Hints and answers

3.9 HINTS AND ANSWERS
3.27
A closed barrel has as its curved surface the surface obtained by rotating about
the x-axis the part of the curve
y = a[2 − cosh(x/a)]
lying in the range −b ≤ x ≤ b, where b < a cosh−1 2. Show that the total surface
area, A, of the barrel is given by
A = πa[9a − 8a exp(−b/a) + a exp(−2b/a) − 2b].
3.28
The principal value of the logarithmic function of a complex variable is defined
to have its argument in the range −π < arg z ≤ π. By writing z = tan w in terms
of exponentials show that
1
1 + iz
.
tan−1 z = ln
2i
1 − iz
Use this result to evaluate
tan−1
√
2 3 − 3i
.
7
3.9 Hints and answers
3.1
3.3
3.5
3.7
3.9
3.11
3.13
3.15
3.17
3.19
3.21
3.23
3.25
(a) 5 + 3i; (b) −1 − 5i; (c) 10 + 5i; (d) 2/5 + 11i/5; (e) 4; (f) 3 − 4i;
(g) ln 5 + i[tan−1 (4/3) + 2nπ]; (h) ±(2.521 + 0.595i).
√
√
Use sin π/4 = cos
√ π/4 = 1/ 2, sin π/3 = 1/2 and cos π/3 = 3/2.
cot π/12 = 2 + 3.
√
√
(a) exp(−2y)
√ 2y sinh 2x)/2; (c) 2 exp(πi/3) or 2 exp(4πi/3);
√ cos 2x; (b) (sin
(d) exp(1/ 2) or exp(−1/ 2); (e) 0.540 − 0.841i; (f) 8 sin(ln 2) = 5.11;
(g) exp(−π/2 − 2πn); (h) ln 8 + i(6n + 1/2)π.
Starting from |x + iy − ia| = λ|x + iy + ia|, show that the coefficients of x and y
are equal, and write the equation in the form x2 + (y − α)2 = r2 .
(a) Circles enclosing z = −ia, with λ = exp c > 1.
(b) The condition is that arg[(z −ia)/(z +ia)] = k. This can be rearranged to give
a(z + z ∗ ) = (a2 − |z|2 ) tan k, which becomes in x, y coordinates the equation
of a circle with centre (−a cot k, 0) and radius a cosec k.
All three conditions are satisfied in 3π/2 ≤ θ ≤ 7π/4, |z| ≤ 4; area = 2π.
Denoting exp[2πi/(2m + 1)] by Ω, express x2m+1 − a2m+1 as a product of factors
like (x − aΩr ) and then combine those containing Ωr and Ω2m+1−r . Use the fact
that Ω2m+1 = 1.
The roots are 21/3 exp(2πni/3) for n = 0, 1, 2; 1 ± 31/4 ; 1 ± 31/4 i.
Consider (1 + i)n . (b) S2 (n) = 2n/2 sin(nπ/4). S2 (6) = −8, S2 (7) = −8, S2 (8) = 0.
Use the binomial expansion of (cos θ + i sin θ)4 .
Show that cos 5θ = 16c5 − 20c3 + 5c, where c = cos θ, and correspondingly for
sin 5θ.√Use cos−2 θ = 1√+ tan2 θ. The √
four required values
√ are
[(5 − 20)/5]1/2 , (5 − 20)1/2 , [(5 + 20)/5]1/2 , (5 + 20)1/2 .
Reality of the root(s) requires c2 + b2 ≥ a2 and a + b > 0. With these conditions,
there are two roots if a2 > b2 , but only one if b2 > a2 .
For a2 = c2 + b2 , x = 12 ln[(a − b)/(a + b)].
Reduce the equation to 16 sinh4 x = 1, yielding x = ±0.481.
113
COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS
3.27
Show that ds = (cosh x/a) dx;
curved surface area = πa2 [8 sinh(b/a) − sinh(2b/a)] − 2πab.
flat ends area = 2πa2 [4 − 4 cosh(b/a) + cosh2 (b/a)].
114