Math 311
Fall 2013
Midterm 2
11/12/2013
Lecturer: Jesus Martinez Garcia
Time Limit: 75 minutes
Name (Print):
Teaching Assistant
This exam contains 5 pages (including this cover page) and 4 problems. Check to see if any pages
are missing. Enter all request
aj +1 (z z0 )j +1

aj (z z0 )j
j
5.3.2. lim 
= lim 
j
aj +1
aj z
z0  = Lz z0 
The ratio test implies that the series is convergent for Lz z0  < 1,
yielding a radius of convergence of
1
L
5.3.3.(a)z  = 1(b)z 1 = 1 (c)z = 0(d)z i = 3(e)z
4.2.6.(c) 24i
2
3R
4.2.14.(a) 3 (b) ee 1 2 (c) (d)1
R
4
4
4.2.16.Let z(t) be an admissible parametrization of with z (a) = and z (b) =
.
2
2
b
t2
zdz = a z (t)z (t)dt = [z(2)] b = .
a
2
4.3.6.Let C0 be the portion of C extending from to .
1
1
dz = lim C0
4.4.5.z (s, t) = (2 s)cos2t + i(3 2s)sin2t, 0 s 1, 0 t 1.
4.4.10.(a)C\cfw_5i, 5i
(b)C
(c)C\cfw_3 i, 3 + i
(d)C\cfw_x + iy : x 3, y = 0
(e)C\cfw_(2k + 1), k = 0, 1,
In each case the integral is 0 by Cauchys Integral Theorem
because the functions are analy
4.6.10.f (z ) = i
4.6.14.If f is nonzero then 1/f (z ) is analytic in D and
1/f (z ) attains its maximum value on the boundary of D, so that
f (z ) attains its maximum value on the boundary of D.
example: f(z)=z on the unit disk z  < 1.
f (z ) att
4.1.2.Consider z (t) = t2 + it3 , 1 t 1.
z (t) = 2t + 3it3 vanishes when t = 0.
2
x = t2 , y = t3 , x = y 3 , 1 y 1.
dx
2 1
3 has a cusp (0, 0).
dy = 3 y
4.1.3.x = acost, y = bsint, 0 t 2.
z (t) = acost + ibsint.
z (t) = asint + ibcost = 0.
so its an admi
4.1.11.15
d
4.1.14. c z2 (s)ds =
b
a
d
c
z1 (s) (s)ds =
z1 (t)dt(t = (s)
4.2.2.(1) [f (z )g (z )]dz = lim
lim
n
k=1
n
g (ck )zk =
f (z )dz
d
c
z1 (s)d(s)( (s) > 0) =
n
k=1 (f (ck )g (ck )zk
g (z )dz
= lim
n
n
k=1
f (ck )zk
n
n
lim
(2) cf (z )dz
3.3.5.(c)ez = 1 3i , z = i( 2 + 2k ), i( 4 + 2k ), k = 0, 1,
2
3
3
3.3.9.z = x + iy, 4 + i x iy D x 4, y = 1.
so Log (4 + i z ) is analytic in C\cfw_x + i : x 4
f (z ) = 4+1z (1) = z1 i
i
4
1
3.3.15.w = Logz
3.4.3.Let (x, y ) = A(x + y ) + B, then is ha
(z )
2.3.14 lim f (z)
z z0 g
=
lim f (z)f (z0 )
z z0 g (z )g (z0 )
f (z )f (z0 )
z z0
g (z )g (z0 )
z z0
z z0
= lim
lim
=
f (z )f (z0 )
z z0
lim
g (z )g (z0 )
z z0
z z0
z z0
=
f (z0 )
g (z0 )
2.4.16.(a)x = + , y = 2i
2
1 x
1 y
i y
i
x
= 2 , = 2 , = 2 , =
1.6.17.N o
1.6.18 S and T are domains S and T are open and connected.
(1)S T is open: z0 S T, z0 S or z0 T, suppose z0 S.
S is open, S contains come circular neighborhood z z0  < , so does S T.
S T is open.
(2)S T is connected: z S T.z1 S, z2 T,
z1 and
1
1.1.10. 4225 (253 + 204i)
4
1.1.22.z 16 = (z 2 4)(z 2 + 4) = 0
z = 2, 2, 2i, 2i.
9
1.2.7.(g )(x 9 )2 + y 2 = 64
8
n
n
1.2.17.z0 is a root of the polymonial equation z0 + a1 z0 1 + + an = 0
n = z n (from z + z = z +
since z1 + z2 + + zn = z1 + z2 + + zn
Math 311
Fall 2013
Midterm 1
10/08/12
Lecturer: Jesus Martinez Garcia
Time Limit: 70 minutes
Name (Print):
Teaching Assistant
This exam contains 6 pages (including this cover page) and 4 problems. Check to see if any pages
are missing. Enter all requested