Download More Selected Solutions 7A-B Multiplication by a complex number is

More Selected Solutions
7A-B Multiplication by a complex number is a linear transformation of the complex plane. What is the
matrix corresponding to multiplication by w = a + bi? For which values of w is it an isometry?
Multiplication by w = a + bi sends 1 to a + bi and sends i to −b + ai. Written as
vectors in the
a −b
plane, this means (1, 0) 7→ (a, b) and (0, 1) 7→ (−b, a). Thus the matrix is
.
b
a
Multiplication by w is an isometry if the distance between wz1 and wz2 is equal to the distance
between z1 and z2 for any complex numbers z1 and z2 .
|wz1 − wz2 | = |w(z1 − z2 )| = |w| |z1 − z2 |,
so this equals |z1 − z2 | if and only if |w| = 1, or in other words a2 + b2 = 1. Note this is also the
determinant of the matrix representation of w. (And this makes sense: as you know from linear
algebra, a linear transformation preserves area exactly when it has determinant ±1.)
One final observation about this: multiplication by any complex number is the combination of
a rotation and a dilation. It’s an isometry exactly when the dilation factor is 1—in other words,
exactly when it’s a rotation.
5.7.3 Before the discovery of perspective, artists sometimes attempted to draw a tiled floor by making the
width of each row of tiles a constant fraction e of the one before. Show that this is not correct by
computing the cross ratio of four points separated by the distances 1, e, and e2 .
(Note that e is an arbitrary constant here, not the base of natural log. The artists were hoping
that some value e < 1 would work, making the spacing in between rows get smaller and smaller, as
it clearly should in a perspective image.)
This just amounts to showing that points separated by those distances, like p = 0, q = 1,
r = 1 + e, and s = 1 + e + e2 , are not the projective images of equally spaced points. Projection
preserves the cross-ratio, and the cross ratio of ANY four equally spaced points is 4/3, for instance
6·6
[2, 5; 8, 11] = 3·9
= 43 .
We compute
[0, 1; 1 + e, 1 + e + e2 ] =
1 + 2e + e2
(1 + e)(e + e2 )
=
.
2
e(1 + e + e )
1 + e + e2
Setting this equal to 4/3 would give 3√+ 6e + 3e2 = 4 + 4e + 4e2 , or e2 − 2e + 1 = 0. By the
quadratic formula, the solutions are (2 ± 4 − 4)/2 = 1. So the only solution is when e = 1, which
produces equally spaced rows (in our case, the points would be 0, 1, 2, 3), and this does not solve the
problem of how to put tiles into perspective.
7.5.3 Find a rotation and reflection of R2 that do not commute.
Any ones that you pick will work, as long as the rotation does not fix the axis of the reflection!
For instance, let r = rπ/2 be the rotation about the origin by π/2, and let f be the usual reflection
in the x-axis. Then r ◦ f sends (1, 0) to (0, 1), while f ◦ r sends (1, 0) to (0, −1).
7.7.1 Verify directly that the 16 points ± 21 ± 21 i ± 12 j ± 12 k are all at distance 1 from the origin in R4 .
√
Using the distance formula in R4 , the distanceqof (a, b, c, d) from the origin is a2 + b2 + c + 2 + d2 .
So in this case, the distance from the origin is 14 + 41 + 14 + 14 = 1 for all of the points.
8.1.1 Show that the transformations z 7→ z + l and z 7→ kz (for k > 0) map the upper half plane onto itself
and that they map “lines” to “lines.”
(Note that in this problem, l and k are real numbers.)
A complex number z = x + yi is in the upper half plane H if and only if y > 0. So let’s suppose
we start with such a z. Adding a real number l sends x + yi to (x + l) + yi, keeping the imaginary
part greater than 0. Multiplying by a positive real number k sends x + yi to (kx) + (ky)i, keeping
the imaginary part greater than 0. So both maps preserve H.
1
2
A “line” is an orthocircle: either a vertical line or a semicircle of arbitrary radius centered on the
real axis. The map z 7→ z + l simply shifts everything l units to the right (since the positive real
direction is right). This sends lines to lines and semicircles centered on the real axis to semicircles
whose centers are l units away. The map z 7→ kz is a dilation that simply scales everything up by
a factor of k. The vertical line x = x0 is mapped to the vertical line x = kx0 , and the semicircle
centered at x0 with endpoints at x0 − r and x0 + r is mapped to the semicircle centered at kx0 with
endpoints at kx0 − kr and kx0 + kr. So “lines” are preserved.
8.2.3 Given that z1 + z2 = z1 + z2 , z1 z2 = z1 z2 , and 1/z = 1/z, deduce that for any a, b, c, d ∈ R and
az+b
z ∈ C, the complex conjugate of az+b
cz+d is cz+d .
We’ll break this down completely, using the facts given above. First note that the complex
conjugate of a real number is itself: a = a. Next, az + b = az + b = az + b = az + b.
Then
az + b
1
1
= az + b ·
= az + b ·
= (az + b) 1/cz + d =
cz + d
cz + d
cz + d
az + b
1
=
,
= (az + b) ·
cz + d
cz + d
as desired.
8.6.4 Show that the non-Euclidean distance between the lines x = 0 and x = 1 tends to zero as y tends to
∞.
The points on the line x = 0 are of the form ti and the points on the line x = 1 are of the form
ti + 1. So it is enough to show that dist(ti, ti + 1) −→ 0 as t → ∞. But remember that every dilation
z 7→ kz is an isometry of the hyperbolic plane, so for any two points z, w ∈ H and any k > 0, we
have dist(z, w) = dist(kz, kw).
But then dist(ti, ti + 1) = dist 1t (ti), 1t (ti + 1) = dist(i, i + 1/t). But this is just the distance
between i and i + 1/t, and these points get closer and closer together as t → ∞.