Download QUIZ/MIDTERM #1 SOLUTIONS (1) For each of the following

QUIZ/MIDTERM #1 SOLUTIONS
(1) For each of the following equations, find all complex z which satisfy the equation. Your answer must be in either polar or rectangular form.
(a) z 3 + 7i = 0.
(b) |ez | = e2 . (This is different than the problem you saw on homework!)
(c) 1 + z 2 + z 4 + z 6 = 0.
Solution.
(a) We want to solve z 3 = −7i. Use polar coordinates: −7i = 7e3πi/2 , so if
z = reiθ , then r3 = 7, 3θ = 3π/2 + 2nπ, where n is any integer.√Therefore
θ = π/2, π/2 + 2π/3
= 7π/6,
π/2 +√4π/3 = 11π/6, r = 3 7, which
√
√
3
3
πi/2
give solutions z = 7e
, 7e7πi/6 , 3 7e11π/6 . A fairly common minor
√
error was to let z = − 3 7eπi/6 , say. Strictly speaking, in polar form, the
negative sign should be absorbed into the eiθ term, since r > 0.
(b) If z = x + iy, then ez = ex eiy expresses ez in polar form. Therefore
|ez | = ex (alternately, |ez | = |ex ||eiy | = |ex | = ex , because |eiy | = 1 for
real y, and ex > 0 is always true). Therefore |ez | = e2 if and only if
z = 2 + iy, where y is any real number.
(c) There are a variety of ways to solve this problem. One way is to notice
that the polynomial 1 + z 2 + z 4 + z 6 = (1 + z 4 )(1 + z 2 ) = 0. The solutions
to 1+z 2 = 0 are z = ±i. To solve 1+z 4 = 1, use polar coordinates: if z =
reiθ , then (reiθ )4 = r4 e4iθ = 1 · eπi , so r = 1, θ = π/4, 3π/4, 5π/4, 7π/4.
Therefore z = eπi/4 , e3πi/4 , e5πi/4 , e7πi/4 solve z 4 = 1, and together with
z = ±i give all solutions to this equation. Alternately, one might recall
that we can factor 1 + z 2 + z 4 + z 6 = (1 − z 8 )/(1 − z 2 ). This is equal to 0
if z 8 = 1, z 2 6= 1, which means z can be any 8th root of unity except for
±1. (2) Let x, y be real numbers. Show that cos(x + y) = cos x cos y − sin x sin y and
sin(x + y) = sin x cos y + sin y cos x. (Hint: Think polar!)
Solution. If x, y are real numbers, then eix = cos x + i sin x, eiy = cos y +
i sin y. Then eix · eiy = ei(x+y) = cos(x + y) + i sin(x + y). On the other
hand, we can also express this product as eix · eiy = (cos x + i sin x)(cos y +
i sin y) = (cos x cos y − sin x sin y) + (sin x cos y + cos x sin y)i. Equating real
and imaginary parts of these two expressions proves the two identities in the
problem. (3) Let a, b, c, d ∈ R be real numbers such that ad − bc = 1. Let H be the upper
half-plane {z ∈ C| Im z > 0}. Show that if z ∈ H, then
az + b
∈H
cz + d
1
2
QUIZ/MIDTERM #1 SOLUTIONS
as well. (Hint: Don’t think polar!) This calculation appears in the study of
modular forms, which is a large branch of modern number theory.
Solution. We want to show that the imaginary part of (az + b)/(cz + d) is
greater than 0. To do this, we eliminate complex numbers from the denominator by multiplying the numerator and denominator by the complex conjugate
of the denominator:
az + b
az + b cz + d
(az + b)(cz + d)
=
·
=
,
cz + d
cz + d cz + d
|cz + d|2
where we used various simple properties of complex conjugation in the equation above, together with the fact that c, d are real numbers. Notice that
|cz + d|2 > 0: indeed, we know that |cz + d|2 ≥ 0 is always true, and also that
cz + d 6= 0, since if cz + d = 0, then either z = −d/c or c, d = 0. The former is
impossible since z ∈ H, which means z has positive imaginary part, and the
latter is impossible since ad − bc = 1. (Virtually everyone missed checking
that cz + d 6= 0.)
Therefore to solve the problem it suffices to show that (az + b)(cz + d) has
positive imaginary part. We have (az + b)(cz + d) = ac|z|2 + bd + adz + bcz.
Notice ac|z|2 , bd are both real, so Im (az + b)(cz + d) = Im (adz + bcz). If we
let z = x + yi, then Im (adz + bcz) = ady − bcy = (ad − bc)y = y > 0, as
desired. (4) It is easy to show that a finite union of closed sets in C will also be closed in
C. In this problem you will show that, in general, one cannot say anything
about an infinite union of closed sets.
(a) (15) Give an example of an infinite collection of closed sets in C whose
union is closed.
(b) (15) Give an example of an infinite collection of closed sets in C whose
union is not closed.
Solution. Both parts of this problem have many possible different answers.
We provide a few simple possibilities here. In both parts we use the basic fact
that a set consisting of a single point in C is closed.
(a) The most trivial possible example is to take an infinite collection of closed
sets all equal to the same closed set X. Then their union is X, which
is closed. For a slightly less trivial answer, let X be any closed set, and
then take the collection consisting of single point sets {x}, where x ranges
over points in X. Then the union of these sets is X, which is closed.
(b) In line with the previous solution, perhaps the simplest solution is to let
X be any non-closed set (say, an open disc, although it doesn’t matter),
and take the collection to be the single point sets {x}, x ∈ X. Then the
union of this collection is X, which is not closed.
A fairly common error was to choose, say, closed discs of radius n, as
n → ∞. These sets have union all of C, but C is a closed (as well as
open) set, so is not a correct answer to this problem. Of course, this
example would have worked on the previous part.