Question 1
Let
f
(
z
)
f(z)f, left parenthesis, z, right parenthesis be defined on
C
∖
{
0
}
C∖{0}C, \setminus, left brace, 0, right brace via
f
(
z
)
=
ln
∣
z
∣
+
i
θ
f(z)=ln∣z∣+iθf, left parenthesis, z, right parenthesis, equals, natural log, vertical bar, z, vertical bar, plus, i, theta, where
θ
θtheta is the argument of
z
zz satisfying
0
≤
θ
<
2
π
0≤θ<2π0, is less than or equal to, theta, \lt, 2, pi (note that this is not the principal argument). Which of the following are true: (Check all that apply.)
f
(
i
2
)
−
2
f
(
−
i
)
=
−
2
π
i
f(i
2
)−2f(−i)=−2πif, left parenthesis, i, squared, right parenthesis, minus, 2, f, left parenthesis, minus, i, right parenthesis, equals, minus, 2, pi, i.
The image under
f
ff of the circle of radius
e
ee, centered at the origin, is the entire vertical line
{
x
=
1
}
{x=1}left brace, x, equals, 1, right brace.
f
′
(
z
)
=
1
z
f′(z)=
z
1
f, prime, left parenthesis, z, right parenthesis, equals, start fraction, 1, divided by, z, end fraction for
z
∈
C
∖
[
0
,
∞
)
z∈C∖[0,∞)z, \in, C, \setminus, open bracket, 0, comma, infinity, right parenthesis.
f
ff is continuous at
z
=
2
z=2z, equals, 2.
1 point
2.
Question 2
Let
f
(
z
)
=
z
2
f(z)=z
2
f, left parenthesis, z, right parenthesis, equals, z, squared,
g
(
z
)
=
e
z
g(z)=e
z
g, left parenthesis, z, right parenthesis, equals, e, start superscript, z, end superscript and
h
(
z
)
=
Log
z
h(z)=Log zh, left parenthesis, z, right parenthesis, equals, start text, L, o, g, space, end text, z. Furthermore, let
γ
1
:
[
0
,
π
]
→
C
γ
1
:[0,π]→Cgamma, start subscript, 1, end subscript, colon, open bracket, 0, comma, pi, close bracket, \to, C be given by
γ
1
(
t
)
=
2
e
i
t
γ
1
(t)=2e
it
gamma, start subscript, 1, end subscript, left parenthesis, t, right parenthesis, equals, 2, e, start superscript, i, t, end superscript and
γ
2
:
[
π
/
2
,
5
π
/
2
]
→
C
γ
2
:[π/2,5π/2]→Cgamma, start subscript, 2, end subscript, colon, open bracket, pi, slash, 2, comma, 5, pi, slash, 2, close bracket, \to, C by
γ
2
(
t
)
=
i
t
γ
2
(t)=itgamma, start subscript, 2, end subscript, left parenthesis, t, right parenthesis, equals, i, t, and
γ
3
:
[
−
2
,
2
]
→
C
γ
3
:[−2,2]→Cgamma, start subscript, 3, end subscript, colon, open bracket, minus, 2, comma, 2, close bracket, \to, C by
γ
3
(
t
)
=
t
γ
3
(t)=tgamma, start subscript, 3, end subscript, left parenthesis, t, right parenthesis, equals, t. Which of the following are true: (Check all that apply.)
The set
{
h
(
γ
1
(
t
)
)
∣
0
≤
t
≤
π
}
{h(γ
1
(t))∣0≤t≤π}left brace, h, left parenthesis, gamma, start subscript, 1, end subscript, left parenthesis, t, right parenthesis, right parenthesis, \mid, 0, is less than or equal to, t, is less than or equal to, pi, right brace is the circle of radius
ln
2
ln2natural log, 2, centered at the origin.
f
,
g
f,gf, comma, g and
h
hh are conformal mappings on the open set bounded by the curves
γ
1
γ
1
gamma, start subscript, 1, end subscript and
γ
3
γ
3
gamma, start subscript, 3, end subscript.
f
∘
γ
1
f∘γ
1
f, circle, gamma, start subscript, 1, end subscript and
f
∘
γ
2
f∘γ
2
f, circle, gamma, start subscript, 2, end subscript intersect at the point
−
4
−4minus, 4 at an angle of
π
2
2
π
start fraction, pi, divided by, 2, end fraction.
The set
{
f
(
γ
2
(
t
)
)
∣
π
/
2
≤
t
≤
5
π
/
2
}
{f(γ
2
(t))∣π/2≤t≤5π/2}left brace, f, left parenthesis, gamma, start subscript, 2, end subscript, left parenthesis, t, right parenthesis, right parenthesis, \mid, pi, slash, 2, is less than or equal to, t, is less than or equal to, 5, pi, slash, 2, right brace is the interval
[
π
2
/
4
,
25
π
2
/
4
]
[π
2
/4,25π
2
/4]open bracket, pi, squared, slash, 4, comma, 25, pi, squared, slash, 4, close bracket on the real axis.
1 point
3.
Question 3
Let
f
(
z
)
=
1
+
z
1
−
z
f(z)=
1−z
1+z
f, left parenthesis, z, right parenthesis, equals, start fraction, 1, plus, z, divided by, 1, minus, z, end fraction. Which of the following are true: (Check all that apply.)
f
ff maps the circle of radius
2
22, centered at the origin, to a circle centered at
1
11.
f
ff maps the real axis to a circle of radius
<
100
<100\lt, 100 through the point
1
11.
f
ff maps
0
00,
1
11 and
∞
∞infinity to
1
11,
∞
∞infinity, and
−
1
−1minus, 1, respectively.
f
ff maps the vertical line through
z
=
0
z=0z, equals, 0 to the circle of radius
1
11, centered at the origin.
1 point
4.
Question 4
Let
f
ff be the Möbius transformation that maps
0
00 to
−
i
2
2
−i
start fraction, minus, i, divided by, 2, end fraction,
i
2
2
i
start fraction, i, divided by, 2, end fraction to
0
00 and
−
2
−2minus, 2 to
∞
∞infinity. Which of the following are true? (Check all that apply.)
f
(
3
)
=
6
i
5
f(3)=
5
6i
f, left parenthesis, 3, right parenthesis, equals, start fraction, 6, i, divided by, 5, end fraction.
f
(
−
2
+
i
)
=
1
−
4
i
f(−2+i)=1−4if, left parenthesis, minus, 2, plus, i, right parenthesis, equals, 1, minus, 4, i.
f
(
−
2
+
i
/
2
)
=
8
i
f(−2+i/2)=8if, left parenthesis, minus, 2, plus, i, slash, 2, right parenthesis, equals, 8, i.
f
(
1
)
=
2
−
i
3
f(1)=
3
2−i
f, left parenthesis, 1, right parenthesis, equals, start fraction, 2, minus, i, divided by, 3, end fraction.
1 point
5.
Question 5
Suppose that
G
GG is a simply connected domain in the complex plane and that
f
ff is a Riemann map of
G
GG, i.e. an analytic function that maps
G
GG conformally onto the unit disk
B
1
(
0
)
B
1
(0)B, start subscript, 1, end subscript, left parenthesis, 0, right parenthesis. Which of the following are true? (Check all that apply.)
g
(
z
)
=
f
(
z
)
+
1
f
(
z
)
+
i
g(z)=
f(z)+i
f(z)+1
g, left parenthesis, z, right parenthesis, equals, start fraction, f, left parenthesis, z, right parenthesis, plus, 1, divided by, f, left parenthesis, z, right parenthesis, plus, i, end fraction also maps
G
GG conformally onto the unit disk
B
1
(
0
)
B
1
(0)B, start subscript, 1, end subscript, left parenthesis, 0, right parenthesis.
g
(
z
)
=
1
3
f
(
z
)
g(z)=
3
1
f(z)g, left parenthesis, z, right parenthesis, equals, start fraction, 1, divided by, 3, end fraction, f, left parenthesis, z, right parenthesis maps
G
GG conformally onto the disk
B
1
/
3
(
0
)
B
1/3
(0)B, start subscript, 1, slash, 3, end subscript, left parenthesis, 0, right parenthesis.
g
(
z
)
=
e
i
π
/
10
f
(
z
)
g(z)=e
iπ/10
f(z)g, left parenthesis, z, right parenthesis, equals, e, start superscript, i, pi, slash, 10, end superscript, f, left parenthesis, z, right parenthesis also maps
G
GG conformally onto the unit disk
B
1
(
0
)
B
1
(0)B, start subscript, 1, end subscript, left parenthesis, 0, right parenthesis.
g
(
z
)
=
f
(
z
)
+
5
i
g(z)=f(z)+5ig, left parenthesis, z, right parenthesis, equals, f, left parenthesis, z, right parenthesis, plus, 5, i maps
G
GG conformally onto the disk of radius
5
55, centered at
i
ii.
1 point
MathBot Answer:
MathBot is working on a solution to your problem.
