for η = 0.
(a) For a quarter-circular contour enclosing the ﬁrst quadrant, the change in the
argument of the function is 0 + 8(π/2) + 0 (since y 8 + 5 = 0 has no real roots);
(b) one negative real zero; a conjugate pair in the second and third quadrants,
− 32 , −1 ± i.
Evaluate
π cot πz
dz
1
+
z 14 + z
2
around a large circle centred on the origin; residue at z = −1/2 is 0; residue at
z = −1/4 is 4π cot(−π/4).
The behaviour of the integrand for large |z| is |z|−2 exp [ (2α − π)|z| ]. The residue
at z = ±m, for each integer m, is sin2 (mα)(−1)m /(mα)2 . The contour contributes
nothing.
Required summation = [ total sum − (m = 0 term) ]/2.
Note that f̄(s) has no pole at s = 0. For t < 0 close the Bromwich contour in the
right half-plane, and for t > 1 in the left half-plane. For 0 < t < 1 the integrand
has to be split into separate terms containing e−s and s − 1 and the completions
made in the right and left half-planes, respectively. The last of these completed
contours now contains a second-order pole at s = 0. f(t) = 1 − t for 0 < t < 1,
but
is 0otherwise.
and γ tend to 0 as R → ∞ and ρ → 0. Put s = r exp iπ and s = r exp(−iπ) on
Γ
∞
the two sides of the cut and use 0 exp(−t2 x) dt = 12 (π/x)1/2 . There are no poles
inside the contour.
925
APPLICATIONS OF COMPLEX VARIABLES
25.17
25.19
25.21
25.23
Use the binomial theorem to expand, in inverse powers of z, both the square
root in the exponent and the fourth root in the multiplier, working to O(z −2 ).
2
2
The leading terms are y1 (z) = Ce−z /4 z ν and y2 (z) = Dez /4 z −(ν+1) . Stokes lines:
arg z = 0, π/2, π, 3π/2; anti-Stokes lines: arg z = (2n + 1)π/4 for n = 0, 1, 2, 3. y1
is dominant on arg z = π/2 or 3π/2.
√
√
2
(a) i πe−z , valid for all z, including i π exp(β 2 ) in case (iii).
(b) The same values as in (a). The (only) saddle point, at t0 = z, is traversed in
the direction θ = + 21 π in all cases, though the path in the complex t-plane varies
with each case.
(c) The same values as in (a). The level lines are v = ±u. In cases (i) and (ii) the
contour turns through a right angle at the saddle point.
All three methods give exact answers in this case of a quadratic exponent.
Saddle points at t1 = −z and t2 = 2z with f1 = −18z and f2 = 18z.
Approximation is
π 1/2 cos(7νz 3 − 1 π) cos(20νz 3 − 1 π)
4
4
+
.
9zν
1 + z2
1 + 4z 2
Saddle point at t0 = cos−1 (ν/z) is traversed in the direction θ = − 14 π. Fν (z) ≈
(2π/z)1/2 exp [ i(z − 12 νπ − 14 π) ].
926
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