Homework #3 Solutions
Math 128, Fall 2013
Instructor: Dr. Doreen De Leon
1
p. 44: 3(a), 7
π
3(a) Sketch the region onto which the sector r ≤ 1, 0 ≤ θ ≤ is mapped by the transformation w = z 2 .
4
Solution: Since we are considering a sector of a circle in the z plane, we consider the exponential
representation of z, z = reiθ . Then,
w = z 2 becomes w = r2 ei(2θ) .
So, if we represent w in exponential notation as w = ρeiϕ , we get
ρ = r2 and ϕ = 2θ.
So, the points z = r0 eiθ0 on a circle r = r0 in the z plane are transformed into points w = r02 ei(2θ)
π
on the circle ρ = r02 . Since we consider only the wedge in which 0 ≤ θ ≤ , this is transformed into
4
π
2
the wedge of radius ρ = r , with 0 ≤ ϕ ≤ , for 0 < r ≤ 1. The point z = 0 maps onto w = 0.
2
z
w
π
4
1
1
7. Find the image of the semi-infinite strip x ≥ 0, 0 ≤ y ≤ π, under the transformation w = ez , and
label corresponding portions of the boundaries.
Solution: Recall: We can write w = ez as w = ex eiy . So, if w = ρeiϕ , then ρ = ex and ϕ = y.
So, the vertical line x = 0 will be mapped to
w = eiy , 0 ≤ y ≤ π,
1
or the arc of a circle of radius 1, 0 ≤ ϕ ≤ π.
All other verical lines x = c will be mapped to w = ec eiy , 0 ≤ y ≤ π. So,
ρ = ec and 0 ≤ ϕ ≤ π.
The horizontal rays y = c1 are mapped to rays ϕ = c1 originating from the arc w = eiϕ , 0 ≤ ϕ ≤ π.
So, the ray y = 0, x ≥ 0 is mapped to the horizontal ray v = 0, π ≤ u and the ray y = π, x ≥ 0 is
mapped to the vertical ray u = 0, v ≥ π.
B π
A
2
B’
A’
p. 55-56: 3, 5, 10
3. Let n be a positive integer and let P (z) and Q(z) be polynomials, where Q(z0 ) ̸= 0. Use Theorem
2 in Section 16, as well as limits appearing in that section, to find
1
(z0 ̸= 0)
z→z0 z n
Solution:
(a) lim
lim
z→z0
1
limz→z0 1
1
=
= n.
n
n
z
limz→z0 z
z0
iz 3 − 1
z→i z + i
Solution:
(b) lim
limz→i (iz 3 − 1)
iz 3 − 1
=
z→i z + i
limz→i (z + i)
i limz→i z 3 − limz→i 1
=
limz→i z + limz→i 1
i(i3 ) − 1
= 0.
=
i+1
lim
P (z)
Q(z)
Solution:
(c) lim
z→z0
lim
z→z0
P (z)
limz→z0 P (z)
P (z0 )
=
=
.
Q(z)
limz→z0 Q(z)
Q(z0 )
2
5. Show that the limit of the function f (z) =
( z )2
as z tends to 0 does not exist. Do this by letting
z
nonzero points z = (x, 0) and z = (x, x) approach the origin.
Solution:
( z )2
f (z) =
z
(
)
x + iy 2
=
x − iy
2
x − y 2 + 2xyi
= 2
.
x − y 2 − 2xyi
Along z = (x, 0), x ̸= 0:
f (z) =
x2
= 1 =⇒ lim f (z) = 1.
z→0
x2
Along z = (x, y), x ̸= 0:
f (z) =
x2 − x2 + 2x2 i
= −1 =⇒ lim f (z) = −1.
z→0
x2 − x2 − 2x2 i
Since a limit is unique, we must conclude that the limit of f (z) as z goes to 0 does not exist.
Note: It is not sufficient to simply consider points z = (x, 0) and z = (0, y), because when z = (0, y),
f (z) = 1.
10. Use the theorem in Section 17 to show that
4z 2
=4
z→∞ (z − 1)2
(a) lim
4z 2
(z − 1)2
( )2
( )
4 z1
1
=⇒ f
= (( )
)2
1
z
z −1
4
)
= 2( 1
z z 2 − z2 + 1
4
=
.
1 − 2z + z 2
f (z) =
Then,
( )
1
lim f (z) = lim f
z→∞
z→0
z
4
= lim
= 4. X
z→0 1 − 2z + z 2
1
=∞
z→1 (z − 1)3
(b) lim
f (z) =
=⇒
1
(z − 1)3
1
= (z − 1)3 .
f (z)
3
Then,
1
= lim (z − 1)3 = 0.
z→1 f (z)
z→1
=⇒ lim f (z) = ∞. X
lim
z→1
z2 + 1
=∞
z→∞ z − 1
(c) lim
z2 + 1
z−1
( ) ( 1 )2
+1
1
=⇒ f
= z1
z
z −1
f (z) =
=
1 + z2
.
z − z2
Then,
1
z − z2
( 1 ) = lim
=0
z→0 f
z→0 1 + z 2
z
lim
=⇒ lim f (z) = ∞. X
z→∞
3
p. 62-3: 3, 1, 5, 7
3. Apply definition (3), Section 19, of derivative to give a direct proof that
dw
1
1
= − 2 when w = (z ̸= 0).
dz
z
z
dw
△w
= lim
△z→0 △z
dz
= lim
△z→0
= lim
1
z+△z
−
1
z
△z
z−(z+△z)
z(z+△z)
△z
−△z
= lim
△z→0 △zz(z + △z)
−1
= lim
△z→0 z(z + △z)
1
= − 2.X
z
△z→0
1. Use results in Section 20 to find f ′ (z) when
4
(a) f (z) = 3z 2 − 2z + 4
d 2
d
d
z −2 z+ 4
dz
dz
dz
= 3(2z) − 2.
f ′ (z) = 3
So, f ′ (z) = 6z − 2 .
(b) f (z) = (1 − 4z 2 )3
d
(1 − 4z 2 )
dz
= 3(1 − 4z 2 )2 (−8z).
f ′ (z) = 3(1 − 4z 2 )2 ·
So, f ′ (z) = −24z(1 − 4z 2 )2 .
(
)
z−1
1
(c) f (z) =
z ̸=
2z + 1
2
[
]
[d
]
− 1) [2z + 1] − (z − 1) dz
(2z + 1)
f (z) =
(2z + 1)2
1(2z + 1) − (z − 1)2
.
=
(2z + 1)2
′
So, f ′ (z) =
(d) f (z) =
d
dz (z
3
.
(2z + 1)2
(1 + z 2 )4
(z ̸= 0)
z2
[
′
f (z) =
=
=
=
So, f ′ (z) =
]
[ d 2]
+ z 2 )4 z 2 − (1 + z 2 )4 dz
z
2
2
(z )
2
3
4(1 + z ) (2z)(z 2 ) − (1 + z 2 )4 (2z)
z4
3
2
3
8z (1 + z ) − 2z(1 + z 2 )4
z4
2
2
3
8z (1 + z ) − 2(1 + z 2 )4
.
z3
d
dz (1
2(1 + z 2 )3 (3z 2 − 1)
.
z3
5
5. Derive formula (3), Section 20, for the derivative of the sum of two functions.
d
[(f + g)(z + △z) − (f + g)(z)]
[f (z) + g(z)] = lim
△z→0
dz
△z
[(f (z + △z) + g(z + △z) − (f (z) + g(z))]
= lim
△z→0
△z
f (z + △z) + g(z + △z) − f (z) − g(z)
= lim
△z→0
△z
(f (z + △z) − f (z)) + (g(z + △z) − g(z))
= lim
△z→0
△z
(
)
f (z + △z) − f (z) g(z + △z) − g(z)
= lim
+
△z→0
△z
△z
f (z + △z) − f (z)
g(z + △z) − g(z)
= lim
+ lim
△z→0
△z→0
△z
△z
′
′
= f (z) + g (z). X
7. Prove that the expression (2), Section 20, for the derivative of z n remains valid when n is a negative
integer, provided z ̸= 0.
Let f (z) = z n , where n is a negative integer.
Let m = −n. Then, m is a positive integer and
z n = z −m =
So, f (z) =
1
. Then,
zm
1
.
zm
(
)
1
zm
] m [d m ]
dz (1) z − dz (z ) (1)
=
(z m )2
0(z m ) − mz m−1
=
z 2m
1
= −m · 2m−m+1
z
−m
= m+1
z
= −mz −(m+1)
d
df
=
dz
dz
[d
= −mz −m−1 .
Writing this expression in terms of n gives
df
= nz n−1 .
dz
So,
d n
z = nz n−1 if n is a negative integer and z ̸= 0.
dz
6