SOLUTION OF TEST 1
MINGFENG ZHAO
September 22, 2012
1. [10 Points] Find the real and imaginary parts of z =
3−i 3+i
−
. Show all steps.
1+i 1−i
Proof. Since we know that
z
=
3−i 3+i
−
1+i 1−i
=
(3 − i)(1 − i) (3 + i)(1 + i)
−
(1 + i)(1 − i) (1 − i)(1 + i)
=
3 − 3i − i − 1 3 + 3i + i − 1
−
2
2
=
2 − 4i 2 + 4i
−
2
2
=
2 − 4i − 2 − 4i
2
=
−8i
2
= −4i.
Hence we know that
Re z = 0,
and
Im z = −4.
1
2
MINGFENG ZHAO
2. [10 Points] Write the polar representation of the complex number z = 1 − i. Use it to find a simple
formula for z 6 .
Proof. Since z = 1 − i, then |z| =
p
√
1 + (−1)2 = 2 and z is located in the 4th quadrant. Hence we
get
z
=
=
=
=
1−i
√
√
2·
1
1
√ −√ i
2
2
π
π
2 · cos − i sin
4
4
1
π
2 2 · e−i 4 .
Then we get
z 6 = 23 · e−i
6π
4
= 8 · e−i
3π
2
= 8 · e−i
3π
2 +i2π
π
= 8 · ei 2 = 8i.
SOLUTION OF TEST 1
3
3. [10 Points] Find two square roots of z = 1 − i in polar representation. (Can you find all three 3rd
roots of z = 1 − i?)
Proof. By the proof of the Problem 2, then
π
1
z = 1 − i = 2 2 · e−i 4 .
Then the two square roots of z are:
1
π
z1 = 2 4 · e−i 8 ,
1
and z2 = 2 4 · ei
− π +2π
4
2
1
= 2 4 · ei
7π
8
.
The three 3rd roots of z are:
1
π
w1 = 2 6 · e−i 12 ,
1
w2 = 2 6 · ei
− π +2π
4
3
1
7π
= 2 6 · ei 12 ,
1
and w3 = 2 6 · ei
− π +4π
4
3
1
= 2 6 · ei
15π
12
1
= 2 6 · ei
5π
4
.
4
MINGFENG ZHAO
4. [10 Points] Find all four 4th roots of z = 16.
Proof. Since z = 16 = 24 , then all four 4th roots of z are: z1 = 2 and
z2 = 2 · ei
2π
4
π
= 2 · ei 2 = 2i,
z3 = 2 · ei
4π
4
= 2 · eπi = −2,
and z4 = 2 · ei
6π
4
=2·e
3π
2 i
= −2i.
SOLUTION OF TEST 1
5
5. [10 Points] Sketch the set {z : 1 < Re z < 2}. Is it open or closed set? Is it convex? Explain.
Proof. The graph of the set {z : 1 < Re z < 2} is:
It is open and convex. Since every point in {z : 1 < Re z < 2} is an interior point, then {z : 1 <
Re z < 2} is open. For any two points p, q in {z : 1 < Re z < 2}, the line segment connecting p
and q lies in the set {z : 1 < Re z < 2}, then {z : 1 < Re z < 2} is path connected, in particular,
{z : 1 < Re z < 2} is connected.
6
MINGFENG ZHAO
6. [10 Points] Sketch the set {z : 1 ≤ Im z ≤ 2}. Is it open or closed set? Is it convex? Explain.
Proof. The graph of the set {z : 1 ≤ Im z ≤ 2} is:
It is closed and convex. Since {z : 1 ≤ Im z ≤ 2} contains its boundary points, then {z : 1 ≤
Im z ≤ 2} is closed. For any two points p, q in {z : 1 ≤ Im z ≤ 2}, the line segment connecting p
and q lies in the set {z : 1 ≤ Im z ≤ 2}, then {z : 1 ≤ Im z ≤ 2} is path connected, in particular,
{z : 1 ≤ Im z ≤ 2} is connected.
SOLUTION OF TEST 1
7
7. [20 Points] Among these four sequences, three are convergent and one is divergent. Identify which
are which and explain.
an =
1+i
n2 + 1
bn =
i
n+1
cn =
4 − 3i
5
n
dn =
1+i
4 − 3i
n
.
Proof. a. an converges to 0, as n → ∞, since
√
1+i 2
=
|an | = 2
→ 0,
n + 1 n2 + 1
as n → ∞.
b. bn converges to 0, as n → ∞, since
|bn | = 1
i =
→ 0,
n + 1 n + 1
as n → ∞.
c. cn diverges, as n → ∞. Otherwise, if cn converges, as n → ∞, than cn is Cauchy, in particular,
there exists some N > 0 such that for all n ≥ N , we have
√
|cn+1 − cn | <
10
.
10
4 − 3i n 4 − 3i n
Since |cn | = = 5 = 1, then we know that
5
√
4 − 3i n+1 4 − 3i n 4 − 3i
−1 − 3i 10
10
|cn+1 − cn | = −
− 1 = =
>
.
=
5
5
5
5
5
10
Contradiction.
d. dn converges to 0, as n → ∞, since
1+i n
|dn | = =
4 − 3i √ !n
2
→ 0,
5
as n → ∞.
8
MINGFENG ZHAO
8. [20 Points] Among these four series, two are convergent and two are divergent. Identify which are
which and explain.
A.
Proof. A.
∞
X
1+i
2+1
n
n=1
B.
∞
X
i
n
+
1
n=1
C.
n
∞ X
4 − 3i
n=1
5
D.
n
∞ X
1+i
.
4 − 3i
n=1
∞
X
1+i
is convergent. Since for any n ≥ 1, we have
2+1
n
n=1
√
1+i 2
=
n2 + 1 n2 + 1 .
√
∞
X
2
1+i
is
convergent,
by
M
-test,
then
is convergent
2
n +1
n2 + 1
n=1
n=1
∞
∞
∞
∞
X
X
X
X
i
i
i
1
B.
is divergent. Otherwise, if
is convergent, then −i ·
=
n
+
1
n
+
1
n
+
1
n
+
1
n=1
n=1
n=1
n=1
∞
X 1
diverges.
is also convergent, which contradicts with
n+1
n=1
n
n
∞ X
4 − 3i
4 − 3i
C.
is divergent. By the result of Problem 7, then
does not converge to
5
5
n=1
n
∞ X
4 − 3i
is divergent.
0, as n → ∞. By the divergent test, we know that
5
n=1
n
∞ X
1+i
D.
is convergent. Since for any n ≥ 1, we have
4 − 3i
n=1
Since
∞
X
∞ X 1 + i n 1 + i n
=
= 4 − 3i 4 − 3i n=1
Since
√
2
5
< 1, then
∞
X
n=1
√ !n
2
.
5
√ !n
n
∞ X
1+i
2
converges. By the M -test, we know that
is con5
4 − 3i
n=1
vergent.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu