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Jeffrey Hellrung Thursday, April 13, 2006 Math 246A, Homework 01
- (a) Prove that 2 has no square root in Q.
(b) Without using trigonometric or exponential functions, and without appealing to the fundamental theorem of algebra, prove that every complex number has a square root. Where do you use completeness of R? (c) Prove that 3 does not have a square root in Q+
2 Q. Does 1+
2 have a square root in Q+
2 Q?
Solution
(a) Suppose, for the sake of contradiction, there exists some r ∈ Q such that r^2 = 2. We can write r = p/q for p, q ∈ Z with p, q relatively prime. We find then that p^2 = 2q^2 , hence p^2 is even, hence p is even, say, p = 2p′. But then 2p′^2 = q^2 , hence q^2 is even, hence q is even, contradicting the choice of p, q being relatively prime. It follows that no such r can exist. (b) Let a + bi ∈ C; then we seek x + yi ∈ C such that
a + bi = (x + yi)^2 = (x^2 − y^2 ) + 2xyi ⇔ a = x^2 − y^2 , b = 2xy;
this system of equations solves to
x^2 =
a +
a^2 + b^2
, y^2 =
−a +
a^2 + b^2
both of which are nonnegative. Though these equations independently give 4 solutions of (x, y), we need to ensure that xy has the same sign as b, giving
(x, y) = ±
a +
a^2 + b^2
, sign b
−a +
a^2 + b^2
We implicitly use the completeness of R by assuming the existence of the square roots. (c) Suppose, for the sake of contradiction, there exists some x ∈ Q +
2 Q such that x^2 = 3. We can write x = r + s
2 for r, s ∈ Q; then 3 = r^2 + 2s^2 + 2rs
2, hence r = 0 or s = 0. In the former case, 2s^2 = 3, and in the latter case, r^2 = 3; neither have solutions in Q. It follows that no such x can exist. From above, we desire
1 +
2 = r^2 + 2s^2 + 2rs
2 ⇔ r^2 + 2s^2 = 1, 2 rs = 1;
thus, (^) ( r^2 − 2 s^2
r^2 + 2s^2
− 8 r^2 s^2 = − 1 ,
which is impossible for r, s ∈ Q, so 1 +
2 has no square root in Q +
2 Q.
- The Gaussian integers are the complex numbers of the form a + bi with integers a and b.
(a) Determine all Gaussian integers whose inverse is again a Gaussian integer (invertible elements). (b) A Gaussian prime number z is a non-invertible element such that for every factorization z = z 1 z 2 into Gaussian integers either z 1 or z 2 is invertible. List all prime numbers of modulus less than 10 in the sector ℜ(z) > 0, −ℜ(z) < ℑ(z) ≤ ℜ(z) (why this sector?). (c) Make a conjecture about which primes in Z are Gaussian primes.
Solution
(a) Given a + bi ∈ Z[i], we desire c + di ∈ Z[i] such that
(a + bi)(c + di) = 1 ⇔ ac − bd = 1, ad + bc = 0;
multiplying the first equation by a and substituting using the second equation gives
c(a^2 + b^2 ) = a,
and since |a^2 + b^2 | ≥ a, we conclude that a^2 + b^2 = 1. It is then easy to verify that 1, − 1 , i, −i are invertible with inverses 1, − 1 , −i, i, respectively. (b) Given z = x + iy, we wish to find a + bi, c + di ∈ Z[i] such that
x + iy = (a + bi)(c + di) ⇔ x = ac − bd, y = ad + bc;
it follows that x^2 + y^2 = (ac − bd)^2 + (ad + bc)^2 =
a^2 + b^2
c^2 + d^2
from which we immediately see that x^2 + y^2 prime implies x + yi prime. Primes with modulus less than 10 in the given sector using this test are
- 1 + i
- 2 + i, 2 − i
- 3 + 2i, 3 − 2 i
- 4 + i, 4 − i
- 5 + 2i, 5 − 2 i
- 6 + i, 6 − i
- 5 + 4i, 5 − 4 i
- 7 + 2i, 7 − 2 i
- 6 + 5i, 6 − 5 i
- 8 + 3i, 8 − 3 i
- 8 + 5i, 8 − 5 i Note that this wouldn’t necessarily account for all the primes in the given sector. Indeed, if p ∈ Z were prime and could not be expressed as a sum of 2 squares, then p + 0i would be a Gaussian prime. It turns out that if p ≡ 1 (mod 4), p can be expressed a sum of 2 squares, whereas if p ≡ 3 (mod 4), p cannot be expressed as a sum of 2 squares. This gives that 3 + 0i and 7 + 0i are also primes. I claim (without proof) that these are all the primes in the given sector with modulus less than
We need only consider the given sector since all other Gaussian primes may be reached by multi- plications by i of the Gaussian primes in the given sector. (c) p ∈ Z prime is a Gaussian prime iff p ≡ 3 (mod 4).
- Let T be a non-zero linear transformation of R^2 given in the natural way by a 2 × 2 matrix.
Prove that T preserves angles, i.e., 〈x, y〉 ‖x‖‖y‖
〈T x, T y〉 ‖T x‖‖T y‖ for all x, y ∈ R^2 , if and only if the matrix T is of the form ( a b −b a
or
a b b −a
Identify a subalgebra of the algebra of all 2 × 2 complex matrices which is isomorphic to the quaternion algebra, and give the isomorphism. (This is a way to establish that indeed the above product is an algebra.) Use the matrix algebra representation to show that every element x in the quaternion algebra has an inverse element y, i.e., yx = xy is the identity element. Solution We can make the identification
(a, b, c, d) ↔
a + bi c + di −c + di a − bi
It is straightforward but tedious to verify that this is an isomorphism. The easiest way seems to be to show that this holds for (1, 0 , 0 , 0),... , (0, 0 , 0 , 1), then by linearity it must hold for all quaternions. We can compute ( a + bi c + di −c + di a − bi
a^2 + b^2 + c^2 + d^2
a − bi −c − di c − di a + bi
to show that (a, b, c, d) has an inverse.
- Prove that for every complex 2 × 2 matrix A there is a complex number λ and a vector v ∈ C^2 such that Av = λv. Do not just quote the corresponding theorem from linear algebra. Solution We note that if there exists a λ ∈ C such that det(λI − A) = 0, then λI − A has a nontrivial null space, i.e., there exists a v ∈ Cn, v 6 = 0 (corresponding to λ) such that (λI − A)v = 0, which is precisely the desired relation. But note that p(λ) = det(λI − A) is a polynomial in λ (of degree n) with complex coefficients depending only on the coefficients of A, hence there indeed does exist a λ such that p(λ) = 0 by the Fundamental Theorem of Algebra.
- (a) Assume that C is a circle in the complex plane C not containing 0. Prove that { 1 /z | z ∈ C} is again a circle. (b) More generally, consider a linear fractional transformation
f (z) = az + b cz + d
Let C be some circle in the complex plane. Prove that the set
{f (z) | z ∈ C}
is again a circle in the plane, provided cz + d 6 = 0 for all z ∈ C. (One possibility is to prove this first for certain special cases of a, b, c, d, and then deduce the general case by writing the map f as a composition of maps of the special types.) (c) What happens if cz + d = 0 for some z ∈ C? (d) For which values of a, b, c, d is the unit circle C = {z | |z| = 1} mapped to itself? Solution
(a) Let C = {z ∈ C | |z − z 0 | = r} for some z 0 ∈ C, r ∈ R. Then for z ∈ C,
1 z
z 0 r^2 − |z 0 |^2
r^2 − |z 0 |^2 + zz 0 z (r^2 − |z 0 |^2 )
r^2 − |z − z 0 |^2 + |z|^2 − zz 0 z (r^2 − |z 0 |^2 )
z(z − z 0 ) z (r^2 − |z 0 |^2 )
has modulus equal to r |r^2 − |z 0 |^2 |
which is constant, hence { 1 /z | z ∈ C} is also a circle.
(b) Clearly, dilations and translations preserve circles, and by (a), inversions do as well. If c = 0, then, f certainly preserves circles (it is a composition of a dilation followed by a translation). If c 6 = 0, we can express f as f (z) =
az + b cz + d
a c
b − ad/c cz + d
hence it is a composition of the maps
f = (z 7 → z + a/c) ◦ (z 7 → b − ad/c) ◦ (z 7 → 1 /z) ◦ (z 7 → z + d) ◦ (z 7 → cz).
(c) Consider first the map in (a) for z 7 → 1 /z. z = 0 somewhere on C iff r = |z 0 |, and in this special case, z ∈ C satisfies the reduced equation
zz 0 + zz 0 = |z|^2.
For z 6 = 0, then, dividing both sides by |z|^2 gives the equation
z 0 /z + z 0 /z = 1,
hence { 1 /z | z ∈ C} is just a line. This generalizes to the more general case of z 7 → (az+b)/(cz+d), since lines are preserved under dilations and translations.
(d) We require that, for |z| = 1,
az + b cz + d
∣ =^
|az + b| |cz + d|
|a|^2 |z|^2 + |b|^2 − 2 ℜ(abz) |c|^2 |z|^2 + |d|^2 − 2 ℜ(cdz)
which reduces to |a|^2 + |b|^2 − |c|^2 − |d|^2 = 2ℜ((ab + cd)z). Since this must be satisfied for all z with |z| = 1, we choose, in particular, z = 1 and z = −1 to get 2 ℜ(ab + cd) = − 2 ℜ(ab + cd), and we conclude that ℜ(ab + cd) = 0, so |a|^2 + |b|^2 = |c|^2 + |d|^2. Now if we had instead chosen z = i and z = −i, then we would conclude that ℑ(ab + cd) = 0 as well, hence we have ab + cd = 0.
