Download 2016-Complex-Numbers_Specimen_Exercise

Oxford University Mathematical Institute
Introduction to Complex Numbers, Michaelmas Term 2016:
Specimen Solutions to Exercise Sheet
Q1. Which of the following quadratic equations have real roots, which do not?
3x2 + 2x − 1 = 0;
2x2 − 6x + 9 = 0;
−4x2 + 7x − 9 = 0;
4x2 − 9x + 5 = 0.
A1. Let D be the discriminant of the equation. For the first equation D = 22 − 4 × 3 × −1 = 16.
Since 16 > 0 the equation has real roots. In fact , of course, 3x2 + 2x − 1 = (3x − 1)(x + 1) and so
the roots are −1, 1/3.
For the second equation D = (−6)2 − 4 × 2 × 9 = −36, which is negative, so 2x2 − 6x + 9 = 0
has no real roots. For the third equation D = 49 − 144 < 0 so there are no real roots. And for the
fourth equation D = 1, so it has two real roots: and in fact 4x2 − 9x + 5 = (x − 1)(4x − 5) so the
roots are 1, 5/4.
Q2. Write a careful proof of the theorem that if z, w ∈ C then z + w = z + w and z w = z w.
A2. Write z = a + bi and w = c + di. Then
z + w = (a + c) + (b + d)i = (a + c) − (b + d)i = (a − bi) + (c − di) = z + w
and
z w = (ac − bd) + (ad + bc)i = (ac − bd) − (ad + bc)i;
z w = (a − bi)(c − di) = (ac − bd) − (ad + bc)i ,
and so z w = z w, as the theorem states.
Q3. Put each of the following complex numbers into standard form a + bi:
√
7 − 2i
(1 + 2i)(3 − i); (2 + i)(1 − 2i); (1 + i)4 ; (1 − 3i)3 ;
;
5 + 12i
A3.
i
.
1−i
(1 + 2i)(3 − i) = 5 + 5i ;
(2 + i)(1 − 2i) = 4 − 3i ;
(1 + i)4
√
(1 − 3i)3
7 − 2i
5 + 12i
i
1−i
= ((1 + i)2 )2 = (2i)2 = −4 ;
√
√
√
√
= (1 − 3i)2 (1 − 3i) = (−2 − 2 3i)(1 − 3i) = −8 ;
(7 − 2i)(5 − 12i)
11
94
=
=
−
i;
169
169 169
i(1 + i)
1 1
=
= − + i.
2
2 2
Q4. Find the modulus and argument of each of the following complex numbers:
√
1 + 3i; (2 + i)(3 − i); (1 + i)5 .
A4.
√
√
|1 + 3i| = 1 + 3 = 2 ,
p
√
|(2 + i)(3 − i)| = (4 + 1)(9 + 1) = 5 2 ,
√
√
|(1 + i)5 | = ( 2)5 = 4 2 ,
arg(1 +
√
3i) = tan−1
√
3 = π/3 ;
arg((2 + i)(3 − i)) = arg(7 + i) = tan−1 17 ;
arg(1 + i)5 = 5 arg(1 + i) =
1
5π
4
.
Q5. On separate Argand diagrams sketch each of the following subsets of C:
A := {z |z| < 1};
B := {z Rez = 3};
C := {z − 14 π < arg z < 14 π};
D := {z arg(z − i) = 12 π}; E := {z |z − 3 − 4i| = 5}; F := {z |z − 1| = |z − i|} .
A5.
Figure A: circular disc, not
including boundary circle
Figure B: line “x = 3”
Figure C:
right-angled sector
not including 0 or
boundary rays
Figure D: “vertical” ray
from i, not including point i
Figure E: circle with centre
at 3 + 4i, radius 5
Figure F : line
“x = y”
Q6. For each of the following complex numbers w, what transformation of the Argand diagram
does multiplication by w represent?
i;
(1 + i);
(1 − i);
(3 + 4i).
A6. Multiplication by i is a rotation about 0 anticlockwise through angle
multiplication by (1 + i) is scaling (magnification) by a factor
through angle 14 π about 0;
√
1
2
π;
2 and rotation anticlockwise
multiplication by (1 − i) is scaling (magnification) by a factor
through angle 14 π about 0;
√
2 and rotation clockwise
multiplication by (3 + 4i) is scaling (magnification) by a factor 5 and rotation anticlockwise
through angle tan−1 34 about 0.
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Q7. Use De Moivre’s Theorem to show that if θ ∈ R then
cos 5θ = 16 cos5 θ − 20 cos3 θ + 5 cos θ;
sin 5θ = (16 cos4 θ − 12 cos2 θ + 1) sin θ .
A7. By De Moivre’s Theorem and a small case of the Binomial Theorem,
cos 5θ + i sin 5θ = (cos θ + i sin θ)5
= cos5 θ + 5i cos4 θ sin θ − 10 cos3 θ sin2 θ − 10i cos2 θ sin3 θ
+ 5 cos θ sin4 θ + i sin5 θ .
Focussing on real parts we see that
cos 5θ = cos5 θ − 10 cos3 θ sin2 θ + 5 cos θ sin4 θ
= cos5 θ − 10 cos3 θ (1 − cos2 θ) + 5 cos θ (1 − cos2 θ)2
= 16 cos5 θ − 20 cos3 θ + 5 cos θ ,
as required. And from the imaginary parts we find that
sin 5θ = 5 cos4 θ sin θ − 10 cos2 θ sin3 θ + sin5 θ
= 5 cos4 θ − 10 cos2 θ (1 − cos2 θ) + (1 − cos2 θ)2 sin θ
= (16 cos4 θ − 12 cos2 θ + 1) sin θ ,
also as required.
Q8. Write down the primitive 6th roots of unity and the primitive 8th roots of unity in standard
form a + bi.
2 (which is −ω ), where ω is a primitive
A8. The primitive 6th roots of unity √
are −ω and −ω
√
cube-root of 1. Therefore they are 12 + 23 i and 12 − 23 i.
The primitive 8th roots of unity are the four complex numbers ± √12 ± √12 i. (Note that these all
have absolute value 1 and their arguments are π/4, 3π/4, 5π/4, 7π/4, respectively.)
Q9. Is 3 + 4i a root of unity? What about
3
5
+ 45 i ?
A9. No, 3 + 4i is not a root of unity. Its absolute value is 5, so it hasn’t a hope. And no, 35 + 45 i is
not a root of unity either, even although its absolute value is 1. There are many ways to see this.
Here is one. Consider complex numbers of the form (3 + 4i) + 5(a + bi), where a, b are integers.
Notice that
((3 + 4i) + 5(a + bi)) × ((3 + 4i) + 5(c + di))
= ((3 + 5a)(3 + 5c) − (4 + 5b)(4 + 5d)) + ((3 + 5a)(4 + 5d) + (4 + 5b)(3 + 5c))i
= (3 + 4i) + 5(A + B i) ,
where
A := −2 + 3a + 3c + 5ac − 4b − 4d − 5bd and B := 4 + 4a + 3d + 5ad + 3b + 4c + 5bc .
Thus the product of two numbers of this special form is again of our special form. Therefore
3
4 n
+
i
= 51n ((3 + 4i) + 5(an + bn i)) ,
5
5
where an , bn are integers. Clearly, 5n never divides 3 + 5an if n > 1, and therefore no power of
3
4
3
4
5 + 5 i is 1. That is, 5 + 5 i is not a root of unity.
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Q10. Let φ := cos(2π/5) + i sin(2π/5), a primitive 5th root of 1. Define α := φ + φ4 , β := φ2 + φ3 .
(i) Show that α and β are real numbers.
(ii) Show that α + β = −1 and αβ = −1 (so that α, β are the roots of x2 + x − 1 = 0).
√
(iii) Deduce that cos(2π/5) = 14 ( 5 − 1).
A10. (i)
Since cos(2π − x) = cos x and sin(2π − x) = − sin x we find that
φ = cos(2π/5) − i sin(2π/5) = cos(8π/5) + i sin(8π/5) = φ4 .
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Therefore α = φ + φ = φ4 + φ = α, so α is a real number. A similar calculation shows that β ∈ R.
(ii) Since φ5 − 1 = 0 and φ5 − 1 = (φ − 1)(φ4 + φ3 + φ2 + φ + 1) we have φ4 + φ3 + φ2 + φ + 1 = 0
(since φ − 1 6= 0). Thus α + β = φ + φ4 + φ2 + φ3 = −1. Similarly, αβ = φ3 + φ4 + φ6 + φ7 =
φ3 + φ4 + φ + φ2 = −1, as required.
(iii) Now α, β satisfy the equation (x−α)(x−β) = 0, that is, x2 −(α+β)x+αβ
√ = 0, which we
know to be x2 + x − 1 = 0. Completing the square we see that α and β are 21 (−1 ± 5), only one of
√
√
which is positive. Therefore 2 cos(2π/5) = φ+φ = α = 12 (−1+ 5), that is, cos(2π/5) = 14 ( 5−1).
Q11. Find the square roots of −7 + 24i. Now solve the equation z 2 − (2 + 2i)z + (7 − 22i) = 0.
A11. Let z = reiα and suppose that z 2 = −7 + 24i.√ Then r2 = √
|−7 + 24i| and 2α is one of the
possible values of arg(−7 + 24i). Now |−7 + 24i| = 72 + 242 = 625 = 25 and therefore r = 5.
2t
Also, tan(2α) = −24/7. If we write t := tan α then − 24
7 = tan(2α) = 1−t2 . This leads to the
quadratic equation 24t2 − 24 = 14t or 12t2 − 7t − 12 = 0. Completing the square we find that
t = 4/3 or t = −3/4. Thus z = ±5ei τ where τ := tan−1 43 , that is, z = ±(3 + 4i).
To solve the equation z 2 − (2 + 2i)z + (7 − 22i) = 0 we again complete the square (rather
than citing from memory—imperfect in my case—the standard formula). It may be rewritten
(z − (1 + i))2 − (1 + i)2 + (7 − 22i) = 0, that is, (z − (1 + i))2 = −7 + 24i. We have just found the
two square roots of −7 + 24i: thus z = (1 + i) ± (3 + 4i), that is, z = 4 + 5i or z = −2 − 3i.
Q12. Show that σ 8 = −1 if and only if σ is a primitive 16th root of 1. Find the roots of the
equation x8 = −1, and use them to write x8 + 1 as the product of four quadratic factors with real
coefficients.
A12. Since arg(−1) = π + 2k π for any integer k, from Euler’s formula we know that
−1 = cos(π + 2k π) + i sin(π + 2k π) = e(1+2k)π i . Then σ 8 = −1 if and only if |σ| = 1 and
1+2k
8 arg σ = (1 + 2k)π for some integer k, that is, σ = e 8 π i . The primitive 16th roots of 1, however,
m
are the complex numbers e 16 2π i for integers m that are co-prime with 16 (and, without loss of
generality, < 16). These are precisely the odd integers 1 + 2k < 16, and therefore σ 8 = −1 if and
only if σ is a primitive 16th root of 1; moreover, the roots of the equation x8 = −1 are ψ m where
ψ = eπ i/8 and m is one of 1, 3, 5, 7, 9, 11, 13, 15.
From the Factor Theorem it follows that
7
3 Y
Y
2k+1
) ,
x8 + 1 =
(x − ψ 2k+1 ) =
(x − ψ 2k+1 )(x − ψ
k=0
k=0
m
2k+1
since ψ 16−m = ψ −m = ψ . Now (x − ψ 2k+1 )(x − ψ
) = x2 − 2 cos( 2k+1
8 π)x + 1 and so
x8 +1 = x2 − 2 cos( 18 π)x + 1 x2 − 2 cos( 38 π)x + 1 x2 − 2 cos( 58 π)x + 1 x2 − 2 cos( 78 π)x + 1 ,
a product of four quadratic factors with real coefficients.
ΠMN: Queen’s and Mathematical Institute: October 2016
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