Download Honors problem 7: complex series.

Honors problem 7: complex series.
A complex series is a series
∞
X
cn
n=0
where each cn is a complex number (real series are a special case of complex series). We will write
cn = an + ibn , where an and bn are the real an imaginary parts of cn (and hence are real numbers).
The real and imaginary parts of the series above are then defined to be
∞
X
∞
X
an and
n=0
bn
n=0
respectively. We say that a complex series converges if and only if both the real and imaginary
parts converge. In this case, the sum is defined to be the sum of the real part plus i times the sum
of the imaginary part:
∞
∞
∞
X
X
X
cn =
an + i
bn
n=0
n=0
1. Prove that if the complex series
n=0
∞
X
cn
n=0
converges, then
lim |cn | = 0.
n→∞
(this is the same as saying
lim cn = 0
n→∞
which means the real and imaginary parts limit to zero.)
Observe that this gives an nth term test for divergence for complex series: if lim |cn | =
6 0, then
n→∞
the series diverges.
P
P
Solution.
We write cn = an + ibn , then to say that
cn converges means that
an and
P
th
bn converge.
But
then
lim
a
=
0
and
lim
b
=
0
by
the
n
term
test
for
real
series.
Since
n
n
p
|cn | = a2n + b2n , it follows that
q
p
√
lim |cn | = lim
a2n + b2n =
lim (a2n + b2n ) = 0 + 0 = 0.
n→∞
n→∞
n→∞
2. The complex series
∞
X
cn
n=0
is said to converge absolutely if the real series
∞
X
n=0
|cn |
converges. Prove that if a complex series converges absolutely, then it converges. Hint: observe
that |an | ≤ |cn | and |bn | ≤ |cn |.
P
P
P
Solution. Since |cn | converges, and |an | ≤ |cn | and |bn | ≤ |cn |, it follows that |an | and |bn |
convergesP(these arePreal series). Therefore, because absolutely convergent
real series converge, we
P
see that
an and
bn both converge. But then by definition,
cn also converges.
From this we see that all the tests for absolute convergence of real series also give tests for absolute
convergence of complex series.
A complex power series centered at a is an expression of the form
∞
X
cn (z − a)n
n=0
where cn and a are all allowed to be complex numbers and we think of z as a variable. For
simplicity, we just consider complex power series centered at 0.
We say that the complex power series
∞
X
cn z n
n=0
converges at w ∈ C if the power series converges when the complex number w is substituted in for
z:
∞
X
cn wn .
n=0
3. Prove that the complex geometric power series
∞
X
zn
n=0
converges absolutely for complex numbers z with |z| < 1 and diverges for |z| ≥ 1. Hint: make sure
you use both 1 and 2 above.
Solution. If |z| < 1, then
∞
X
|z n | =
n=0
∞
X
|z|n
n=0
converges (since
is geometric). Therefore, by 2 we
P n |z| is a real number less than 1 nand the series
know that
z converges. If |z| ≥ 1, then lim |z | = lim |z|n is not 0, so by 1, the series diverges.
4. Suppose that the power series
∞
X
cn z n
n=0
converges at w0 6= 0. By part (a) it follows that
lim |cn w0n | = 0.
n→∞
Prove that if |w| < |w0 |, then the power series converges absolutely at w. Hint: use the same idea
as for real power series.
P
Solution. This is identical to the proof for real power series. Assuming cn w0n converges, number
1 above implies lim |cn w0n | = 0, and hence there is some constant C > 0 so that |cn w0n | < C for all
n. Then if |w| < |w0 |, we have |w/w0 | < 1, and
n
n
n
w
n w nw n
|cn w0 | = cn w0 n = |cn w0 | ≤ C w0
w0
w0
P
P
Since
C|w/w0 |n converges by 3, it follows that
|cn wn | converges, and hence the power series
converges absolutely at w.
[[nothing to do here]] From this we obtain the analogue of Theorem 3 from Section 11.8 of the book:
Theorem Given a power series
∞
X
cn z n
n=0
exactly one of the following holds
1. the series converges only for z = 0 and diverges for all other z ∈ C,
2. the series converges for every z ∈ C,
3. there is a positive real number R > 0 so that the series converges absolutely for |z| < R and
diverges for |z| > R.
We can therefore define the radius of converges as 0 ≤ R ≤ ∞ (with R = 0 and R = ∞ in cases
(1) and (2), respectively) just as for real power series.
5. Prove that the following power series have infinite radius of convergence.
(a.)
∞
X
zn
n!
n=0
(b.)
∞
X
(−1)n z 2n+1
(2n + 1)!
n=0
(c.)
∞
X
(−1)n z 2n
(2n)!
n=0
We can now define ez , sin(z) and cos(z) to be the sums of these series, respectively for every
complex number z. From what we have done in class, we know that this agrees with the definitions
when z is a real number.
Solution. Taking absolute values, this is identical to the proof for the corresponding real power
series.
6. Using the previous problem and definitions, prove that for every complex number z we have
cos(z) =
eiz + e−iz
2
and
sin(z) =
eiz − e−iz
.
2i
and conclude that
eiz = cos(z) + i sin(z).
(As a comparison, recall that in the first honors problem, we defined eix = cos(x) + i sin(x) for a
real number x. Now we see that this equation is valid not just for real numbers, but for complex
numbers, as a consequence of the more natural, power series definitions.)
Solution. We have
eiz + e−iz
2
=
=
=
∞
∞
X
(iz)n X (−iz)n
+
n!
n!
n=0
n=0
1
2
∞ X
1
2
n
nz
(1 + (−1) )i
n=0
n
n!
!
!
1
=
2
∞ X
n=0
1
=
2
zn
(i + (−i) )
n!
n
n
!
!
∞ 2k
X
2k z
2i
(2k)!
k=0
X
∞ ∞ X
z 2k
z 2k
(i2 )k
=
(−1)k
= cos(z)
(2k)!
(2k)!
k=0
k=0
The computation for sin(z) is similar.
Finally, just compute
cos(z) + i sin(z) =
eiz + e−iz
eiz − e−iz
eiz + e−iz + eiz − eiz
+i
=
= eiz .
2
2i
2
7. We define the hyperbolic cosine and sine just as in the real case
cosh(z) =
ez + e−z
2
and
sinh(z) =
ez − e−z
.
2
Using these definitions and the previous problem, prove that the following relations hold between
the trig functions and the (simpler) hyperbolic trig functions.
cosh(iz) = cos(z)
cos(iz) = cosh(z)
sinh(iz) = i sin(z)
sin(iz) = i sinh(z).
and
Solution. These are simple substitutions
cosh(iz) =
eiz + e−iz
= cos(z)
2
ei(iz) + e−i(iz)
e−z + ez
=
= cosh(z)
2
2
eiz − e−iz
eiz − e−iz
=i
= i sin(z)
sinh(iz) =
2
2i
ei(iz) − e−i(iz)
e−z − ez
1 ez − e−z
ez − e−z
sin(iz) =
=
=−
=i
= i sinh(z).
2i
2i
i
2
2
cos(iz) =