part 3CV tutorial 4 part A
Surname: Nkele
StudentID: nklsip004
May 10, 2020
1
(a)
w = z + 1 − 2i
where z = x + iy
and w = u + iv
so u + iv = x + iy + 1 − 2i = (x + 1) + (y − 2)i

u = x + 1
v = y − 2
so the rectangular region is then given by the figure below x = 0, y = 0, y = 1, x = 2
make up the rectangular region
Figure 1: z-plane
x=0:u=0+1=1
x=2:v =2+1=3
for the y values we get that
y = 0 : v = 0 − 2 = −2
1
y = 1 : v = 1 − 2 = −1
so the on the w-plane the we have the figure below.
Figure 2: phase plot
(b)
√ iπ
w = 2 4 z, where z = x + iy
√
π
= 2(cos(
+ isin( π4 ))z
√ √2 4 √2
= 2( 2 + i 2 )(x + iy)
w = (1 + i)(x + iy)
u + iv = (x − y) + (x + y)i

u = x − y
v = x + y
(
)
u = −y
x =0→
→u = -v
v=y
(
)
u=2−y
x=2→
→u = -v+4
v =2+y
(
)
u=x
y=0→
→v =u
v=x
)
(
u=x−1
y=1→
→v = u+2
v =x+1
(1)
the w-plane below is then as follows
2
Figure 3: phase plot
2
Proof. below is the figure of two concentric circles C1 and C2 , with r1 > r2
Figure 4: phase plot
so if f (z) is analytic within the region R, then we have the Cauchy integral formula as
1
f (z0 ) =
2iπ
I
R
3
f (z)
dz
z − z0
(2)
doing a crosscut on the above figure then gives us the following figure
Figure 5: phase plot
where R" = C1 U L2 U#C2 U L2
H f (z)
1
f (z0 ) = 2πi
R z−z0 dz
"
R f (z)
H
H f (z)
1
dz
+
dz
−
= 2πi C1 z−z
z−z
L
C2
0
0
1
since
f (z)
dz
L2 z−z
"0
R
f (z0 ) =
1
2πi
→−
f (z)
C1 z−z0
H
R
f (z)
L1 z−z0 dz
dz −
dz −
R
f (z)
L2 z−z0 dz
then they cancel each other out to get the following results
#
f (z)
C2 z−z0
H
#
f (z)
z−z0
dz
3
(a)
Proof. if the curve C has the length L and |f (z)| ≤ M on C then
R
R
R
C f (z)dz ≤ C |f (z)||dz| ≤ C M |dz| = M L
R
C
f (z)dz ≤ M L
(b)
Proof. if f and |f (z)| < M is analytic then its derivative with respect to a is given by the
Cauchy integral formula
4
n!
f (a) =
2πi
I
n!
2πi
Z
n
|f n (a)| ≤
≤
M n!
2πirn+1
C
|f (z)|
|dz|
|z − a|n+1
C
|dz|
rn+1
Z
n!M
2πi
≤
f (z)
dz
(z − a)n+1
C
Z
M n!
M n!
(2πr) = n
n+1
2πir
r
|dz| =
C
so
|f n (a)| ≤
M n!
rn
4
(a)
f (z) =
sin(πz 2 )+cos(πz 2 )
(z−1)(z−2)
since z=1 and z=2 are on the interior of |z| = 3 then I is not 0.
I
I=
C1
I
=
C1
f (z)
dz +
z−1
sin(πz 2 )+cos(πz 2 )
z−2
z−1
I
C1
I
+
f (z)
dz
z−2
sin(πz 2 )+cos(πz 2 )
z−1
C2
z−2
= 2πif (1) + 2πif (2)
0−1
0+1
= 2πi
+ 2πi
−1
1
2πi + 2πi = 4πi
(b)
f n (a) =
n!
2πi
I
C
f (z)
dz
(z − a)n+1
n = 3 and z = -1
f (z) = e2z
0
f (z) = 2e2z , f 2(z) = 4e2z , f 3 (z) = 8e2z
I
C
f (z)
2πif 3 (z)
2πi(8e2(−1) )
8πie−2
dz
=
=
=
(z + 1)4
3!
3!
3
5