MATH 1700 Homework
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Homework 10 Solution
Section 4.6 ∼ 4.7.
• For every problem, explain your answer. It is not acceptable to write the answer
only.
1. Let z1 = 3 + 7i and z2 = −2 + i. Compute z1 + z2 , z1 − z2 , z1 z2 ,
z1
, z1 , and |z1 |.
z2
z1 + z2 = (3 + 7i) + (−2 + i) = (3 + (−2)) + (7 + 1)i = 1 + 8i
z1 − z2 = (3 + 7i) − (−2 + i) = (3 − (−2)) + (7 − 1)i = 5 + 6i
z1 z2 = (3 + 7i)(−2 + i) = (3 · (−2) − 7 · 1) + (3 · 1 + 7 · (−2))i = −13 − 11i
z1
3 + 7i
(3 + 7i)(−2 − i)
(−6 + 7) + (3 · (−1) + 7 · (−2))i
1 17
=
=
=
= − i
2
2
z2
−2 + i
(−2 + i)(−2 − i)
(−2) + 1
5
5
z1 = 3 + 7i = 3 − 7i
p
√
|z1 | = |3 + 7i| = 32 + 72 = 58
2. By using complex plane and algebraic manipulations, solve the following equations for z.
(a) z 5 = 1.
Let z = reiθ . Then z 5 = r5 ei5θ .
1 = |1| = |z 5 | = |z|5 = r5 ⇒ r = 1
ei5θ = 1 ⇒ cos(5θ) + sin(5θ)i = 1 ⇒ cos(5θ) = 1, sin(5θ) = 0
⇒ 5θ = 2kπ for some integer k.
θ = 0,
Therefore
2π 4π 6π 8π
, , ,
5 5 5 5
2π
2π
4π
4π
+ sin i, cos
+ sin i,
5
5
5
5
6π
6π
8π
8π
cos
+ sin i, cos
+ sin i.
5
5
5
5
z = 1, cos
(b) z 2 − z + 1 = 0.
By quadratic formula,
p
√
√
1 ± (−1)2 − 4 · 1 · 1
3
1 ± −3
1
=
= ±
i
z=
2
2
2
2
1
MATH 1700 Homework
Han-Bom Moon
3. Find the polar coordinates.
(a) 4i
r=
So
p
√
02 + 42 = 16 = 4
π
θ=
2
π
(r, θ) = (4, ).
2
(b) −2 + 2i
r=
p
√
√
(−2)2 + 22 = 8 = 2 2
θ=
Therefore
(c)
√
3 − i.
7π
4
√ 7π
(r, θ) = (2 2, ).
4
q√
√
r = ( 3)2 + (−1)2 = 4 = 2
θ=
11π
6
So
(r, θ) = (2,
√
1
3
4. Let z = +
i. Compute z 2016 .
2
2
v
u 2
u 1
+
|z| = t
2
11π
).
6
√ !2
√
3
= 1=1
2
So z = cos θ + sin θi. Because
√
1
3
cos θ = , sin θ =
,
2
2
θ=
π
. Therefore
3
π
π
+ sin i.
3
3
2016π
2016π
= cos
+ sin
i = cos 672π + sin 672πi = 1
3
3
z = cos
π
π 2016
z 2016 = cos + sin i
3
3
2
MATH 1700 Homework
Han-Bom Moon
5. Find the (real or complex) eigenvalues of
"
#
1 2
.
−1 3
Let A be the given matrix.
"
A − rI =
1−r
2
−1 3 − r
#
det(A − rI) = (1 − r)(3 − r) − 2 · (−1) = r2 − 4r + 5
p
√
4 ± (−4)2 − 4 · 1 · 5
4 ± −4
2
r − 4r + 5 = 0 ⇔ r =
=
=2±i
2
2
Therefore there are two complex eigenvalues, 2 + i and 2 − i.
6. Determine whether the origin is a sink or a source of xn+1 = Axn where
"
#
0 −2
;
(a) A =
1/6 −1
"
#
−r
−2
A − rI =
1/6 −1 − r
1
1
det(A − rI) = (−r)(−1 − r) − (−2) · = r2 + r +
6
3
√
√
1
−3 ± 32 − 4 · 3 · 1
1
3i
2
2
r + r + = 0 ⇔ 3r + 3r + 1 = 0 ⇔ r =
=− ±
3
6
2
6
v
u 2
√ !2 r
u
1
3
1
1
t
|r| =
−
+
=
= √ <1
2
6
3
3
Therefore the origin is a sink.
"
#
5
4
(b) A =
.
−1/5 1
"
#
5−r
4
A − rI =
−1/5 1 − r
29
1
det(A − rI) = (5 − r)(1 − r) − 4 · −
= r2 − 6r +
5
5
p
√
30 ± (−30)2 − 4 · 5 · 29
29
320
2
2
r −6r+ = 0 ⇔ 5r −30r+29 = 0 ⇔ r =
= 3±
5
10
10
√
√
320
320
3+
> 1, 3 −
≈ 1.211 > 1.
10
10
Therefore the origin is a source.
3
MATH 1700 Homework
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7. Determine all real values of the constant t for which the origin is a sink for the
system
xn+1 = txn − 0.18yn
yn+1 = 2xn + tyn .
The transition matrix is
"
#
t −0.18
A=
.
2
t
"
#
t − r −0.18
A − rI =
2
t−r
det(A − rI) = (t − r)2 − (−0.18) · 2 = r2 − 2tr + t2 + 0.36
p
2t ± (−2t)2 − 4(t2 + 0.36)
2
2
r − 2tr + t + 0.36 = 0 ⇔ r =
= t ± 0.6i
2
p
|r| < 1 ⇔ t2 + 0.62 < 1 ⇔ t2 < 1 − 0.36 = 0.64 ⇔ −0.8 < t < 0.8
Therefore the origin is a sink if and only if −0.8 < t < 0.8.
8. Find the general form of the solution of
xn+1 = 2xn − yn
yn+1 = xn + yn ,
where x0 = y0 = 1000.
The transition matrix is
"
A=
"
A − zI =
2 −1
1 1
#
.
2 − z −1
1
1−z
#
det(A − zI) = (2 − z)(1 − z) − (−1) · 1 = z 2 − 3z + 3
p
√
3 ± (−3)2 − 4 · 3
3 ± 3i
2
z − 3z + 3 = 0 ⇔ z =
=
2
2
v
u 2
√ !2
u 3
√
3
t
r = |z| =
+
= 3
2
2
So if z = r cos θ + r sin θi,
√
3
1
, sin θ = .
2
2
cos θ =
4
MATH 1700 Homework
Han-Bom Moon
π
.
6
The general solution of the linear system is
Therefore θ =
xn = rn cos(nθ)v + rn sin(nθ)w
for some two vectors v and w.
"
#
"
#
1000
1000
= x0 = v ⇒ v =
1000
1000
"
#
#
√
√
π 1000
π
1000
+ 3 sin w
=A
= x1 = 3 cos
6 1000
6
1000
"
#
"
#
"
#
"
#
√
√
− 1000
3
3 1000
1000
−500
3
w=
−
=
⇒w=
1000
√
2
2 1000
2000
500
3
"
1000
2000
#
"
Therefore the general solution is
"
#
" 1000 #
√ n
√ n
− √3
nπ
nπ
1000
+ 3 sin( )
.
xn = 3 cos( )
1000
√
6
6
1000
3
5