Homework #1 Solutions
Math 128, Fall 2013
Instructor: Dr. Doreen De Leon
1
p. 5: 2, 3, 8, 11
2. Show that
(a) Re (iz) = −Im z
(b) Im (iz) = Re z
Since z = x + iy, Re z = x and Im z = y.
Since iz = i(x + iy) = −y + ix, we know that Re (iz) = −y and Im (iz) = x.
Therefore, Re (iz) = −Im z and Im (iz) = Re z.
3. Show that (1 + z)2 = 1 + 2z + z 2 .
(1 + z)2 = (1 + z)(1 + z)
= 1(1 + z) + z(1 + z)
= 1 + z + z + z2
= 1 + 2z + z 2 .
8. (a) Write (x, y) + (u, v) = (x, y) and point out how it follows that the complex number (0, 0) is
unique as an additive identity.
(b) Likewise, write (x, y)(u, v) = (x, y) and show that the number 1 = (1, 0) is a unique multiplicative identity.
(a) Let z = (x, y) and w = (u, v).
z + w = z if, and only if, (x, y) + (u, v) = (x, y)
(x + u, y + v) = (x, y).
Since (x + u, y + v) = (x, y) if, and only if, x + u = x and y + v = y, we obtain u = 0 and
v = 0 as the only solution. Therefore, (0, 0) is the unique additive identity.
(b) Let z = (x, y) and w = (u, v).
zw = z if, and only if, (x, y)(u, v) = (x, y)
(xu − yv, xv + yu) = (x, y),
1
which is true if, and only if u, v satisfy
xu − yv = x
xv + yu = y
or
(xu − yv = x)x
(yu + xv = y)y.
This gives
x2 u − xyv = x2
2
2
y u + xyv = y .
(1)
(2)
If we add (1) and (2), we obtain
x2 u + y 2 u = x2 + y 2
(x2 + y 2 )u = x2 + y 2
u = 1.
Since xu − yv = x and u = 1,
xu − yv = x =⇒ −yv = 0
=⇒ v = 0.
Therefore, w = (1, 0) is the only solution.
11. Solve the equation z 2 + z + 1 = 0 for z = (x, y) by writing
(x, y)(x, y) + (x, y) + (1, 0) = (0, 0)
and then solving a pair of simultaneous equations in x and y.
(x, y)(x, y) + (x, y) + (1, 0) = (0, 0)
(x − y , xy + xy) + (x, y) + (1, 0) = (0, 0)
2
2
(x2 − y 2 + x + 1, 2xy + y) = (0, 0),
which is true if, and only if,
x2 − y 2 + x + 1 = 0
(3)
2xy + y = 0.
(4)
The left-hand side of Equation (4) may be factored to give
y(2x + 1) = 0.
1
So, there are two possibilities: y = 0 or x = − , y ̸= 0.
2
2
Case 1: y = 0
Then (3) gives x2 + x + 1 = 0, which has no real solution. This is not possible, since x is a
real number. Therefore, y ̸= 0.
1
Case 2: x = − , y ̸= 0
2
1
Plugging x = − into (3) gives
2
( )2
( )
1
1
2
−
−y + −
+1=0
2
2
3
− y2 = 0
4
3
y2 =
4√
3
y=±
,
2
So, there are two solutions:
(
z=
2
(
√ )
√ )
3
1 3
1
,z= − ,
.
− ,−
2
2
2 2
p. 8: 1, 6
1. Reduce each of these quantities to a real number.
(a)
1 + 2i 2 − i
+
3 − 4i
5i
1 + 2i
(1 + 2i)(3 + 4i)
=
3 − 4i
(3 − 4i)(3 + 4i)
(3 − 8) + i(6 + 4)
=
9 + 16
−5 + 10i
1 2
=
= − + i,
25
5 5
2−i
(2 − i)(−i)
=
5i
5i(−i)
−1 − 2i
=
5
1 2
= − − i.
5 5
So,
1 + 2i 2 − i
+
=
3 − 4i
5i
(
) (
)
1 2
1 2
2
− +
+ − − i = − .
5 5
5 5
5
3
(b)
5i
(1 − i)(2 − i)(3 − i)
(1 − i)(2 − i)(3 − i) = (1 − i)[(6 − 1) + i(−3 − 2)]
= (1 − i)(5 − 5i)
= (5 − 5) + i(−5 − 5)
= −10i.
So,
5i
5i
1
=
= − .
(1 − i)(2 − i)(3 − i)
−10i
2
(c) (1 − i)4
(1 − i)4 = (1 − i)2 (1 − i)2
(1 − i)2 = (1 − i)(1 − i)
= (1 − 1) + i(−1 − 1) = −2i.
So,
(1 − i)4 = (−2i)(−2i) = −4.
6. With the aid of relations (10) and (11) in Section 3, derive the identity
( )( )
z2
z1 z2
z1
=
(z3 ̸= 0, z4 ̸= 0).
z3
z4
z3 z4
z1
= z1
z2
So,
3
(
z1
z3
(
)(
1
z3
z2
z4
)
)
z2
and
= z2
z4
(
1
z4
)
.
( ( )) ( ( ))
1
1
z1
z2
z3
z4
( )( )
1
1
= z1 z2
z3
z4
(
)
1
= (z1 z2 )
z3 z4
z1 z2
.
=
z3 z4
=
p. 12: 2, 3
2. Verify inequalities (4), Section 4, involving Re z, Im z, and |z|.
(i) Re z ≤ |Rz| ≤ |z|
Let z = x + iy. Then Re z = x. Since x is a real number, we know that x ≤ |x|. So,
Re z ≤ |Re z|.
√
√
Then, |z| = x2 + y 2 , and x ≤ x2 + y 2 . So,
x ≤ |z| =⇒ |Re z| ≤ |z|.
Therefore, Re z ≤ |Re z| ≤ |z|,
4
(ii) Im z ≤ |Im z| ≤ |z|
Let z = x + iy. Then Im z = y. Since y is a real number, we know that y ≤ |y|. So,
Im z ≤ |Im z|.
√
√
Then, |z| = x2 + y 2 , and y ≤ x2 + y 2 . So,
y ≤ |z| =⇒ |Im z| ≤ |z|.
Therefore, Im z ≤ |Im z| ≤ |z|,
3. Use established properties of moduli to show that when |z3 | ̸= |z4 |,
Re (z1 + z2 )
|z1 | + |z2 |
≤
.
|z3 + z4 |
||z3 | − |z4 ||
To do this problem, we must show two things,
(i) Re (z1 + z2 ) ≤ |z1 | + |z2 |, and
(ii) |z3 + z4 | ≥ ||z3 | − |z4 ||.
(i)
Re (z1 + z2 ) ≤ |Re (z1 + z2 )|
≤ |z1 + z2 |
≤ |z1 | + |z2 |.
(ii) |z3 + z4 | ≥ ||z3 | − |z4 || was demonstrated in class.
Therefore,
4
|z1 | + |z2 |
Re (z1 + z2 )
≤
.
|z3 + z4 |
||z3 | − |z4 ||
p. 14-15: 1, 9
1. Use properties of conjugates and moduli established in Section 5 to show that
(a) z + 3i = z − 3i
z + 3i = z + 3i
= z − 3i.X
(b) iz = −iz
iz = iz
= −iz.X
(c) (2 + i)2 = 3 − 4i
(2 + i)2 = (2 + i)2
= (2 − i)2
= (4 − 1) + i(−2 − 2)
= 3 − 4i.
5
Alternately,
(2 + i)2 = (4 − 1) + i(2 + 2)
= 3 + 4i
= 3 − 4i.
√
√
(d) (2z + 5)( 2 − i) = 3|2z + 5|
√
√
(2z
+
5)(
2
−
i)
= |2z + 5|| 2 − i|
|2z + 5|2 = (2z + 5)(2z + 5)
= (2z + 5)(2z + 5)
= (2z + 5)(2z + 5)
= (2z + 5)(2z + 5)
= (2z + 5)(2z + 5)
= |2z + 5|2
=⇒ |2z + 5| = |2z + 5|.
√√
√
√
| 2 − i| = ( 2)2 + (−1)2 = 3.
√
√
So, (2z + 5)( 2 − i) = 3|2z + 5|. X
9. By factoring z 4 − 4z 2 + 3 into two quadratic factors and using inequality (8), Section 4, show that
if z lies on the circle |z| = 2, then
1
1
z 4 − 4z 2 + 3 ≤ 3 .
1
1
=
, we need to show that |z 4 − 4z 2 + 3| ≥ 3.
Since 4
2
z − 4z + 3 |z 4 − 4z 2 + 3|
z 4 − 4z 2 + 3 = (z 2 − 1)(z 2 − 3).
So, |z 4 − 4z 2 + 3| = |(z 2 − 1)(z 2 − 3)|
= |z 2 − 1| · |z 2 − 3|.
Now, |z 2 − 1| ≥ ||z 2 | − |1||
= ||z|2 − 1|
and |z 2 − 3| ≥ ||z 2 | − |3||
= ||z|2 − 3|.
So, |z 2 − 1| · |z 2 − 3| ≥ ||z|2 − 1| · ||z|2 − 3|.
If z lies on the circle |z| = 2, then
||z|2 − 1| · ||z|2 − 3| = |22 − 1| · |22 − 3| = 3.
Therefore,
1
1
≤ .X
|z − 4z + 3| ≥ 3 =⇒ 4
2
z − 4z + 3 3
4
2
6
5
p. 22-23: 1, 5
1. Find the principal argument Arg z when
(a) z =
i
−2 − 2i
i
z1
=
=⇒ arg z = arg z1 − arg z2 .
−2 − 2i
z2
π
z1 = i =⇒ arg z1 = .
2
( )
3π
−1 −2
=−
z2 = −2 − 2i =⇒ arg z2 = tan
(since the angle is in the third quadrant
−2
4
(
)
π
3π
So, arg z = − −
2
4
5π
=
.
4
z=
Since π <
5π
5π
3π
, Arg z =
− 2π =⇒ Arg z = − .
4
4
4
√
(b) z = ( 3 − i)6
√
arg z = 6(arg( 3 − i)).
(
)
√
π
−1 −1
√
arg( 3 − i) = tan
= − (since the angle is in the fourth quadrant).
6
3
( π)
So, arg z = 6 −
= −π.
6
Since −π ∈
/ (−π, π], Arg z = −π + 2π =⇒ Arg z = π.
5. By writing the individual factors on the left in exponential form, performing the needed operations,
and finally changing back to rectangular coordinates, show that
√
√
√
(a) i(1 − 3i)( 3 + i) = 2(1 + 3i)
π
1−
√
3i :
π
i = 1ei 2 = ei 2 .
√
√
r = 12 + (− 3)2 = 2
( √ )
− 3
π
θ = tan−1
=−
1
3
√
π
3 = 2ei(− 3 )
√√
r = ( 3)2 + 12 = 2
(
)
1
π
θ = tan−1 √
=
6
3
√
i π6
=⇒ 3 + i = 2e .
=⇒ 1 −
√
3+i:
7
Therefore,
i(1 −
√
√
π
π
π
3i)( 3 + i) = ei 2 · 2ei(− 3 ) · 2ei 6
π
π
π
= 4ei( 2 − 3 + 6 )
π
= 4ei 3
(
(π )
( π ))
= 4 cos
+ i sin
3
√ 3
= 2 + 2 3i
√
= 2(1 + 3i).X
(b)
5i
= 1 + 2i
2+i
π
5i = 5ei 2 .
√
√
2 + i : r = 22 + 1 2 = 5
( )
−1 1
θ = tan
.
2
π
5ei 2
5i
=√
So,
−1 1
2+i
5ei(tan ( 2 ))
(
(
( ))
(
( )))
√
π
π
−1 1
−1 1
= 5 cos
− tan
+ i sin
− tan
2
2
2
2
(
( )))
( (
( ))
√
1
1
+ i cos tan−1
= 5 sin tan−1
2
2
(
)
√
1
2
= 5 √ + i√
5
5
= 1 + 2i.X
(
( ))
(
( ))
−1 1
−1 1
The values for sin tan
and cos tan
come from the right triangle below
2
2
and trigonometric properties.
√
5
1
θ
2
8
(c) (−1 + i)7 = −8(1 + i)
√
√
(−1)2 + 12 = 2
( )
1
3π
−1
θ = tan
=
(since θ is in the second quadrant).
−1
4
√ 3π
=⇒ −1 + i = 2ei 4 .
(√
)7
3π
So, (−1 + i)7 =
2ei( 4 )
√
21π
= ( 2)7 ei 4
√
3π
= 8 2ei(− 4 )
)
(
))
(
(
√
3π
3π
= 8 2 cos −
+ i sin −
4
4
(
)
√
1
1
= 8 2 −√ − i√
2
2
= −8(1 + i).X
−1 + i :
(d) (1 +
r=
√ −10
√
3i)
= 2−11 (−1 + 3i)
√
1 + 3i :
√
√
12 + ( 3)2 = 2
(√ )
π
3
−1
θ = tan
=
1
3
r=
√
π
=⇒ 1 + 3i = 2ei 3 .
( π )−10
√
So, (1 + 3)−10 = 2ei 3
= 2−10 e−i
10π
3
2π
10π
2π
= 2−10 ei 3 (since −
+ 4π =
)
3
(
( )
( ))3
2π
2π
= 2−10 cos
+ i sin
3
3
(
√ )
3
1
= 2−10 − + i
2
2
√
= 2−11 (−1 + 3i).X
9