SOLUTIONS OF HW3
MINGFENG ZHAO
February 27, 2011
1. [Page 35, 1] Split the function w = z 2 into real and imaginary parts and investigate:
(a) How straight lines in the z-plane are mapped are mapped into the w-plane;
(b) How straight lines in the w-plane are mapped into the z-plane.
Proof. Let z = x + iy ∈ C, and w = u + iv ∈ C. Since w = z 2 , which implies that u + iv = (x + iy)2 =
x2 − y 2 + 2xyi, that is, u = x2 − y 2 , and v = 2xy.
a. For any line in z-plane, we can think the line has the form ax + by = c for some a, b, c ∈ R, and
a2 + b2 > 0.
Case I: a = 0, b 6= 0, c = 0, that is y =
c
b
= 0.
So we can get u = x2 − y 2 = x2 , v = 2xy = 0, that is, the graph of image of y =
c
b
= 0 is a ray.
Case II: a = 0, b 6= 0, c 6= 0, that is y = cb .
So we can get u = x2 − y 2 = x2 −
us x2 =
b2 v 2
4c2
= u+
c2
b2 .
c2
b2 , v
= 2xy = 2x cb =
Therefore, we know u =
b2 2
4c2 v
2c
b x.
c2
b2 .
−
Then we can get x =
bv
2c ,
which gives
So the graph of imapge of y =
c
b
is a
parabola.
Case III: a 6= 0, b = 0, c = 0, that is x =
c
a
= 0.
So we can get u = x2 − y 2 = −y 2 , v = 2xy = 0, that is, the graph of image of x =
c
a
= 0 is a ray.
Case IV: a 6= 0, b = 0, c 6= 0, that is x = ac .
So we can get u = x2 − y 2 =
2
a
2
that is, u = − 4c
2v +
c2
a2 .
c2
a2
− y 2 , v = 2xy = 2 · zc · y =
So the graph of image of x =
c
a
2c
a y.
Then y =
av
2c .
So y 2 =
a2 v 2
4c2
=
c2
a2
− u,
is a parabola.
Case V: a 6= 0, b 6= 0, c = 0, that is y = − ab x.
So we can get u = x2 − y 2 = x2 −
u=
b2 −a2
b2
bv
· − 2a
=
a2 −b2
2ab v.
a2 2
b2 x
=
b2 −a2 2
b2 x ,
bv
2
2
and v = 2xy = − 2a
b x . So x = − 2a . Then
So the graph of image of y = − ab x is a ray.
1
2
MINGFENG ZHAO
Figure 1. a = 0, b 6= 0, c = 0
Figure 2. a = 0, b 6= 0, cb > 0
SOLUTIONS OF HW3
Figure 3. a = 0, b 6= 0, cb < 0
Figure 4. a 6= 0, b = 0, c = 0
3
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MINGFENG ZHAO
Figure 5. a 6= 0, b = 0, ac > 0
Figure 6. a 6= 0, b = 0, ac < 0
SOLUTIONS OF HW3
Figure 7.
a
b
< 0, c = 0, a2 = b2
Figure 8.
a
b
> 0, c = 0, a2 = b2
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MINGFENG ZHAO
Figure 9.
Figure 10.
a
b
a
b
< 0, c = 0, a2 > b2
< 0, c = 0, a2 < b2
SOLUTIONS OF HW3
Figure 11.
a
b
> 0, c = 0, a2 > b2
Figure 12.
a
b
> 0, c = 0, a2 < b2
Case VI: abc 6= 0.[Very Complicated]
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MINGFENG ZHAO
b. For any lines in the w-plane, we have the form au + bv = c for some a, b, c ∈ R, and a2 + b2 > 0.
Since u = x2 − y 2 , v = 2xy, then we can get
a(x2 − y 2 ) + 2bxy = c.
That is ax2 − ay 2 + 2bxy − c = 0. Look at the determent, we have


 a
det 

b
b 
 = −a2 − b2 < 0.

−a
By the property of quadratic curves, we know the graph of x, y should like a hyperbola, or two
intersecting lines.
Case I: a = 0, b 6= 0, c = 0, we get v = 0, and xy =
c
2b
=0
That is, x = 0 or y = 0, two axises.
Case II: a = 0, b 6= 0, c 6= 0, we get v = cb , and xy =
c
2b .
That is, we get a hyperbola.
Case III: a 6= 0, b = 0, c = 0, we get u = 0, and x2 = y 2 .
That is, we get y = x or x = −x.
Case III: a 6= 0, b = 0, c 6= 0, we get u = ac , and x2 − y 2 = ac .
That is, we get a hyperbola.
Case IV: abc 6= 0. [Complicated, but any way it is a hyperbola.]
SOLUTIONS OF HW3
Figure 13. a = 0, b 6= 0, c = 0
Figure 14. a = 0, cb > 0
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10
MINGFENG ZHAO
Figure 15. a = 0, cb < 0
Figure 16. a 6= 0, b = 0, c = 0
SOLUTIONS OF HW3
11
Figure 17. b = 0, ac > 0
Figure 18. b = 0, ac < 0
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MINGFENG ZHAO
2. [Page 36, 3] What values does the function
√
4
z posses at the point z = i? Write these values in
trigonometric and in algebraic form. In what sequence do these values come up, when starting at
z = 1 with the function value 1, z describes the unit circle once in the negative direction (that is, in
the direction in which arg z decreases) ?
π
Proof. Since z = i = ei 2 , so we can find that
π
w1
= ei 8
w2
= ei
5π
8
w3
= ei
−5π
8
w4
= ei
−π
8
√
Now we compute cos π8 . Since 0 < π8 < π2 , then A > 0. Since 2 cos2 π8 = 1 + cos π4 = 1 + 22 =
√ √
√ √
q
√
p
2
2− 2
2+ 2
π
2 π =
1
−
cos
1
−
So cos π8 = 2+
.
And
sin
=
=
. Therefore, we get
2
8
8
4
2
w1
=
w2
=
w3
=
w4
=
√
2+ 2
2 .
p
p
√
√
2+ 2
2− 2
+i
e =
2
2
p
p
√
√
2− 2
2+ 2
i 5π
8
e
=−
+i
2
2
p
p
√
√
2+ 2
2− 2
i −5π
8
e
−i
=−
2
2
p
p
√
√
−π
2− 2
2+ 2
ei 8 =
−i
2
2
iπ
8
So if considering in the negative direction, we should have w4 , w3 , w2 , w1 , and we can describe
w1 , w2 , w3 , w4 as below:
Figure 19. w1 , w2 , w3 , w4
SOLUTIONS OF HW3
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MINGFENG ZHAO
3. [Page 36, 5] Decompose the functions
z2
z4
,
−1
1
,
+1
1
z(z + 1)2 (z + 2)3
and
z3
into partial fractions.
Proof. a. Since z 2 + 1 = (z + i)(z − i), we have
1
z2 + 1
=
=
=
=
1
(z + i)(z − i)
1
1
1
·
−
2i
z−i z+i
1
1
1
1
·
−
·
2i z − i 2i z + i
−i
2
z−i
+
i
2
z+i
b. Since
3
√ !
1
3
z+ −i
·
2
2
2
z − 1 = (z − 1)(z + z + 1) = (z − 1) ·
√ !
1
3
z+ +i
,
2
2
so we have
z4
z3 − 1
=
z(z 3 − 1) + z
z3 − 1
z
z3 − 1
= z+
z
= z+
(z − 1) · z +
1
2
√
−i
3
2
· z+
1
2
+i
√ 3
2
B
C
A
√ +
√
+
3
z−1 z+ 1 −i
z + 21 + i 23
2
2
= z+
Assume
z
(z − 1) · z +
1
2
√
− i 23 · z +
1
2
√
+i
3
2
=
A
B
C
√ +
√
+
3
1
1
z−1 z+ −i
z + 2 + i 23
2
2
for some A, B, C ∈ C.
4. [Page 75, 1] Calculate the value of the function ez at the points z = −i π2 , i 3π
4 , 1 − 2i.
Proof. For any z = x + iy ∈ C, we know
ez
= ex+iy
SOLUTIONS OF HW3
15
= ex eiy
= ex cos y + iex sin y.
a. When z = −i π2 , we get x = 0, y = − π2 . So
π
π
π
+ i sin −
= −i.
e−i 2 = cos −
2
2
b. When z = i 3π
4 , we get x = 0, y =
e
i 3π
4
3π
4 .
= cos
So
3π
4
+ i sin
3π
4
√
√
2
2
=−
+i
.
2
2
c. When z = 1 − 2i, we get x = 1, y = −2. So
e1−2i = e cos(−2) + ie sin(−2) = e cos 2 − ie sin 2.
5. [Page 75, 2] Change the following numbers into the form reiθ :
−3,
2i,
i
− ,
2
√
3
1
− +i
,
2
2
1 + 2i,
a + ib
,
a − ib
where a, b ∈ R, and a − ib 6= 0.
Proof. We think any z ∈ C as z = x + iy, moreover, we know z = r cos θ + ir sin θ such that
r=
p
x2 + y 2 , and x = r cos θ, y = r sin θ.
a. When z = −3, that is x = −3, and y = 0, so r = 3, cos θ = −1, and sin θ = 0, so we get θ = π + 2kπ
for some k ∈ Z. Hence
−3 = 3eiπ .
b. When z = 2i, that is x = 0, and y = 2, so r = 2, cos θ = 0, and sin θ = 1, so we get θ =
π
2
+ 2kπ
for some k ∈ Z. Hence
π
2i = 2ei 2 .
c. When z = − 2i , that is x = 0, and y = − 12 , so r = 12 , cos θ = 0, and sin θ = 1, so we get θ = − π2 +2kπ
for some k ∈ Z. Hence
−
π
i
1
= e−i 2 .
2
2
16
MINGFENG ZHAO
d. When z =
− 12
√
+i
3
2 ,
that is x =
√
sin θ =
3
2 ,
so we get θ =
2π
3
− 12 ,
r
√
3
2 ,
and y =
− 12
so r =
2
+
√ 2
3
2
= 1, cos θ = − 21 , and
+ 2kπ for some k ∈ Z. Hence
√
2π
1
3
− +i
= 1 · ei 3 .
2
2
e. When z = 1 + 2i, that is x = 1, and y = 2, so r =
√
12 + 2 2 =
√
5, cos θ =
we get θ = arctan 2 + 2kπ for some k ∈ Z. Hence
1 + 2i =
√
5ei arctan 2 .
Figure 20. z = −3, 2i, − 2i , − 12 + i
f. When z =
a+ib
a−ib ,
3
2 ,1
then
z
That is, x =
√
a2 −b2
a2 +b2 , y
=
2ab
a2 +b2 .
=
a + ib
a − ib
=
(a + ib)(a + ib)
(a − ib)(a + ib)
=
a2 − b2 + 2abi
a 2 + b2
=
a 2 − b2
2ab
+i 2
a 2 + b2
a + b2
So
s
r
=
a 2 − b2
a 2 + b2
2
+
2ab
a2 + b2
2
+ 2i
√1 ,
5
and sin θ =
√2 ,
5
so
SOLUTIONS OF HW3
a4 + b4 − 2a2 b2
4a2 b2
+ 2
2
2
2
(a + b )
(a + b2 )2
s
a4 + 2a2 b2 + b4
(a2 + b2 )2
s
(a2 + b2 )2
(a2 + b2 )2
=
=
=
=
So we have cos θ =
a2 −b2
a2 +b2 ,
17
s
1.
and sin θ =
2ab
a2 +b2 .
Case I: If a = b 6= 0, we get cos θ = 0, sin θ = 1. So θ =
π
2
+ 2kπ for some k ∈ Z. Hence
π
a + ib
a + ia
=
= i = ei 2 .
a − ib
a − ia
Case II: If a = −b 6= 0, we get cos θ = 0, sin θ = −1. So θ = − π2 + 2kπ for some k ∈ Z. Hence
π
a + ib
a − ia
=
= −i = e−i 2 .
a − ib
a + ia
Figure 21. |a| = |b| =
6 0
Recall that arctan(R) = (− π2 , π2 ).
Case III: If |a| > |b|, we get cos θ =
a2 −b2
a2 +b2
> 0, and tan θ =
2ab
a2 −b2 .
So θ = arctan a22ab
−b2 + 2kπ for some
k ∈ Z. Hence
a + ib
a2 − b2
2ab
i arctan a22ab
−b2 .
= 2
+i 2
=e
2
2
a − ib
a +b
a +b
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MINGFENG ZHAO
Figure 22. y = arctan x
Figure 23. |a| > |b|
Case IV: If |a| < |b|, we get cos θ =
a2 −b2
a2 +b2
< 0, and tan θ =
2ab
a2 −b2 .
So θ = arctan a22ab
−b2 − π + 2kπ for
some k ∈ Z. Hence
a + ib
a2 − b2
2ab
i
= 2
+i 2
=e
a − ib
a + b2
a + b2
arctan
2ab
a2 −b2
−π
.
SOLUTIONS OF HW3
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Figure 24. |a| < |b|
6. [Page 75, 8] How does ez change when the point z goes off to infinity along a ray from the origin?
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu