SOLUTION OF TEST 2
MINGFENG ZHAO
October 13, 2012
1. [10 Points] Find ln(2i) (write real and imaginary parts).
π
Proof. Since 2i = 2ei 2 , then
ln(2i) = ln 2 + i(
π
+ 2kπ),
2
k ∈ Z.
1
2
MINGFENG ZHAO
2. [10 Points] Find (2i)i (write real and imaginary parts).
Proof. Since (2i)i = ei ln(2i) , by the result of the Problem 1, then
(2i)i
π
= ei[ln 2+i( 2 +2kπ)]
π
= ei ln 2−( 2 +2kπ)
π
= e−( 2 +2kπ) ei ln 2
π
π
= e−( 2 +2kπ) cos(ln 2) + ie−( 2 +2kπ) sin(ln 2),
∀k ∈ Z
SOLUTION OF TEST 2
3
3. [10 Points] Find sin(2i) and cos(2i) (write real and imaginary parts).
Proof. Since sin(2i) =
ei·2i − e−i·2i
, then
2i
sin(2i)
=
ei·2i − e−i·2i
2i
=
e−2 − e2
2i
= i
e2 − e−2
2
= i sinh 2.
Since cos(2i) =
ei·2i + e−i·2i
, then
2
cos(2i)
=
ei·2i + e−i·2i
2
=
e−2 + e2
2
=
cosh 2.
4
MINGFENG ZHAO
Z
4. [10 Points] Compute
|z|2 dz where γ is the straight line segment from 0 to 2i.
γ
Proof. For γ, we know that z(t) = t(2i) = 2ti for all t ∈ [0, 1], then dz = 2i dt and
Z
2
|z| dz
Z
1
|2ti|2 · 2i dt
=
0
γ
Z
=
1
4t2 dt
2i
0
Z
=
8i
1
t2 dt
0
=
8i
3
SOLUTION OF TEST 2
5
Z
5. [10 Points] Compute
z dz where γ is the straight line segment from 0 to 2i.
γ
Proof. For γ, we know that z(t) = t(2i) = 2ti for all t ∈ [0, 1], then dz = 2i dt and
Z
Z
z dz
1
2ti · 2i dt
=
0
γ
Z
=
−4
1
t dt
0
=
−2
6
MINGFENG ZHAO
6. [20 Points] As usual, we denote z = x + iy and f (z) = u + iv where x, y are real variables and u, v
are real valued functions, that is, x = Re z and y = Im z, u = Re f and v = Im f . Find a formula
for v if u(z) =
x2
y
and f is analytic in C\{0}. Can you find a simple formula for f ?
+ y2
Proof. Since u(z) =
x2
y
, then
+ y2
ux =
−2xy
,
+ y 2 )2
(x2
and uy =
x2 + y 2 − 2y 2
x2 − y 2
= 2
2
2
2
(x + y )
(x + y 2 )2
If f is analytic in C\{0}, then
ux = vy ,
and uy = −vx .
Consider the function f (z) = zi , then
i
i
x − iy
y + ix
=
=i 2
= 2
.
2
z
x + iy
x +y
x + y2
Then v(x, y) =
x2
x
+ C.
+ y2
SOLUTION OF TEST 2
7. [15 Points] Find a power series for
Proof. For
7
1
. What is its radius of convergence?
2i − z
1
, we have
2i − z
1
2i − z
=
1
1
·
z
2i 1 − 2i
=
∞
1 X z k
·
2i
2i
If |z| < 2
k=0
=
∞
X
k=0
zk
.
(2i)k+1
The radius of convergence is 2.
8
MINGFENG ZHAO
8. [15 Points] Find a power series for
Proof. For
1
1−z ,
1
. What is its radius of convergence?
(1 − z)2
then
∞
X
1
=1+
zk ,
1−z
∀|z| < 1.
k=1
Now differentiating the above identity, then
∞
X
1
=
kz k−1 ,
2
(1 − z)
∀|z| < 1.
k=1
The radius of convergence is 1.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu