MATH 185A HW 8-SOLUTIONS
JACOB MILLER
Problem 3.3.3
See solutions in back of text.
Problem 3.3.4-GRADED
(5 points each) Note using partial fractions that
1
1
1
1
=
+
.
z(z 1)(z 2)
2z z 1 2(z 2)
a) Since 0 < |z| < 1 we also have that 0 < |z/2| < 1 so using
MATH 185A HW 5-SOLUTIONS
JACOB MILLER
Problem 2.1.10-GRADED
On the arc given |z| = 2 so by the reverse triangle inequality:
1
1
1
1
= 2
2
= 2
2
z +1
|z + 1|
|z (1)|
|z | | 1|
=
1
1
1
=
= .
1|
|4 1|
3
|z|2
Alternatively on the arc given z = 2ei , 0 /2
MATH 185A HW 6-SOLUTIONS
JACOB MILLER
Problem 2.4.2-GRADED
part (a) is worth 6 points, part (b) is worth 4 points
a) We have using that f (z) = 1 is analytic
Z
Z 2
1/i
1/i
z 1
1
dz =
+
dz
2
zi z+i
z +1
Z
Z
Z
1/i
1/i
= dz
+
dz
z i
z +i
then using tha
MATH 185A HW 7-SOLUTIONS
JACOB MILLER
Problem 3.1.1
a) Does not converge, note for all n and for all L C that
|(1)n Re(L)| = |Re(zn L)| |zn L|,
and thus if zn converged to L that would imply that lim (1)n = Re(L), and we have
from real analysis that (1)n
MATH 185A HW 2-SOLUTIONS
1.2.2
(Note and 3 denote postive real roots)
a) z = 2 cos(/6 + k/3)
+
i
2 sin(/6 + k/3) for 0 k 5.
3
3
b) z = 4 cos(2k/3) + i 4 sin(2k/3) for 0 k 2.
1.2.9
Since w is an nth root of unity this means by defintion that wn = 1. Then (
Solutions to HW 3
Karol Koziol
February 8, 2011
p. 313
Note that there are two (essentially equivalent) ways of computing the residue of a meromorphic function f at a point z0 : one can use the method using limits and derivatives (Theorem 1 in Sa/Snider),
MATH 185A HW 3-SOLUTIONS
JACOB MILLER
1.4.3
Proof. Assume f is continuous and f (z0 ) 6= 0. Then |f (z0 ) 0| > 0 so define = |f (z0 ) 0|
and notice that by the definition of continuity there exists a > 0 such that |z z0 | <
since
= |f (z) f (z0 | < . In
MATH 185A HW 4-SOLUTIONS
JACOB MILLER
1.5.17
See solution in back of text.
1.5.28
If u and v are harmonic contugates then it suffices to find a v that satisfies:
u
v
u
v
= 3x2 3y 2 =
, and
= 6xy =
.
x
y
y
x
Thus
v(x, y) = 3x2 y y 3 + f (x), and v(x, y) =
Complex Analysis
Math 185A, Winter 2010
Sample Final Exam Questions
1. (a) Consider the change of variables from (x, y ) to (z, z ) given by
z = x iy.
z = x + iy,
(1)
If z denotes the partial derivative with respect to z keeping z xed and z
denotes the pa
185A HW7 Solutions
2.5.10) Show that u(x, y ) = log
x2 + y 2 is harmonic but that it has no harmonic conjugate on
C \ cfw_0.
Solution: u = log
x2 + y 2 = 1 log(x2 + y 2 ) ux =
2
x
2x2
1
uxx = 2
+2
.
2 + y2
2 )2
x
(x + y
x + y2
Thus, by symmetry, we have
185A HW2 Solutions
1.5.2) Determine the sets on which the following functions are analytic, and compute their
derivatives:
a) 3z 2 + 7z + 5 is analytic on C and its derivative is 6z + 7.
b) (2z + 3)4 is analytic on C and its derivative is 8(2z + 3)3 .
c)
185A HW4 Solutions
2.1.8) Evaluate
z 2 dz along two paths joining (0, 0) and (1, 1) as follows:
a) is the straight line joining (0, 0) to (1, 1).
Solution: Let (t) = (1 + i)t for 0 t 1. Then
1
[(1 i)t]2 (1+ i)dt
0
=
= (1 i)2 (1+ i)
z 2 dz =
12
t dt
0
1
0
185A HW9 Solutions
4.1.2) Find the residues of the following functions at the indicated points:
2
a) f (z ) =
ez
, z0 = 1
z1
2
ez0
Solution: By Prop 4.1.2, Res(f ; 1) =
= e.
1
2
ez
b) f (z ) =
, z0 = 0
(z 1)2
Solution: f is analytic near z0 = 0, so Res(f
Complex Analysis
Math 185A, Winter 2010
Midterm: Solutions
1. If a, b are complex numbers such that |a| < 1, |b| < 1, prove that
ab
< 1.
1 ab
Solution.
We have
ab
1 ab
2
=
(a b)( )
ab
|a|2 (a + ab) + |b|2
b
=
.
(1 ab)(1 a)
b
1 (a + ab) + |a|2 |b|2
b
(1)
185A HW8 Solutions
3.3.8) Prove, using the Taylor series, the following complex version of lHpitals rule: Let f (z )
o
and g (z ) be analytic, both having zeros of order k at z0 . The f (z )/g (z ) has a removable
singularity and
limzz0
f (z )
f (k) (z0 )
MATH 185A HW 9-SOLUTIONS
JACOB MILLER
Problem 4.2.3
The denominator is zero at z = 1 2i which is not contained in , hence the integral is
zero by Cauchys theorem.
Problem 4.2.4
The singularities of ez11 are at z = 2in, and Res( ez11 ; 2in) = 1 for all n Z