Download WRITTEN HOMEWORK #1 SOLUTIONS

WRITTEN HOMEWORK #1 SOLUTIONS
(1) For each of the following equations, give a geometric description of the set of
complex numbers (ie, describe how this set looks in the complex plane) which
solve that equation. The numbers z1 , z2 , . . . refer to arbitrary, distinct, fixed
complex numbers.
(a) |z − z1 | = |z − z2 |. Your description should involve z1 , z2 .
(b) zc = z, where c > 0 is a fixed real number.
(c) |z − z1 | + |z − z2 | = c, where c > 0 is a fixed real number. (Hint: the
shape depends on whether c < |z1 − z2 |, c = |z1 − z2 |, or c > |z1 − z2 |.)
(d) |z| = Im z + 1.
Solution.
(a) The equation |z − z1 | = |z − z2 | exactly expresses the fact that z is the
same distance to z1 as it is to z2 . From geometry, this set of points is just
the perpendicular bisector of the segment connecting z1 and z2 (the line
which is perpendicular to this segment and passes through the midpoint
of this segment).
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(b) The equation
√ is equivalent to c = zz = |z| , so its solutions form a circle
of radius c centered at 0.
(c) First, notice that the triangle inequality can also be expressed in the
form |z1 | + |z2 | ≥ |z1 − z2 |. (Take the form of the triangle inequality we
used, and replace z2 with −z2 .) Then applying this form of the triangle
inequality gives |z − z1 | + |z − z2 | ≥ |z1 − z2 |.
Therefore, if c < |z1 − z2 |, it is impossible for |z − z1 | + |z − z2 | = c
to have any solutions, so the set in question is the empty set. Suppose
c = |z1 − z2 |. Then z must lie on the line segment connecting z1 and
z2 . Indeed, if z is not on the line connecting z1 , z2 , then z, z1 , z2 are not
collinear, and since z, z1 , z2 form the vertices of an actual triangle, the
triangle inequality yields |z − z1 | + |z − z2 | > |z1 − z2 |. Also, if z is on the
line connecting z1 , z2 but not on the segment between them, then one of
|z − z1 |, |z − z2 | is greater than z1 − z2 , so |z − z1 | + |z − z2 | > c would
be impossible.
Finally, if c > |z1 − z2 |, then the set of points in question form an ellipse.
This is actually one of the possible definitions of an ellipse: as the set of
points whose sum of distances from two fixed points is constant. Also,
z1 , z2 are the foci of this ellipse.
p
(d) Let z = x + yi. Then the equation in question reads x2 + y 2 = y + 1,
2
or, squaring both sides, x2 + y 2 = (y +
p1) , together with y ≥ −1. (Notice
that y < −1 is impossible because x2 + y 2 ≥ 0 is always true). But
x2 + y 2 = (y + 1)2 reduces to x2 = 2y + 1, so the points form part of a
parabola (namely, the parabola x2 = 2y + 1 with y ≥ −1).
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WRITTEN HOMEWORK #1 SOLUTIONS
(2) For each of the following equations in z, find all complex solutions. You may
leave your answers in either rectangular or polar form.
(a) z 2 + iz − 2 = 0.
(b) z 4 = −1.
√
(c) z 3 = − 3 − i.
(d) ez = e2 . (Hint: the answer is not just z = 2.)
Solution.
(a) Use the quadratic formula. (Why can you use the quadratic formula?
The derivation of the quadratic formula holds true even if coefficients are
complex rather than real. The only potential sticking point is if the term
discriminant b2 − 4ac is not real, but fortunately this does not occur in
this problem.) The quadratic equation yields
z=
−i ±
p
√
−1 − 4(1)(−2)
7 i
=±
− .
2
2
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(b) Use the polar form of complex numbers. If z = reiθ , then z 4 = r4 e4iθ =
−1 = 1 · eiπ . If we take the absolute value of both sides, we get |r4 e4iθ | =
|r4 | = r4 = | − 1| = 1. Since r > 0, the equation r4 = 1 has the single
solution r = 1. Also, e4iθ = eiπ if and only if 4θ = π + 2nπ, for some
integer n (since eit is periodic with fundamental period 2π). In other
words, θ = π/4 + nπ/2 give all solutions to e4iθ = −1.
This yields the numbers eiπ/4 , e3iπ/4 , e5iπ/4 , e7iπ/4 = ± √12 ± √12 i. Notice
that ei9π/4 = eiπ/4 , etc., so there really are only four solutions to z 4 = −1.
(c) Use the√polar form of complex numbers again. First, one quickly checks
that − 3 − i = 2e7πi/6 , so if z = reiθ , then√we must have r3 = 2, 3θ =
7π/6+2πn for some integer
n. Therefore
r√
= 3 2 and θ = 7π/18, 19π/18, 31π/18,
√
√
so solutions are z = 3 2e7πi/18 , 3 2e19πi/18 , 3 2e31πi/18 .
(d) Let z = x+yi. If ez = e2 , then ex+yi = ex eyi = e2 . Taking absolute values,
and using the fact that x, y ∈ R, we see that |ex eyi | = |ex | = ex = e2 .
Since x is a real number, this only has solution x = 2. Plugging in x = 2
into ex+yi = e2 , we get eyi = 1, but this only has solutions y = 2nπ for
n ∈ Z, so the solutions of ez = e2 are z = 2 + 2πni, n ∈ Z.
(3) (a) Recall that an nth root of unity is any solution to z n = 1. If ζn is an nth
root of unity not equal to 1, show that 1 + ζn + ζn2 + . . . + ζnn−1 = 0.
(b) Let P be a regular n-gon inscribed in the unit circle. Fix one of the vertices, and consider the n−1 diagonals obtained by connecting that vertex
to the remaining n − 1 vertices. Show that the product of the lengths
of these diagonals is equal to n. This is a rather remarkable geometric
fact which is perhaps most easily proved using complex numbers! (Hint:
there is a way to use the previous part of this problem...)
Solution.
(a) One checks from polynomial algebra that z n − 1 = (z − 1)(z n−1 + z n−2 +
. . . + z + 1). If we plug in ζn , we get ζnn − 1 = (ζn − 1)(ζnn−1 + . . . + ζn + 1).
However, we also know that ζnn = 1, and also that ζn 6= 1, so we get
WRITTEN HOMEWORK #1 SOLUTIONS
3
0 = (ζn − 1)(ζnn−1 + . . . + ζn + 1),
and since we know that ζn − 1 6= 0, we can divide both sides by ζn − 1 to
obtain 1 + ζn + . . . + ζnn−1 = 0.
(b) Let ζn = e2πi/n . After a suitable rotation, we can assume that the vertices
of our polygon are given by the points 1, e2πi/n , e4πi/n , . . . , e2πi(n−1)/n =
1, ζn , . . . , ζnn−1 in the complex plane. Since rotations do not alter lengths
of line segments, we can perform this rotation without affecting the value
of the product of the lengths of the diagonals in the question. Also, notice
that ζnn = 1, and furthermore, every ζnk is also an nth root of unity.
The previous part of this problem guarantees that ζnk satisfies 1 + z + z 2 +
. . . + z n−1 = 0, and the number of such solutions, namely n − 1, is equal
to the degree of this polynomial. Therefore we can factor the polynomial
on the left hand side as c(z − ζn )(z − ζn2 ) . . . (z − ζnn−1 ). By comparing
leading coefficients we see that c = 1. So in summary, we have shown
that
1 + z + z 2 + . . . + z n−1 = (z − ζn )(z − ζn2 ) . . . (z − ζnn−1 ).
Plug in z = 1 into this equality of polynomials to get
1 + 1 + . . . + 1 = n = (1 − ζn )(1 − ζn2 ) . . . (1 − ζnn−1 ).
Notice that |1 − ζnk | is the distance from the vertex 1 to the vertex ζnk .
We can let 1 be the vertex we fix in the original problem. Then the
product of the lengths of the n − 1 diagonals containing this vertex is
just |1 − ζn ||1 − ζn2 | . . . |1 − ζnn−1 |. Going back to the equation above, take
absolute values of both sides to obtain
n = |1 − ζn ||1 − ζn2 | . . . |1 − ζnn−1 |,
as desired.
(4) Let z1 , z2 , z3 be distinct complex numbers. Show that z1 , z2 , z3 are the vertices
of an equilateral triangle in the complex plane if and only if z12 + z22 + z32 =
z1 z2 + z1 z3 + z2 z3 . (Hint: one approach is to solve the problem in the special
case where z1 = 0, and then use this to solve the general case.)
Solution. First we solve the problem in the special case where z1 = 0.
Suppose z2 , z3 , 0 form the vertices of an equilateral triangle. Let r = |z2 | =
|z3 |. Then z2 = reiθ2 , z3 = reiθ3 , and θ3 − θ2 = ±π/3. Therefore
z3
reiθ3
= iθ2 = ei(θ3 −θ2 ) = e±iπ/3 .
z2
re
(Notice division by z2 is fine since we assumed z1 = 0, z2 , z3 were distinct.)
However, one immediately checks that e±iπ/3 are the roots of the polynomial
z 2 − z + 1. Therefore,
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WRITTEN HOMEWORK #1 SOLUTIONS
2
z3
z3
− + 1 = 0.
z2
z2
2
Multiplying this expression by z2 gives z32 − z3 z2 + z22 = 0, which is exactly
the same thing as the equation z12 + z22 + z32 = z1 z2 + z1 z3 + z2 z3 when z1 = 0.
Conversely, if z1 = 0, and z2 , z3 satisfy z22 + z32 = z2 z3 , then reversing all
the calculations above shows that z3 /z2 = e±iπ/3 , which means that |z3 | = |z2 |
and the angle between z3 , z2 is π/3. Then Euclidean geometry tells us that
0, z2 , z3 must form the vertices of an equilateral triangle. So we have solved
the original problem in the special case where z1 = 0.
For the general case, if z1 , z2 , z3 are the vertices of an equilateral triangle,
then so are z10 = z1 − z1 = 0, z20 = z2 − z1 , z30 = z3 − z1 . Apply what we have
already proven to obtain z202 + z302 = z20 z30 , or
(z2 − z1 )2 + (z3 − z1 )2 = (z2 − z1 )(z3 − z1 ).
If we expand all the products out, and then collect terms, we obtain z12 + z22 +
z32 = z1 z2 + z1 z3 + z2 z3 , as desired.
For the other direction, if z12 + z22 + z32 = z1 z2 + z1 z3 + z2 z3 is true, then of
course we can reverse the algebra we did above to see that (z2 − z1 )2 + (z3 −
z1 )2 = (z2 − z1 )(z3 − z1 ) is true. But then based on the special case we proved,
this shows that 0, z2 − z1 , z3 − z1 form the vertices of an equilateral triangle.
We can then translate by z1 to see that z1 , z2 , z3 also form the vertices of
an equilateral triangle, since translation does not affect whether a triangle is
equilateral or not.