Math 185 – Homework 11 Solutions
Note: All problems refer to the 9th edition of Brown and Churchill. The corresponding
problems in the 8th edition are in parentheses.
1. Let C denote the unit circle |z| = 1 parametrized in the positive sense. Determine
∆C arg f (z) for (a) f (z) = z n , (b) f (z) = (3z + 1)4 /z 3 , and (c) f (z) = (z − 5)10 /(2z − 1).
2. Suppose that f is analytic inside and on a positively oriented simple closed contour C
and that it has no zeros on C. Show that that if f has zeros at z1 , . . . , zn inside C with
multiplicities m1 , . . . , mn , then
Z
C
n
X
zf 0 (z)
dz = 2πi
mk z k .
f (z)
k=1
Proof. Consider the zero zk of multiplicity mk . We can write
f (z) = (z − zk )mk ϕ(z),
where ϕ is analytic at zk and ϕ(zk ) 6= 0. Therefore
f 0 (zk ) = mk (z − zk )mk −1 ϕ(z) + (z − zk )mk ϕ0 (z),
and we have
zmk
ϕ0 (z)
zk mk
ϕ0 (z)
zf 0 (z)
=
+
=
+ mk +
,
f (z)
z − zk
ϕ(z)
z − zk
ϕ(z)
in a punctured disk 0 < |z − zk | < ε for small enough ε > 0. The expression mk +
ϕ0 (z)/ϕ(z) is analytic at zk , hence zf 0 (z)/f (z) has a simple pole at zk with residue
zf 0 (z)
= mk zk .
z=zk f (z)
Res
The result now follows from Cauchy’s Residue Theorem.
3. Determine the number of zeros, counting multiplicities, of the polynomial
z 6 + z 5 + 7z 4 + z 3 + z 2 + z + 1
inside the circle |z| = 1.
Solution. Let f (z) = 7z 4 and g(z) = z 6 + z 5 + z 3 + z 2 + z 1 + 1. Notice that for |z| = 1,
|f (z)| = 7 and |g(z)| ≤ 6 < |f (z)|.
Therefore, f and f + g have the same number of zeros, counting multiplicities, inside
|z| = 1. Since f has only one zero at z = 0 with multiplicity m = 4, the polynomial in
question has 4 zeros, counting multiplicity, inside the circle |z| = 1.
4. Find the image of the strip
n
o
S = (x, y) : 0 < x ≤ 2 and y > 0
under the mapping w = iz + 1. Sketch the strip and the image in the complex domain.
5. Show that when a circle is transformed into a circle under the transformation w = 1/z,
the center of the original circle is never mapped to the center of the image circle.
Solution. Consider a circle
A(x2 + y 2 ) + Bx2 + Cy 2 + D = 0,
where A =6= 0, and B 2 +C 2 > 4AD. The center of the circle is at z0 = (−B/2A, −C/2A).
Under the mapping w = 1/z, this circle is mapped to the circle (or line)
D(u2 + v 2 ) + Bu2 − Cv 2 + A = 0
in the uv plane. For the image to be a circle, we need also assume that D 6= 0. The
center of the image circle is w0 = (−B/2D, C/2D).
We have two cases. 1: Assume at least one of B or C is non-zero (so that the circle is
not centered at the origin).The center of the original circle is mapped to the point
B
C
− 2A
+ 2A
i
1
z0
−2AB
2AC
=
=
= 2
+ 2
i.
2
2
B
C2
z0
|z0 |2
B
+
C
B
+ C2
2 +
2
4A
4A
If B 6= 0, then using B 2 + C 2 6= 4AD we have
−2AB
−2AB
B
6=
=−
.
B2 + C 2
4AD
2D
which shows that the x coordinate of the center of the original circle is not mapped to
the u coordinate of the center of the image circle, verifying the claim. If C 6= 0, then we
have
2AC
2AC
C
6=
=
,
2
2
B +C
4AD
2D
which verifies the claim.
2: Both B and C are zero. Then the original circle is centered at z0 = 0. The point
z = 0 is mapped to the point at infinity by w = 1/z, and not the center of the image
circle, which is w0 = 0.
6. Show that the transformation
1
z
maps circles of the form |z| = R onto ellipses when R 6= 1. What happens when R = 1?
[Hint: Use the polar form z = reiθ . ]
w=z+
2
Solution. Points on the circle |z| = R can be expressed in polar form z = Reiθ for
0 ≤ θ ≤ 2π. Thus
1
1 −iθ
1
iθ
w = Re + e
cos(θ) +i R −
sin(θ) .
= R+
R
R
R
{z
} |
{z
}
|
u
v
When R 6= 1, this is the expression for an ellipse in the uv-plane. When R = 1, we have
v = 0 and u = 2 cos(θ) for 0 ≤ θ ≤ 2π. Thus, the circle |z| = 1 is mapped onto the
segment of the u-axis −2 ≤ u ≤ 2.
7. Find the linear fractional transformation mapping distinct points z1 , z2 , z3 onto the
points w1 = 0, w2 = 1, w3 = ∞.
Solution. Recall the implicit form:
(z − z1 )(z2 − z3 )
(w − w1 )(w2 − w3 )
=
.
(w − w3 )(w2 − w1 )
(z − z3 )(z2 − z1 )
Taking the limit as w3 → ∞ and noting that w1 = 0 and w2 − w1 = 1 we have
w=
(z − z1 )(z2 − z3 )
az + b
=
,
(z − z3 )(z2 − z1 )
cz + d
where
a = z2 − z3 , b = −z1 (z2 − z3 ), c = z2 − z1 , and d = −z3 (z2 − z1 ).
8. Show that the composition of two linear fractional transformations is a linear fractional
transformation. [Hint: You need to verfiy the ad − bc 6= 0 property for the composition.]
Solution. Let
S(z) =
az + b
,
cz + d
and T (z) =
ez + f
,
gz + h
where ad − bc 6= 0 and eh − f g 6= 0. Composing S and T we have
S(T (z)) =
a·
c·
ez+f
gz+h
ez+f
gz+h
+b
+d
=
(ae + gb)z + af + bh
aez + af + gbz + bh
=
.
cez + cf + dgz + dh
(ce + dg)z + cf + dh
This is certainly in the form of an LFT. We just need to verify it is not constant. We
have
(ae + gb)(cf + dh) − (af + bh)(ce + dg)
= aecf + gbcf + aedh + gbdh − af ce − bhce − af dg − bhdg
= (ad − bc)eh + (bc − ad)f g
= (ad − bc)(eh − f g).
3
Since ad − bc 6= 0 and eh − f g 6= 0, we see that S ◦ T is an LFT.
A more clever way to do this problem is to associate to an LFT S(z) = (az + b)/(cz + d)
the matrix
a b
AS =
.
c d
The condition ad − bc 6= 0 is then recast as det(AS ) 6= 0. Inspecting the formula for the
composition of S and T , we see that the matrix for the LFT S ◦ T is given by
AS◦T = AS · AT .
Then we have
det(AS◦T ) = det(AS · AT ) = det(AS )det(AT ) 6= 0,
which shows that S ◦ T is a bona fide LFT.
9. (a) Show that the linear fractional transformation
w=i
1−z
1+z
maps the disk |z| ≤ 1 onto the half plane Im(w) ≥ 0. [Hint: Show that the inverse
of the linear fractional transformation above is
i−w
iπ w − i
=e
z=
.]
i+w
w−i
Solution. To check that the transformation above is indeed the inverse, we compute
1−z
i − i 1+z
i+
i 1−z
1+z
=
2iz
i(1 + z) − i(1 − z)
=
= z.
i(1 + z) + i(1 − z)
2i
The mapping w 7→ z above maps the upper half of the w plane onto the unit disk
|z| < 1, and maps the real axis in the w plane onto the boundary of the circle
|z| = 1. Being the inverse of this transformation, the mapping z 7→ w maps the
disk |z| ≤ 1 onto the closed upper half plane.
(b) Show that the linear fractional transformation
w=
z−2
z
maps the disk |z − 1| ≤ 1 onto the left half plane Re(w) ≤ 0. [Hint: Observe that
this linear fractional transformation can be expressed as the composition of
Z = z − 1, W = i
and appeal to part (a).]
4
1−Z
, w = iW,
1+Z
Solution. To check the decomposition above, write
w = iW =
z−1−1
z−2
Z −1
=
=
.
Z +1
z−1+1
z
Using this decomposition, we have that the disk |z − 1| ≤ 1 is mapped onto the
disk |Z| ≤ 1, which is then mapped onto the half plane Im(W ) ≥ 0, by part (a). If
W = U + iV , then the final mapping
w = iW = −V + iU,
maps the half plane V = Im(W ) ≥ 0 to the left half plane Re(w) ≤ 0.
10. Let z1 , z2 , z3 be distinct points in the complex plane, and let w1 , w2 , w3 be another set
of distinct points in the complex plane. Show that there is exactly one linear fractional
transformation T satisfying wi = T (zi ) for i = 1, 2, 3. [Hint: We showed existence
of such a linear fractional transformation in class. To show uniqueness, let T1 and T2
be two linear fractional transformations satisfying the property above. Observe that
S(z) := T1−1 (T2 (z)) satisfies S(zi ) = zi for i = 1, 2, 3, i.e., S has three distinct fixed
points. Then show that any linear fractional transformation that has three distinct fixed
points is the identity, i.e., S(z) = z. ]
Proof. Note: It is not enough to use the implicit form to deduce that S(z) = z. This only
shows that the identity is an LFT with three (actually infinitely many) distinct fixed
points, which is already obvious. We never showed that every LFT can be expressed in
the implicit form, and this is what the problem is getting at.
Let T1 and T2 be linear fractional transformations mapping zi to wi for i = 1, 2, 3. Let
S(z) = T1−1 (T2 (z)).
By Problem 8, S is a linear fractional transformation, and we can see that S(zi ) = zi
for i = 1, 2, 3.
Let
S(z) =
az + b
,
cz + d
and consider the fixed point equation
z = S(z) =
az + b
.
cz + d
Simplifying, we see that any fixed point of S satisfies
cz 2 + (d − a)z − b = 0.
(1)
Assume first that c 6= 0. Then (1) is a quadratic equation, which has at most two distinct
roots. This contradicts the fact that each of the three distinct fixed points of S are roots
of (1). Therefore c = 0.
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Suppose now that a 6= d. Then (1) becomes
(d − a)z = b,
which has the unique solution z = b/(d − a), contradicting again the fact that S has
three distinct fixed points. Therefore a = d and c = 0. Equation (1) and the existence
of a fixed point of S shows that b = 0, which yields
S(z) =
a
b
z + = z.
d
d
Hence S is the identity tranformation and therefore T2 (z) = T1 (S(z)) = T1 (z).
Suggested problems: 94.1 (87.1), 94.2 (87.3), 94.9 (87.9), 96.2 (90.2), 96.5 (90.4), 98.2
(92.2), 98.5 (92.5), 98.10 (92.10), 98.11 (92.11), 100.1 (94.1), 100.8 (94.8), 100.11 (94.11),
100.12 (94.12)
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