SOLUTIONS OF HW4
MINGFENG ZHAO
March 11, 2011
1. [Problem 3, in Page 55] Find the linear transformation which carries
a. The points z = 0, 1, i into the points w = −1, −i, 0;
b. The points z = −1, 1, i into the points w = 0, 3, ∞.
Proof. a. Since w takes the points z = 0, 1, i into the points w = −1, −i, 0, then [z, 0, 1, i] =
[w, −1, −i, 0], that is
w + 1 −i + 0
z−0 1−i
·
=
·
z−i 1−0
w − 0 −i + 1
That is, we have
z
w + 1 −i
· (1 − i) =
·
.
z−i
w
1−i
So
w+1
−i
·
w
(1 − i)2
w + 1 −i
·
=
w
−2i
w+1
=
2w
Hence we have 2zw = (w + 1)(z − i) = (z − i)w + z − i, which implies that
z
z−i
=
w=
z−i
z+i
b. Since w takes the points z = −1, 1, i into the points w = 0, 3, ∞, then [z, −1, 1, i] = [w, 0, 3, ∞],
that is,
z+1 1−i
w−0
·
=
z−i 1+1
3−0
That is, we have
z+1
w
2
2w
= ·
=
.
z−i
3 1−i
3 − 3i
So (z + 1)(3 − 3i) = 2(z − i)w. Hence
w=
z + 1 3 − 3i
·
.
z−i
2
2. [Problem 5, in Page 55] Which linear transformations w =
itself?
az+b
cz+d
map the plane congruently onto
Proof. Since w = az+b
cz+d : C −→ C, then c = 0, and d 6= 0, that is, w =
since w is a congruent map of the plane, so |a| = |d|.
So all such maps are: w(z) = az + b, where |a| = 1.
1
a
dz
+ db . On the other hand,
2
MINGFENG ZHAO
3. [Problem 6, in Page 55] What linear transformation corresponds to a rotation of the z-plane through
90◦ about the point z = 2?
Proof. Let w be the linear transformation corresponds to a rotation of the z-plane through 90◦ about
the point z = 2. So w has two fixed points 2 and ∞. So we have w = λ(z − 2) + 2 for some λ ∈ C.
Figure 1. w(4) = 2 + 2i
We consider z = 4, if we rotate z = 4 through 90◦ about the point z = 2, then we gt w(4) = 2 + 2i.
So 2λ + 2 = 2 + 2i, that is, λ = i. Therefore, we have w = iz − 2i + 2.
4. [Problem 12, in Page 56] Determine the type of the following linear transformations:
w=
z
,
2z − 1
w=
2z
,
3z − 1
w=
3z − 4
,
z−1
w=
z
.
2−z
Proof. Let T (w) be the trace of w.
z
−iz
a. If w = 2z−1
= −2iz+i
, then T (w) = i − i = 0, so w is elliptic.
b. If w =
c. If w =
d. If w =
2
√
z
2z
2i
i 2
√
− √i2 = √i2 , so w is loxodromic.
3
1 , then T (w) =
√
3z−1 = √
−
2
i 2
i 2
3z−4
z−1 , then T (w) = 3 − 1 = 2, so w is parabolic.
z
√
z
2
=
, then T (w) = √12 + √22 = √32 > 2, so w is hyperbolic.
z
√
2−z
− 2 + √22
5. [Problem 13, in Page 56] Show that every involutory linear transformation is elliptic.
Proof. Let w(z) =
az+b
cz+d ,
and ad − bc = 1, and w(w(z)) = z, w(z) 6≡ z, then
z
=
=
=
a·
c·
az+b
cz+d
az+b
cz+d
+b
+d
a2 z + ab + bcz + bd
acz + bc + cdz + d2
(a2 + bc)z + ab + bd
(ac + cd)z + bc + d2
SOLUTIONS OF HW4
3
So we have ac + cd = ab + bd = 0, a2 + bc = bc + d2 . If c = 0, then a2 = d2 . Since ad − bc = 1, so
ad = 1, that is, a = d = 1 or a = d = −1. Since ab + bd = 0, so b = 0. Hence w(z) = z or w(z) = −z.
Therefore w(z) = z. So w is elliptic. If c 6= 0, then a + d = 0, so w is elliptic.
Therefore, w is elliptic.
6. [Problem 4, in Page 75] Show that the real and imaginary parts of the function ez satisfy both the
Cauchy-Riemann equations and Laplace’s equation.
Proof. Since w(z) = ez = ex+iy = ex cos y + iex sin y = u(x, y) + iv(x, y), so we have
ux (x, y)
= ex cos y
uy (x, y)
= −ex sin y
vx (x, y)
= ex sin y
vy (x, y)
= ex cos y.
Hence ux = vy , and uy = −vx , that is, u, v satisfy the Cauchy-Riemann equations. So ∆u =
uxx + uyy = vxy − vxy = 0, and ∆v = −uxy + uxy = 0, that is, u, v satisfy the Laplace’s equation.
7. [Problem 9, in Page 75] Determine which curves in the w-plane correspond to straight lines in the
z-plane under the mapping w = ez , and which curves in the z-plane correspond to straight lines in
the w-plane under the mapping.
Proof. a. Case I: x = a for some a ∈ R, then
w(x, y)
= ea+iy
= ea (cos y + i sin y)
Figure 2. w(x, y) = ea (cos y + i sin y)
which is a circle whose center is 0, and radius is ea .
Case II: y = kx + b for some k, b ∈ R, then
w(x, y)
= ex+i(kx+b)
4
MINGFENG ZHAO
= ex (cos(kx + b) + i sin(kx + b))
If k = 0, which is a ray.
Figure 3. w(x, y) = ex (cos b + i sin b)
If k 6= 0, we get a spiral curve.
Figure 4. w(x, y) = ex (cos(kx + b) + i sin(kx + b))
b. Conversely, if we want cos θu(x, y) + sin θv(x, y) = c for some θ, c ∈ R, then
cos θex cos y + sin θex sin y = c
c
So we get cos θ cos y + sin θ sin y = ce−x , that is, cos(θ − y) = ce−x . So x = ln cos(θ−y)
.
8. [Problem 10, in Page 75] Let z = z(τ ) 6= 0 be a continuous function of τ ∈ [α, β], and denote the
curve z = z(τ ) by γ. Show that arg z has unique value at every point of γ, once a definite value
for arg z at the point z(α) has been established and the curve is described in a continuous manner,
starting at that point.
SOLUTIONS OF HW4
5
Proof. Let τ (t) = (β − α)t + α, t ∈ [0, 1], then we consider w(t) = z(τ (t)) = γ(t), and w(t) 6= 0 for all
t ∈ [0, 1], we can define
w(t)
v(t) =
= x(t) + iy(t).
|w(t)|
Our problem becomes: If arg z(α) = θ0 , we want to find a unique continuous function θ : [0, 1] −→
∞ such that θ(0) = θ0 , and eiθ(t) = v(t).
Existence: Since v(t) is continuous on [0, 1, then v is uniformly continuous, that is, for any > 0 such
that there exists δ > 0, whenever |t − t0 | < δ, we have |v(t) − v(t0 )| < . Take = 1, then there exists
δ > 0 such that, whenever |t − t0 | < δ, we have |v(t) − v(t0 )| < 1.
If k 6= 0, we get a spiral curve.
Figure 5. |z − z 0 | < 1
Let n ≥ 1 such that n1 < δ. Define t0 = 0, t1 = n1 , · · · , tn = nn = 1. For any 0t ∈ [t0 , t1 ], we can
let θ0 (t) = θ0 + arcsin Im e−iθ0 v(t) , so θ0 (t) is continuous at [t0 , t1 ], and θ0 (t0 ) = θ0 (0) = θ0 .
And for any t ∈ [t1 , t2 ], we define θ1 (t) = θ0 (t1 ) + arcsin Im e−iθ(t1 ) v(t) , so θ1 (t) is continuous at [t1 , t2 ], and θ1 (θ0 (t1 )). By induction, for any t ∈ [tk−1 , tk ], we define θk (t) = θk−1 (tk−1 ) +
arcsin Im e−iθk−1 (tk−1 ) v(t) , so θk (t) is continuous at [tk , tk+1 ]. Therefore, we can define
θ(t) = θk (t),
if t ∈ [tk , tk+1 ]
Therefore, θ(t) is continuous on [0, 1], and θ(0) = θ0 (t0 ) = θ0 .
Uniqueness: If we have two continuous functions f, g : [0, 1] −→ ∞ such that f (0) = g(0) = θ0 , and
eif (t) = eig(t) = v(t). So we get h(t) = f (t) − g(t) is continuous, and h(t) = 2πk(t), k(t) ∈ Z for all
t ∈ [0, 1]. So k(t) has to be constant. On the other hand, we know h(0) = f (0) − g(0) = θ0 − θ0 = 0.
So h(t) = 0, i.e, f (t) ≡ g(t) for all t ∈ [0, 1].
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu