Download HOMEWORK ASSIGNMENT 3 Exercise 1 ([1, Exercise 6.2]). Let u

HOMEWORK ASSIGNMENT 3
ACCELERATED PROOFS AND PROBLEM SOLVING [MATH08071]
Each problem will be marked out of 4 points.
Exercise 1 ([1, Exercise 6.2]). Let u and v be two complex numbers. Prove that
|u + v| 6 |u| + |v|.
Solution. We have
2
|u + v| 6 |u| + |v| ⇐⇒ |u + v|2 6 |u| + |v| ,
since |u + v| and |u| + |v| are positive. On the other hand, we have
|u + v|2 = (u + v)(ū + v̄) = uū + vv̄ + ūv + uv̄ = |u|2 + |v|2 + ūv + uv̄,
and (|u| + |v|)2 = |u|2 + |v|2 + 2|u||v|. Thus, to prove that |u + v| 6 |u| + |v|, it is enough
to prove that
ūv + uv̄ 6 2|u||v|
holds. But ūv + uv̄ = 2Re(uv̄). We know that
Re(uv̄) = |uv̄| cos θ 6 |uv̄|| cos θ| 6 |uv̄|,
where θ is an argument of uv̄. Then
ūv + uv̄ = 2Re(uv̄) 6 2|u||v|,
which exactly what we had to prove to prove that |u + v| 6 |u| + |v|.
√
Exercise 2 ([1, Exercise 6.4]). (a) What is i?
(b) Find all the tenth roots of i. Which one is nearest
to i in the Agrand diagram.
√
7
(c) Find the seven roots of the equation x − 3 + i = 0. Which one of these roots is
closest to the imaginary axis.
√
√
Solution. What is i? We know that i is a complex number z such that z 2 = i. There
are two such complex numbers. Let us find them. Put z = r(cos θ + i sin θ), where r = |z|
and θ ∈ R such that 0 6 θ < 2π. Then
z 2 = r2 cos(2θ) + i sin(2θ) = i = cos(π/2) + sin(π/2)i,
which implies that r = 1 and θ = (π/2 + 2πk)/2 for any integer k by [1, Proposition 6.2].
Since we assume that 0 6 θ < 2π, we see that k is either 0 or 1. Then
cos(π/4) + i sin(π/4) and cos(5π/4) + i sin(5π/4)
are two possible solutions of z 2 = i. These are the same complex numbers as
√
√
√
√
2
2
2
2
+i
and −
−i
,
2
2
2
2
√
√
which we can shortcut as ±( 2/2 + i 2/2).
Now let us find all the tenth roots of i and find the one that is closest to i in the Agrand
diagram. Let z be a tenth roots of i. Then z 10 = i. There are 10 such complex numbers.
This assignment is due on Thursday 15th October 2015.
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Let us find them. Put z = r(cos θ + i sin θ), where r = |z| and θ ∈ R such that 0 6 θ < 2π.
Then
z 10 = r10 cos(10θ) + i sin(10θ) = i = cos(π/2) + sin(π/2)i,
which implies that r = 1 and
π
+ 2πk
π
2πk
π + 4πk
1 + 4k
θ= 2
=
+
=
=
π
10
20
10
20
20
for any integer k by [1, Proposition 6.2]. Since we assume that 0 6 θ < 2π, we see that
k ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Then
!
!
!
!
!
!
1
1
5
5
37
37
cos
π + i sin
π , cos
π + i sin
π , . . . , cos
π + i sin
π
20
20
20
20
20
20
are all possible solutions of z 10 = i (all tenth root of i). Which one of them is is closest
to i in the Agrand diagram? The one that minimize the number |z − i|. This is the same
root that minimize |z − i|2 . But
|z − i|2 = (z − i)(z̄ + i) = z z̄ + zi − z̄i = 2 + 2Re(zi) = 2 − 2 sin θ,
since z z̄ = 1 and Re(zi) = −Im(z) = − sin θ. But we have
)
(
5
9
37
1
π, π, π, . . . , π ,
θ∈
20 20 20
20
so we can find the right θ just by plugging in all of them into 2 − 2 sin θ and choosing the
one that makes 2 − 2 sin θ to be as small as possible. But one can find θ without doing
this.
The function sin(x) is increasing x ∈ (0, π/2), the function sin(x) is decreasing for
x ∈ (π/2, π), we have sin(x) < 0 for x ∈ (π, 2π), and sin(π/2 + x) = sin(π/2 − x).
Moreover, sin(x) = 1 only for x = π/2 if x ∈ [0, 2π]. The graph of 2 − 2 cos(x) is shown
on the picture
.
Thus, to get the θ we want, we should minimize
1 + 4k
π |θ − π/2| = π − ,
20
2
2
which is quite easy. This gives us the answer
9
π,
20
θ=
and the corresponding tenth root of i is cos(9π/20) + i sin(9π/20).
√
Now let us find the seven roots of the equation x7 − 3 + i = 0 and to determine
which√one of these roots is closest to the imaginary axis. Let z be a root of the equation
x7 − 3 + i = 0. Put z = r(cos θ + i sin θ), where r = |z| and θ ∈ R such that 0 6 θ < 2π.
Then
√
z 7 = r7 cos(7θ) + i sin(7θ) = 3 − i = 2 cos(11π/6) + sin(11π/6)i ,
which implies that r =
√
7
2 and
11π
6
θ=
+ 2πk
11π + 12πk
11 + 12k
=
=
π
7
42
42
for any integer k [1, Proposition 6.2]. Since we assumed earlier that 0 6 θ < 2π, we see
that k ∈ {0, 1, 2, 3, 4, 5, 6}. Then
√
7
2 cos
!
11
π +i sin
42
11
π
42
√
7
2 cos
47
π
42
!
√
7
2 cos
71
π
42
!
!!
,
√
7
!
23
π +i sin
42
2 cos
!!
+ i sin
47
π
42
!!
+ i sin
71
π
42
,
,
√
7
√
7
23
π
42
!
2 cos
59
π
42
!
2 cos
83
π
42
!!
,
√
7
2 cos
!
35
π +i sin
42
!!
+ i sin
59
π
42
!!
+ i sin
83
π
42
,
,
√
are all possible solutions of x7 − 3 + i = 0. Which one of these complex numbers is
closest to the imaginary axis. The one that minimizes the number | cos θ|. But we have
(
θ∈
)
11 23 35 47 59 71 83
π, π, π, π, π, π, π ,
42 42 42 42 42 42 42
so we can find the θ with smallest | cos θ| just by plugging in all of them into | cos θ| and
choosing the one that makes | cos θ| to be as small as possible. But one can find θ without
doing this.
The function | cos(x)| is decreasing for x ∈ (0, π/2), the function | cos(x)| is increasing
for x ∈ (π/2, π), the function | cos(x)| is increasing for x ∈ (π/2, π), the function | cos(x)|
is decreasing for x ∈ (π, 3π/2), the function | cos(x)| is increasing for x ∈ (3π/2, 2π), and
we have two types of symmetries: | cos(π/2 + x)| = | cos(π/2 − x)| and | cos(3π/2 + x)| =
| cos(3π/2 − x)|. Moreover, we known that cos(x) = 0 only for x = π/2 and x = 3π/2 if
x ∈ [0, 2π]. The graph of | cos(x)| is shown on the picture
3
35
π
42
!!
,
.
Thus, to find θ we should minimize |θ − π/2|, minimize |θ − 3π/2|, then chose the
smallest between these two values and this gives us our θ. Minimizing
π 12k − 10 π
π 11 + 12k
π− =
π = 12k − 10,
θ − = 42
2 42
2 42
we get k = 1. This gives us the first candidate θ1 = 23π/42. Minimizing
11 + 12k
12k − 52 3π
3π
π
π−
π = 12k − 52,
θ −
=
=
42
2 42
2 42
we get k = 4. This gives us the second candidate θ4 = 59π/42. Since
π
π
2π
3π
<
= θ4 −
θ1 − =
,
2 21
21 2 we see that the final answer here is given by θ1 . Thus, we see that
!
!!
√
23
23
7
2 cos
π + i sin
π
42
42
√
is the solution of x7 − 3 + i = 0 that is closest to the imaginary axis.
Exercise 3
√ ([1, Exercise 6.10]). Prove that there is no complex number z such that
|z| = |z + i 5| = 1.
√
Solution. Let z be any complex number. Applying Exercise 1 to −z and z + i 5, we
see that
√
√
√
√
√
√
|z| + |z + i 5| = | − z| + |z + i 5| > | − z + z + i 5| = |i 5| = 5 > 4 = 2 = 1 + 1,
√
which implies that the equalities |z| = 1 and |z + i 5| = 1 can not
√ be satisfied simultaneously. Thus, there is no complex number z such that |z| = |z + i 5| = 1.
Exercise 4 ([1, Exercise 7.2]). Use the method for solving cubic equations given in [1,
Chapter 7] to find the roots of the equation
x3 − 6x2 + 13x − 12 = 0.
4
Now notice that 3 is one of the roots. Reconcile this with the roots you have found.
Solution. Let us solve this problem other way around. We do not need to notice that 3
is one of the roots. Then we see that
x3 − 6x2 + 13x − 12 = (x − 3)(x2 − 3x + 4)
by dividing x3 − 6x2 + 13x − 12 by x − 3. Solving
x2 − 3x + 4 = 0,
√
√
we get two other roots, which are 3/2 + 7i/2 and 3/2 − 7i/2.
Now let us apply the method for solving cubic equations given in [1, Chapter 7]. Put
y = x − 2. Then x = y + 2 and y must be a root of the equation
(y + 2)3 − 6(y + 2)2 + 13(y + 2) − 12 = 0,
which can be rewritten as y 3 + y − 2 = 0. Then the formula in [1, Chapter 7] gives us
r
r
2√
2√
3
3
y = 1+
21 + 1 −
21,
9
9
where we consider y is a real number here (and all cubes roots are real as well). Then The
other two solutions are given by
!
!
r
r
r
r
2√
2√
2√
2√
3
3
3
3
2
2
1+
21ω + 1 −
21 ω and
1+
21ω + 1 −
21 ω,
9
9
9
9
where ω = cos(2π/3) + i sin(2π/3). Then x is one of these three complex numbers:
r
r
2√
2√
3
3
2+ 1+
21 + 1 −
21,
9
9
!
!
r
r
√
√
2
2
3
3
2+
1+
21 ω +
1−
21 ω 2 ,
9
9
!
!
r
r
√
2√
2
3
3
2+
1+
21 ω 2 +
1−
21 ω,
9
9
√
√
which do not looks like 3, 3/2 + 7i/2, and 3/2 − 7i/2. But this is OK. Keeping in mind
that
√
√
1
3
3
1
ω=− +
i and ω = − −
i,
2
2
2
2
one can show (similar to [1, Example 7.1]) that
r
r
2√
2√
3
3
2+ 1+
21 + 1 −
21 = 3,
9
9
!
!
r
r
√
2√
2√
3
7
3
3
2
2+
1+
21 ω +
1−
21 ω = + i
,
9
9
2
2
!
!
r
r
√
√
2√
2
3
7
3
3
2
2+
1+
21 ω +
1−
21 ω = − i
,
9
9
2
2
but this is tricky to see (it is easy to check this approximately using calculator). Let us
use the method suggested by Tony Gilbert. Namely, arguing as in [1, Example 7.1], we
may guess that
s
√
21
3
√
1±2
=p± q
9
5
for some rational numbers p and q (it is not obvious that such p and q exist). Since we
believe that
r
r
2√
2√
3
3
2+ 1+
21 + 1 −
21 = 3,
9
9
we see that p = 1/2. Now we can find q by taking cube of
s
√
21
1 √
3
1+2
= + q,
9
2
which gives us q = 7/12. Then we can deduce our reconciliation.
Exercise 5 ([1, Exercise 7.6]). Let n be a positive integer.
(a) Factorize x2n+1 − 1 as a product of real linear and quadratic polynomials.
(b) Factorize x2n + x2n−1 + · · · + x2 + x + 1 as a product of real quadratic polynomials.
(c) Let ω = e2πi/(2n+1) Show that
2n+1
X
ω i+j = 0.
i=1,j=1,i<j
Solution. It follows from [1, Proposition 6.3] that
x2n+1 − 1 = (x − ω)(x − ω 2 ) · · · (x − ω 2n+1 )
where ω = e2πi/(2n+1) . Since ω 2n+1 = 1, we see that
ω 2n+1−k ω k = 1
for every k ∈ {1, . . . , n}. This implies that ω 2n+1−k is a complex conjugate of ω k for every
k ∈ {1, . . . , n}. Then
(x−ω 2n+1−k )(x−ω k ) = x2 −(ω 2n+1−k +ω k )x+ω 2n+1−k ω i = x2 −2Re(ω k )x+1 = x2 −2 cos(2πk/(2n+1))x+1)
for every for every k ∈ {1, . . . , n}. Then x2n+1 − 1 can be factorized as
(x−1) x2 −2 cos(2π/(2n+1))x+1) x2 −2 cos(4π/(2n+1))x+1) · · · x2 −2 cos(2nπ/(2n+1))x+1) ,
because ω 2n+1 = 1. Since
x2n+1 − 1 = (x − 1) x2n + x2n−1 + · · · + x2 + x + 1
and we know the how to factorize x2n+1 − 1 as a product of real linear and quadratic
polynomials, we see that x2n + x2n−1 · · · + x2 + x + 1 can be factorized as
x2 −2 cos(2π/(2n+1))x+1) x2 −2 cos(4π/(2n+1))x+1) · · · x2 −2 cos(2nπ/(2n+1))x+1) .
Finally, we have to show that
2n+1
X
ω i+j = 0
i=1,j=1,i<j
to complete the solution. Note that
ω 1 , ω 2 , ω 3 , . . . , ω 2n+1
are all roots of the polynomial x2n+1 − 1 by [1, Proposition 6.3]. Since ω i+j = ω i ω j , we
see that
2n+1
2n+1
X
X
ω i+j =
ωiωj ,
i=1,j=1,i<j
i=1,j=1,i<j
6
where the later sum is just the sum of all products of pairs of roots of the polynomial
x2n+1 − 1. By [1, Proposition 7.1] this sum must be equal to the coefficients in front of
x2n−1 in x2n+1 − 1, which is zero. Thus, the sum is zero.
References
[1] M. Liebeck, A concise introduction to pure mathematics
Third edition (2010), CRC Press
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