MH, 30/10/07
M1F — Sheet 4 solutions
1. (a) The 10th roots of i are of the form eiπ/20 times ω where ω is any 10th root
of unity. Hence they are the 10 complex numbers ei(4k+1)π/20 for k = 0, 1, . . . , 9.
eiπ9/20 √
is the root closest to i.
(b) 3 − i = 2e−πi/6 . Hence one root is α = (2)1/7 e−iπ/42 . Hence every root
is of the form αω for one of the 7th roots of unity. Hence it is one of the seven
complex numbers ei(12k−1)π/42 for k = 0, 1, . . . , 6.
(c) By standard trigonometry it is enough to show that the three sides of the
triangle formed by the cube roots of z all have the same length. Let z = reiθ .
Then the three cube roots of z are given by
z1 = r1/3 eiθ/3 , z2 = z1 e2πi/3 , z3 = z1 e4πi/3 = z1 e−2πi/3 .
The lengths of the three sides of the triangle formed by z1 , z2 , z3 are |z1 −
z2 |, |z1 − z3 | and |z2 − z3 |. |z2 − z1 | = |z1 (e2πi/3 − 1)| = |(e2πi/3 − 1)|, while
|z1 −z3 | = |z1 e−2πi/3 (e2πi/3 −1)| = |(e2πi/3 −1) and |z3 −z2 | = |z1 e2πi/3 (e2πi/3 −
1)| = |(e2πi/3 − 1)|. Hence all sides of the triangle do have the same length.
2. (a) Just multiply out the RHS to see that we do get x5 − 1. Since ω 5 − 1 = 0,
then by the first part either ω = 1 or ω 4 + ω 3 + ω 2 + ω + 1 = 0. Clearly ω 6= 1
and hence the latter equality holds.
ˉ implies w
ˉ = ww
ˉ 5 = |w|2 w4 = w4 . Hence
(b) Multiplying w5 = 1 by w
4
w+w = w+w
ˉ = 2Re(w) = α. Similarly multiplying w5 = 1 by w
ˉ 2 implies
3
2
2
3
2
2
2
ˉ , and hence w +w = w + w
ˉ = 2Re(w ) = β. By part (a) α+β +1 =
w =w
(w+w4 )+(w2 +w3 )+1 = 0. Also αβ = (w+w4 )(w2 +w3 ) = w3 +w4 +w6 +w7 =
w3 (1 + w + w3 + w4 ) = −w2 .w3 = −w5 = −1. Clearly a quadratic equation with
roots α and β is given by (x − α)(x − β) = x2 − (α + β)x + αβ = x2 + x − 1,
which has roots
√
1
5
±
.
2
2
√
1
Hence cos 2π
5 = α/2 = 4 ( 5 − 1) (since clearly cos 2π/5 > 0 and the other root
of the above quadratic is negative).
3. (a) Expanding the RHS of cos 4θ + i sin 4θ = (cos θ + i sin θ)4 , equating real
parts, using sin2 θ = 1 − cos2 θ eventually yields
cos 4θ = 8 cos4 θ − 8 cos2 θ + 1.
Let θ = π/12. Then cos 4θ = cos π/3 = 1/2. Then x = cos θ satisfies 8x4 −
8x2 +1 = 12 or equivalently 16x4 −16x2 +1 = 0. The other roots of this equation
are cos(θ + kπ/2) for k = 1, 2 or 3.
(b) wn = 1 implies wn−1 = 1/w and w
ˉ = wn−1 . Hence w
ˉ = 1/w as required.
n
1 − w = (ˉ
1 − w)
ˉ n = (1 − 1/w)n =
since wn = 1.
(w − 1)n
= (w − 1)n ,
wn
To show a complex number is real it is equivalent to show that z = zˉ.
2n
n
(1 − w)2n = (1 − w) = ((1 − w) )2 = ((w − 1)n )2 = (w − 1)2n = (1 − w)2n
and hence (1 − w)2n is real.
4. (a) Any complex
the form z = eiθ .
√number satisfying |z| = 1 can be written in √
The condition |z+ 2| = 1 then becomes the condition |(cos θ+ 2)+i sin θ| = 1.
This holds if and only if
√
(cos θ + 2)2 + sin2 θ = 1,
which after some simple algebra we see is equivalent to cos θ = − √12 . Hence the
two complex numbers satisfying both the conditions are e3πi/4 and e5π/4 , which
are easily seen both to be 8th roots of unity.
iθ
(b) Any complex
√number satisfying |z| = 1 can be written in the
√ form z = e .
The condition |z+i 5| = 1 then becomes the condition | cos θ+i( 5+sin θ)| = 1.
This holds if and only if
√
cos2 θ + (sin θ + 5)2 = 1,
√
which after some simple algebra we see is equivalent to sin θ = − 25 . But since
√
5 > 2, there are no solutions of this equation since the minimum value sin
attains is −1.
5. (a) Let si be the some of all i-tuples of roots of the cubic. Then
s1 = a + (a + d) + (a + 2d) = 3(a + d),
s2 = a(a + d) + a(a + 2d) + (a + d)(a + 2d),
and
s3 = a(a + d)(a + 2d).
From the equation satisfied by the roots we have s1 = −6, s2 = k and s3 = 10.
Using the expression for s1 in terms of a andd, s1 = 6 implies a + d = −2. Then
we can rewrite s2 as s2 = −2a + a(a + 2d) − 2(a + d) = −4(a + d) + a(a + d) =
8 + a(a + d) and s3 as s3 = −2a(a + d). Since s3 = 10, we see a(a + d) = −5.
Hence since s2 = 8 + a(a + d) = k, we find k = 3.
Hence we want to solve the cubic x3 + 6x2 + 3x − 10 = 0. By inspection
we see that x = 1 is a root of this equation. Therefore we can factorize it as a
product of x − 1 and a quadratic. Some simple algebra shows that the quadratic
must be x2 + 7x + 10 = (x + 2)(x + 5). Hence the solutions are x = −5, x = −2,
and x = 1, i.e. a = −5 and d = 3.
(b) From the cubic equation satisfied by α, β and γ we have α + β + γ = 0,
αβ+αγ+βγ = −1 and αβγ = 1. We need to find α2 +β 2 +γ 2 , α2 β 2 +α2 γ 2 +β 2 γ 2
and α2 β 2 γ 2 . The latter is easy: α2 β 2 γ 2 = (αβγ)2 = 1. Also
α2 + β 2 + γ 2 = (α + β + γ)2 − 2(αβ + αγ + βγ) = 0 + 2 = 2.
Some elementary algebra shows that
α2 β 2 + α2 γ 2 + β 2 γ 2 = (αβ + αγ + βγ)2 − 2αβγ(α + β + γ) = 12 − 2.1.0 = 1.
Hence the cubic satisfied by α2 , β 2 and γ 2 is x3 − 2x2 + x − 1 = 0.
We saw in lectures that if a0 + a1 x + . . . an xn = 0 has roots λ1 , . . . , λn , then
an + an−1 x + . . . a0 xn = 0 has roots 1/λ1 , . . . , 1/λn . Hence a cubic equation
with roots 1/α, 1/β, 1/γ is −x3 − x2 + 1 = 0.
6. Let p = x + y + z, q = xy + xz + yz and r = xyz. Then elementary algebra
shows 2q = p2 − (x2 + y 2 + z 3 ) = 32 − 3 = 6. Hence q = 3.
Less obviously, we also have
3r = x3 + y 3 + z 3 − p3 + 3pq.
Hence we get r = 1. Therefore any component x, y or z of a solution of these 3
simultaneous equations must satisfy the cubic equation
t3 − pt2 + qt − r = t3 − 3t2 + 3t − 1 = (t − 1)3 = 0.
The only solution (either real or complex) of this equation is obviously t = 1.
Hence x = y = z = 1 (which obviously is a solution) is the only solution of these
simultaneous equations.