MA671 Assignment #2
Morgan Schreffler
Page 1 of ??
∞
X
Exercise 1 Assume the power series f (z) =
Cn (z − z0 )n converges for 0 < |z − z0 | < R.
n=0
Show that f is continuous at z0 ; that is,
lim f (z) = f (z0 ).
∞
X
Proof. Recall that if a power series f (z) =
Cn (z − z0 )n converges when |z − z0 | < R,
z→z0
then the series f 0 (z) =
∞
X
n=0
nCn (z − z0 )n−1 is the derivative of f and also converges when
n=1
|z − z0 | < R. Thus, f is differentiable when |z − z0 | < R and, in particular, at z0 . Since
differentiability implies continuity, it follows that f is continuous at z0 .
Exercise 2 Verify the following properties of the exponential:
(a)
d
(ez )
dz
= ez
(b) ez1 ez2 = ez1 z2
(c) If z = x + iy then ez = ex (cos y + i sin y)
(d) ez+2πi = ez for all z ∈ C
(e) If z = x + iy, then |ez | = ex
(f ) ez 6= 0 for any z ∈ C
(g) The map t 7→ eit is a continuous one-to-one map of the interval [0, 2π) onto the unit
circle |z| = 1.
Proof.
z
(a) Recall that e =
∞
X
zn
n=0
n!
. It follows that
∞
∞
∞
X z n−1 X z n−1
X zn
d z
(e ) =
n
=
=
= ez .
dz
n!
(n
−
1)!
n!
n=1
n=1
n=0
(b) The nth term of the series for ez1 ez2 is
z2n
z1 z2n−1
z12 z2n−2
z1j z2n−j
z1n−1 z2
z1n
+
+
+ ... +
+ ... +
+ ,
n! (n − 1)! 2!(n − 2)!
j!(n − j)!
(n − 1)! n!
or equivalently,
z2n nz1 z2n−1
+
+
n!
n!
n!
z 2 z n−2
2!(n−2)! 1 2
n!
n
k
+ ... +
1
z1j z2n−j
nz n−1 z2 z1n
+ ... + 1
+ .
n!
n!
n!
MA671 Assignment #2
Morgan Schreffler
Page 2 of ??
(z1 + z2 )n
, so we have
By the Binomial Theorem, this is precisely
n!
! ∞
!
∞
∞
n
X
X zn
X
z
(z1 + z2 )n
1
2
z1 z2
e e =
=
= ez1 +z2 .
n!
n!
n!
n=0
n=0
n=0
(c) Knowing (b) and Euler’s formula eiθ = cos θ + i sin θ, it follows easily that
ez = ex+iy = ex eiy = ex (cos y + i sin y).
(d) Knowing (b) and (c), ez+2πi = ez e2πi = ez (cos 2π + i sin 2π) = ez (1 + i · 0) = ez .
(e) Since |eiy | = | cos y + i sin y| = cos2 y + sin2 y = 1, when z = x + iy we have
|ez | = |ex+iy | = |ex eiy | = |ex ||eiy | = ex .
(f ) Suppose ez = ex+iy = ex eiy = 0. Then either ex = 0 or eiy = 0. We know for all x ∈ R
that ex 6= 0, so eiy = cos y + i sin y = 0. But this would imply either cos y = −i sin y
or cos y = sin y = 0. But both of these options are clearly contradictory, since the first
implies that the sine or cosine function is complex-valued in R, and the second implies
cos2 y + sin2 y = 0 6= 1. Hence, ez = 0 is impossible for all z ∈ C.
(g) We already established that the map f (t) = eit maps to the unit circle in part (e).
Now, for t1 , t2 ∈ [0, 2π), suppose f (t1 ) = f (t2 ). Then cos t1 + i sin t1 = cos t2 + i sin t2 ,
which implies that cos t1 = cos t2 and sint1 = sint2 . The first equation tells us that
either t1 = t2 or, without loss of generality, t1 ∈ [0, π) and t2 ∈ (π, 2π). But this is a
contradiction, since this would imply that sin(t1 ) ≥ 0 and sin(t2 ) < 0. Thus, t1 = t2
and f is one-to-one.
Now, let z be a point on the unit circle. In polar coordinates, this point can be
written z = (1, θ) for some θ ∈ [0, 2π). Converting this into rectangular coordinates
we have x = 1 cos θ and y = 1 sin θ. Placing this point on the complex plane yeilds
z = cos θ + i sin θ = eiθ = f (θ). Therefore, every point on the unit circle |z| = 1
corresponds to a point θ ∈ [0, 2π), which makes f surjective.
To show f is continuous, we will show that the preimage of a basic open set of the
unit circle is open. Note that a basic open set O in the unit circle is the intersection
of the unit circle and an open ball in C. The boundary of this ball can intersect the
unit circle in at most two places. If it intersects at less than two points, O = ∅. If it
intersects at exactly two points (call them cos t1 + i sin t1 and cos t2 + i sin t2 , t1 < t2 ),
then f −1 (O) is either (t1 , t2 ) or [0, t1 ) ∪ (t2 , 2π), both of which are open in [0, 2π).
2
MA671 Assignment #2
Morgan Schreffler
Page 3 of ??
Exercise 3 Find all complex numbers z such that cos z = 0, and sin z = 0.
eiz + e−iz
, in order for cos z = 0
Proof. Since the cosine is defined in C to be cos z =
2
iz
−iz
2iz
iπ+2πin
we need e = −e , or equivalently e = −1. Recalling that e
= −1, n ∈ Z, we
have 2iz = iπ + 2πin. This implies that z =
cos x = 0, x ∈ R.
π
2
+ nπ. Hence, cos z = 0 precisely when
eiz − e−iz
, in order for sin z = 0
2i
= 1. Recalling that e2πi+2πi(n−1) = 1, n ∈ Z, we
Similarly, since the sine is defined in C to be sin z =
we need eiz = e−iz , or equivalently e2iz
have 2iz = 2πi + 2πi(n − 1). This implies that z = nπ. Hence, sin z = 0 precisely when
sin x = 0, x ∈ R.
Exercise 4 Is the identity sin2 z + cos2 z = 1 valid for all z ∈ C? Explain!
Solution. The answer is “yes!” Based on the definitions of sin z and cos z, we have
iz
2 iz
2
e − e−iz
e + e−iz
2
2
sin z + cos z =
+
2i
2
2iz
−2iz
2iz
e −2+e
e + 2 + e−2iz
=
+
−4
4
2+2
= 1.
=
4
Exercise 5 Prove that the series
∞
X
zn
n=1
n
converges at every point of the unit circle |z| = 1
except at the point z = 1.
Proof. Notice that for any k ≥ 0 and any z 6= 1,
zk =
z k (1 − z)
z k − z k+1
z k − 1 + 1 − z k+1
1 − z k+1 1 − z k
=
=
=
−
.
1−z
1−z
1−z
1−z
1−z
It follows that, for any N > 0,
N
X
zn
n=1
n
=
N
X
n=1
1−z n+1
1−z
−
n
1−z n
1−z
=
N
X
1−z n+1
1−z
n=1
n
N
X
−
n=1
1−z n
1−z
n
=
N −1
N −1
1 − z N +1 X 1 − z n+1 X
1 − z n+1
+
−
.
N (1 − z) n=1 n(1 − z) n=0 (n + 1)(1 − z)
a
a
Now, note that the first term in the last sum equals 1. Also, note that for any a, −
=
n n+1
a
. Thus, our formula is reduced to
n(n + 1)
N −1
N −1
N −1
1 − z N +1 X 1 − z n+1 X
1 − z n+1
1 − z N +1 X
1 − z n+1
1+
+
−
=1+
+
.
N (1 − z) n=1 n(1 − z) n=1 (n + 1)(1 − z)
N (1 − z) n=1 n(n + 1)(1 − z)
3
MA671 Assignment #2
Morgan Schreffler
Page 4 of ??
1 − z N +1 2
1 − z N +1 ≤ 1+
≤
1
+
, which tends to 1 as N goes to
Now, 1 +
N (1 − z)
N (1 − z)
N |1 − z|
infinity. Finally, notice that as N → ∞,
N
−1
X
n=1
and
∞
X
1 − z n+1
1 − z n+1
becomes
,
n(n + 1)(1 − z)
n(n + 1)(1 − z)
n=1
∞
∞
∞ X
X
1 − z n+1
2
1 − z n+1
1 X
,
≤
≤
n(n + 1)(1 − z) n=1 n(n + 1)(1 − z)
|1 − z| n=1 n(n + 1)
n=1
which equals
2
due to the telescoping nature of the series. Hence, we have shown that
|1 − z|
∞
X
zn
n=1
n
≤1+
2
,
|1 − z|
which is finite by our assumption that z 6= 1.
Exercise 6 Prove that if a function F and its conjugate F̄ are both analytic in a region Ω
then F ≡ C for come constant C ∈ C.
∞
∞
X
X
n
Proof. Let F (z) =
Cn (z − z0 ) . Then F̄ (z) =
Cn (z − z0 )n . Now, the sum of
n=0
n=0
two analytic functions is also analytic, so (F + F̄ )(z) = 2Re(F (z)) is analytic. But this
is a real-valued function, so F + F̄ is a constant function. This implies that 2Re(Cn ) =
Re(Cn ) = 0 for n ≥ 1. We also know that the difference of two analytic functions, as well as
a constant multiple of an analytic function, is analytic, so now let us consider i(F − F̄ )(z) =
2i2 Im(F (z)) = −2Im(F (z)). This is also a real-valued function, which is necessarily constant
on Ω. Hence, for n ≥ 1, −2Im(Cn ) = Im(Cn ) = 0. We now have that Re(Cn ) = Im(Cn ) = 0
for all n ≥ 1, so F ≡ C0 on Ω.
Exercise 7 Show that if F is analytic in |z| < 1, continuous on |z| ≤ 1, and if F ≡ 0 on
some boundary arc I, then F ≡ 0 in |z| < 1.
Proof.
4