SOLUTIONS OF HW1
MINGFENG ZHAO
February, 12, 2011
1. [Page 19,4] Show that the product x1 x2 in Section 1.7 satisfies the axioms IV. 1-3.
Proof. For any x1 = ξ1 e + η1 i, and x2 = ξ2 e + η2 i, define
x1 x2 = (ξ1 ξ2 − η1 η2 )e + (ξ1 η2 + η1 ξ2 )i.
Hence, for any x1 = ξ1 e + η1 i, and x2 = ξ2 e + η2 i, and x3 = ξ3 e + η3 i, we have
(i) Commutative:
From the definition, we have x1 x2 = (ξ1 ξ2 − η1 η2 )e + (ξ1 η2 + η1 ξ2 )i, and x2 x1 = (ξ2 ξ1 − η2 η1 )e +
(ξ2 η1 + η2 ξ1 )i. Because R is commutative, then
x2 x1
(1)
=
(ξ2 ξ1 − η2 η1 )e + (ξ2 η1 + η2 ξ1 )i
=
(ξ1 ξ2 − η1 η2 )e + (ξ1 η2 + η1 ξ2 )i
= x1 x2
So x1 x2 = x2 x1 .
(ii) Bilinear:
From the definition, we have
x1 (x2 + x3 )
=
(ξ1 e + η1 i)((ξ2 e + η2 i) + (ξ3 e + η3 i))
=
(ξ1 e + η1 i)((ξ2 + ξ3 )e + (η2 + η3 )i)
=
[ξ1 (ξ2 + ξ3 ) − η1 (η2 + η3 )]e + [ξ1 (η2 + η3 ) + η1 (ξ2 + ξ3 )]i
1
2
MINGFENG ZHAO
(2)
=
[ξ1 ξ2 + ξ1 ξ3 − η1 η2 − η1 η3 ]e + [ξ1 η2 + ξ1 η3 + η1 ξ2 + η1 ξ3 ]i
=
(ξ1 ξ2 − η1 η2 )e + (ξ1 ξ3 − η1 η3 )e + (ξ1 η2 + η1 ξ2 )i + (ξ1 η3 + η1 ξ3 )i
= x1 x2 + x1 x3
and
(x1 + x2 )x3
= x3 (x1 + x2 )
By (1)
= x3 x1 + x3 x2
By (2)
= x1 x3 + x2 x3
By (1)
x1 = ξ1 e + η1 i, and x2 = ξ2 e + η2 i, and x3 = ξ3 e + η3 i.
And for any λ ∈ R, we have
(λx1 )x2
=
[λ(ξ1 e + η1 i)](ξ2 e + η2 i)
=
(λξ1 e + λη1 i)(ξ2 e + η2 i)
=
(λξ1 ξ2 − λη1 η2 )e + (λξ1 η2 + λη1 ξ2 )i
= λ[(ξ1 ξ2 − η1 η2 )e + (ξ1 η2 + η1 ξ2 )i]
= λ(x1 x2 ).
(iii) Associative:
From the definition, we have
x1 (x2 x3 )
=
(ξ1 e + η1 i)[(ξ2 e + η2 i)(ξ3 e + η3 i)]
=
(ξ1 e + η1 i)[(ξ2 ξ3 − η2 η3 )e + (ξ2 η3 + η2 ξ3 )i]
=
[ξ1 (ξ2 ξ3 − η2 η3 ) − η1 (ξ2 η3 + η2 ξ3 )]e + [ξ1 (ξ2 η3 + η2 ξ3 ) + η1 (ξ2 ξ3 − η2 η3 )]i
=
[ξ1 ξ2 ξ3 − ξ1 η2 η3 − η1 ξ2 η3 − η1 η2 ξ3 ]e + [ξ1 ξ2 η3 + ξ1 η2 ξ3 + η1 ξ2 ξ3 − η1 η2 η3 ]i
SOLUTIONS OF HW1
3
And
(x1 x2 )x3
=
[(ξ1 e + η1 i)(ξ2 e + η2 i)](ξ3 e + η3 i)
=
[(ξ1 ξ2 − η1 η2 )e + (ξ1 η2 + η1 ξ2 )i](ξ3 e + η3 i)
=
[(ξ1 ξ2 − η1 η2 )ξ3 − (ξ1 η2 + η1 ξ2 )η3 ]e + [(ξ1 ξ2 − η1 η2 )η3 + (ξ1 η2 + η1 ξ2 )ξ3 ]i
=
[ξ1 ξ2 ξ3 − η1 η2 ξ3 − ξ1 η2 η3 − η1 ξ2 η3 ]e + [ξ1 ξ2 η3 − η1 η2 η3 + ξ1 η2 ξ3 + η1 ξ2 ξ3 ]i
=
[ξ1 ξ2 ξ3 − ξ1 η2 η3 − η1 ξ2 η3 − η1 η2 ξ3 ]e + [ξ1 ξ2 η3 + ξ1 η2 ξ3 + η1 ξ2 ξ3 − η1 η2 η3 ]i
= x1 (x2 x3 )
2. [Page 19, 5] What is the geometric interpretation of the multiplication of complex numbers?
Proof. Let z1 = r1 cos θ1 + ir1 sin θ1 , and z2 = r2 cos θ2 + ir2 sin θ2 , then
z1 z2
=
(r1 cos θ1 + ir1 sin θ1 ) (r2 cos θ2 + ir2 sin θ2 )
= r1 r2 cos θ1 cos θ2 − r1 r2 sin θ1 sin θ2 + i(r1 r2 cos θ1 sin θ2 + r1 r2 sin θ1 cos θ2 )
= r1 r2 (cos θ1 cos θ2 − sin θ1 sin θ2 ) + i[(r1 r2 )(cos θ1 sin θ2 + sin θ1 cos θ2 )]
=
r1 r2 cos(θ1 + θ2 ) + ir1 r2 sin(θ1 + θ2 ).
So the geometric interpretation of the multiplication of complex numbers is that for the vector
z1 = (r1 cos θ1 , r1 sin θ1 ), multiple z1 by z2 = (r2 cos θ2 , r2 sin θ2 ), which means the length of vector
should shrink or extend r2 -times, and then rotated the vector z1 with the angle θ2 counterclockwise.
3. [Page 19, 10] Prove that
a−b 1 − ab = 1,
4
MINGFENG ZHAO
if a and b are complex numbers with |a| = 1 or |b| = 1, and ab 6= 1.
Proof. If |a| = 1, then
a−b 1 − ab a−b 1
·
= 1 − ab 1
a−b 1
·
= 1 − ab |a|
a−b = (1 − ab)a a−b = a − aab a−b = a − |a|2 b a − b
= a − b
=
1.
If |b| = 1, so |b| = 1. Then
a−b 1 − ab a−b 1
·
= 1 − ab 1
a−b 1
·
= 1 − ab |b|
a−b = (1 − ab)b a−b = b − abb a−b = b − a|b|1 a − b
= b − a
=
|a − b|
|b − a|
=
|a − b|
|b − a|
=
1.
SOLUTIONS OF HW1
5
4. [Page 20, 11] Prove that
a−b 1 − ab < 1,
if |a| < 1 and |b| < 1.
Proof. Since |a| < 1, and |b| <, then |1 − ab| ≥ 1 − |a||b| > 0. We want to show
a−b 1 − ab < 1,
which is equivalent to show
a − b 2
1 − ab < 1.
So we have
a−b 1 − ab < 1
⇐⇒
a − b 2
1 − ab < 1
⇐⇒
|a − b|2 < |1 − ab|
⇐⇒
(a − b)(a − b) < (1 − ab)(1 − ab)
⇐⇒
(a − b)(a − b) < (1 − ab)(1 − ab)
⇐⇒
aa − ab − ba + bb < 1 − ab − ab + (ab)(ab)
⇐⇒
|a|2 + |b|2 < 1 + |a|2 |b|2
⇐⇒
1 − |a|2 − |b|2 + |a|2 |b|2 > 0
⇐⇒
(1 − |a|2 )(1 − |b|2 ) > 0
Since |a| < 1, |b| < 1, then (1 − |a|2 )(1 − |b|2 ) > 0. Therefore,
a−b 1 − ab < 1.
6
MINGFENG ZHAO
5. [Page 21, 27] Prove that if a real function u(x, y) of two real variables possesses partial derivatives
which are continuous at a point, then the function is differentiable at this point.
Proof. Assume the function u(x, y) possesses partial derivatives which are continuous at point (x0 , y0 ),
we want to show u(x, y) is differentiable at (x0 , y0 ), that is there exists real numbers A, B such that
u(x, y) − u(x0 , y0 ) − A(x − x0 ) − B(y − y0 )
p
.
(x,y)→(x0 ,y0 )
(x − x0 )2 + (y − y0 )2
lim
In fact, we take A = ux (x0 , y0 ), B = uy (x0 , y0 ). From the condition, we know there exists a small
neighborhood U of (x0 , y0 ) such that ux (x, y) and uy (x, y) exists in U .
Consider (x, y) is in this small neighborhood U of (x0 , y0 ). By mean value theorem, we know that
u(x, y) − u(x0 , y) = ux (x0 + θ1 (x, y)(x − x0 ), y),
0 < θ1 (x) < 1
u(x0 , y) − u(x0 , y0 ) = uy (x0 , y0 + θ2 (y)(y − y0 )),
0 < θ2 (x) < 1
and
Since ux (x, y) and uy (x, y) are continuous at (x0 , y0 ), since the
then
ux (x0 + θ1 (x, y)(x − x0 ), y) = ux (x0 , y0 ) + α,
uy (x0 , y0 + θ2 (y)(y − y0 )) = uy (x0 , y0 ) + β
where α → 0 uniformly with respect to 0 < θ1 (x, y) < 1, as x → x0 , and β → 0 uniformly with respect
to 0 < θ2 (y) < 1, as y → y0 . So
u(x, y) − u(x0 , y0 )
= u(x, y) − u(x0 , y) + u(x0 , y) − u(x0 , y0 )
= ux (x0 + θ1 (x − x0 ), y)(x − x0 ) + uy (x0 , y0 + θ2 (y − y0 ))(y − y0 )
= ux (x0 , y0 )(x − x0 ) + α(x − x0 ) + uy (x0 , y0 )(y − y0 ) + β(y − y0 )
SOLUTIONS OF HW1
7
Therefore, we get
u(x, y) − u(x0 , y0 ) − ux (x0 , y0 )(x − x0 ) − ux (x0 , y0 )(y − y0 )
p
(x,y)→(x0 ,y0 )
(x − x0 )2 + (y − y0 )2
lim
=
α(x − x0 ) + β(y − y0 )
p
(x,y)→(x0 ,y0 )
(x − x0 )2 + (y − y0 )2
lim
On the other hand, we have
α(x − x ) + β(y − y ) 0
0 p
(x − x0 )2 + (y − y0 )2 ≤
α(x − x0 )
β(y − y0 )
+ p
p
(x − x0 )2 + (y − y0 )2 (x − x0 )2 + (y − y0 )2 =
|α||x − x0 |
|β||y − y0 |
p
+p
2
2
(x − x0 ) + (y − y0 )
(x − x0 )2 + (y − y0 )2
≤
|α||x − x0 |
|β||y − y0 |
p
+p
(x − x0 )2
(y − y0 )2
≤
|α| + |β|
→ 0
as x → x0 , and y → y0 .
Therefore, we have
u(x, y) − u(x0 , y0 ) − ux (x0 , y0 )(x − x0 ) − ux (x0 , y0 )(y − y0 )
p
(x,y)→(x0 ,y0 )
(x − x0 )2 + (y − y0 )2
lim
=
lim
(x,y)→(x0 ,y0 )
=
α(x − x0 ) + β(y − y0 )
p
(x − x0 )2 + (y − y0 )2
0.
6. [Page 21, 28] Show by means of the example u = x2 y/(x2 + y 2 ) that the continuity of a function
and the existence of partial derivatives is not sufficient for the differentiability of the function.
Proof. Let
u(x, y) =





x2 y
x2 +y 2



 0
if (x, y) 6= (0.0)
if (x, y) = (0, 0)
8
MINGFENG ZHAO
So u(x, y) is infinitely differentiable if (x, y) 6= (0, 0), and for any (x, y) 6= (0, 0), we have
ux (x, y)
=
2xy(x2 + y 2 ) − x2 y · 2x
(x2 + y 2 )2
=
2x3 y + 2xy 3 − 2x3 y
(x2 + y 2 )2
=
uy (x, y)
2xy 3
+ y 2 )2
(x2
=
x2 (x2 + y 2 ) − x2 y · 2y
(x2 + y 2 )2
=
x4 + x2 y 2 − 2x2 y 2
(x2 + y 2 )2
=
x4 − x2 y 2
(x2 + y 2 )2
Now we study the differentiability at (0, 0).
2 y For any (x, y) 6= (0, 0), then x2x+y
2 =
x2
x2 +y 2
· |y| ≤ |y| → 0 as (x, y) → (0, 0). So u(x, y) is
continuous at (0, 0).
For ux (x, y), we know
lim ux (x, y)
x=y→0
=
2x4
x=y→0 (2x2 )2
=
2x4
x=y→0 4x4
=
lim
ux (x, y)
=
=
So ux (x, y) is not continuous at (0, 0).
x=y→0 (x2
=
=
x=0,y→0
2xy 3
+ y 2 )2
lim
lim
lim
lim
x=y→0
1
2
1
2
lim
x=0,y→0
0
0
(y 2 )2
SOLUTIONS OF HW1
9
For uy (x, y), we know
lim
y=0,x→0
uy (x, y)
=
x4 − x2 y 2
y=0,x→0 (x2 + y 2 )2
=
x4 − 0
y=0,x→0 (x2 )2
=
x4
y=0,x→0 x4
=
=
lim
x=0,y→0
ux (x, y)
=
lim
lim
lim
lim
1
lim
0
(y 2 )2
y=0,x→0
1
x=0,y→0
=
0
lim
u(x, 0) − u(0, 0)
x
lim
0−0
x
So uy (x, y) is not continuous at (0, 0).
For ux (0, 0), we know
ux (0, 0)
=
=
=
x→0
x→0
0
For uy (0, 0), we know
uy (0, 0)
=
=
=
lim
u(0, y) − u(0, 0)
y
lim
0−0
y
y→0
y→0
0
Now consider the differentiability of u at (0, 0), since ux (0, 0) = uy (0, 0) = 0, then if u is differentiable at (0, 0), then
lim
x,y→0
u(x, y) − u(0, 0) − ux (0, 0)x − uy (0, 0)y
p
= 0.
x2 + y 2
10
MINGFENG ZHAO
But
lim
x=y→0
u(x, y) − u(0, 0) − ux (0, 0)x − uy (0, 0)y
p
x2 + y 2
=
lim
x=y→0+
u(x, x) − u(0, 0) − ux (0, 0)x − uy (0, 0)x
√
x2 + x2
=
u(x, x)
√
x=y→0+
2x2
=
x2 x
x2 +x2
lim √
x=y→0+
2x2
=
2x
√
x=y→0+
2x
=
x3
2 2 x=y→0+ x3
lim
x3
=
6=
2
lim
1
√
lim
1
√
2 2
0
Therefore, u is not differentiable at (0, 0).
7. [Page 21, 33] Prove that, in a disk on which the derivative of a complex function vanishes identically,
the function is constant.
Proof. Let f (z) be analytic in ∆ = {z ∈ C : |z| < 1}, and f 0 (z) = 0 for all z ∈ ∆. We write
f (z) = f (x, y) = u(x, y) + iv(x, y).
Since f is analytic and f 0 (z) = 0, then
ux (x, y) = vy (x, y) = 0, uy (x, y) = −v(x, y) = 0,
for all x2 + y 2 < 1.
That is ux (x, y) = uy (x, y) = vx (x, y) = vy (x, y) = 0 for all x2 + y 2 < 1, in particular, we know
ux , uy , vx , vy are continuous on ∆. For any x2 + y 2 < 1, by the fundamental theorem, we have
Z
u(x, y) − u(0, 0)
1
=
0
Z
=
du(tx, ty
dt
dt
1
[ux (tx, ty)t + uy (tx, ty)t]dt
0
SOLUTIONS OF HW1
11
1
Z
=
[0t + 0t]dt
0
1
Z
0 dt
=
0
=
0.
Hence u(x, y) = u(0, 0) for all x2 + y 2 < 1. By the similar argument, we get v(x, y) = v(0, 0).
Therefore, we get f (z) = f (x, y) = u(x, y) + iv(x, y) = u(0, 0) + iv(0, 0).
8. [Page 21, 35] What are the most general polynomials, with real coefficients, of second degree
U (x, y)
=
a + bx + cy + dx2 + exy + f y 2 ,
V (x, y)
=
a0 + b0 x + c0 y + d0 x2 + e0 xy + f 0 y 2 ,
for which U (x, y) + iV (x, y) is an analytic function of z = x + iy? Show that the function in question
is a polynomial of second degree in z.
Proof. Let f (x, y) = f (z) = U (x, y) + iV (x, y), if f is analytic, then U and V should satisfy CauchyRiemann equation:
Ux
= Vy
Uy
= −Vx
In fact, we know
Ux (x, y)
=
b + 2dx + ey
Uy (x, y)
=
c + ex + 2f y
Vx (x, y)
=
b0 + 2d0 x + e0 y
Vy (x, y)
=
c0 + e0 x + 2f 0 y
12
MINGFENG ZHAO
Hence, we have for all (x, y), we have
b + 2dx + ey
= c0 + e0 x + 2f 0 y
c + ex + 2f y
= −(b0 + 2d0 x + e0 y)
Therefore,
b = c0 , 2d = e0 , e = 2f 0 , c = −b0 , e = −2d0 , 2f = −e0
So b = c0 , c = −b0 , d =
e0
2
= −f, e = 2f 0 = −2d0 . Therefore, we get
U (x, y)
=
a + bx + cy + dx2 + exy − dy 2
V (x, y)
=
e
e
a0 − cx + by − x2 + 2dxy + y 2 .
2
2
On the other hand, since Ux , Uy , Vx , Vy are continuous, so if Ux = Vy , Uy = −Vx , then f (x, y) =
f (z) = U (x, y) + iV (x, y) should be analytic.
We set z = x + iy, then we get
f (z)
=
U (x, y) + iV (x, y)
=
e
e a + bx + cy + dx2 + exy − dy 2 + i a0 − cx + by − x2 + 2dxy + y 2
2
2
=
e 2
e
y
a + ia0 + (b − ic)x + (c + ib)y + (d − i )x2 + (e + 2di)xy + −d + i
2
2
=
e
e
e
a + ia0 + (b − ic)x + (b − ic)(iy) + (d − i )x2 + d − i
2ixy − (d − i )y 2
2
2
2
=
e 2
a + ia0 + (b − ic)(x + iy) + d − i
(x + 2ixy − y 2 )
2
=
e
a + ia0 + (b − ic)(x + iy) + d − i
(x + iy)2
2
=
e
a + ia0 + (b − ic)z + (d − i )z 2
2
SOLUTIONS OF HW1
13
Hence f (z) = U (x, y) + iV (x, y) is a polynomial of z of degree 2.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu