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3. Power series and elementary analytic functions
§3.1 Recap on convergence and absolute convergence of series
Recall that we have already discussed what it means for an infinite sequence of complex numbers to converge. Recall that if sn ∈ C then we say that sn converges to s ∈ C if for all ε > 0 there exists N ∈ N such that |s − sn| < ε for all n ≥ N. Let zk ∈ C. We say that the series
k=0 zk^ converges if the sequence of partial sums sn =
∑n k=0 zk^ converges. The limit of this sequence of partial sums is called the^ sum^ of the series. A series which does not converge is called divergent.
Remark. One can show (see Exercise 3.1) that
∑∞^ n=0^ zn^ is convergent if, and only if, both n=0 Re(zn) and^
n=0 Im(zn) are convergent. We will need a stronger property than just convergence.
Definition. Let zn ∈ C. We say that
∑∞^ n=0^ zn^ is^ absolutely convergent^ if the real series n=0 |zn|^ is convergent
Lemma 3.1. Suppose that
n=0 zn^ is absolutely convergent. Then^
n=0 zn^ is convergent.
Proof. Suppose that
n=0 zn^ is absolutely convergent.^ Let^ zn^ =^ xn^ +^ iyn.^ Then |xn|, |yn| ≤ |zn|. Hence by the comparison test, the real series
n=0 xn^ and^
n=0 yn are absolutely convergent. As xn, yn are real, we know that ∣∣ ∣∣ ∣
∑^ ∞
n=
xn
∑^ ∞
n=
|xn|,
∑^ ∞
n=
yn
∑^ ∞
n=
|yn|,
so that
n=0 xn^ and^
n=0 yn^ are convergent. By the above remark,^
n=0 zn^ is convergent. 2
Remark. It is easy to give an example of a series which is convergent but not absolutely convergent. In fact, we can give an example using real series. Recall from MATH Sequences and Series that
n=0(−1)
n/n is convergent but ∑∞ n=0 |(−1)
n/n| = ∑∞ n=0 1 /n^ is divergent.
The reason for working with absolutely convergent series is that they behave well when multiplied together. Indeed, two series which converge absolutely may be multiplied in a similar way to two finite sums.
Proposition 3.1. Let an, bn ∈ C. Suppose that
n=0 an^ and^
n=0 bn^ are absolutely convergent. Then ( (^) ∞ ∑
n=
an
n=
bn
∑^ ∞
n=
cn
where cn = a 0 bn + a 1 bn− 1 + a 2 bn− 2 + · · · + anb 0 and
n=0 cn^ is absolutely convergent.
Proof. Omitted. 2
In MATH10242 Sequences and Series you looked at some tests to see whether a real series converged. The same tests continue to hold for complex series, and we state them below as propositions.
Proposition 3.1.3 (The Ratio Test) Let zn ∈ C. Suppose that
lim n→∞
|zn+1| |zn|
If ℓ < 1 then
n=0 zn^ is absolutely convergent. If^ ℓ >^1 then^
n=0 zn^ diverges.
Remark. If ℓ = 1 in (3.1.1) then we can say nothing: the series may converge absolutely, it may converge but not absolutely converge, or it may diverge.
Proposition 3.1.4 (The Root Test) Let zn ∈ C. Suppose that
n^ lim→∞ |zn|^1 /n^ =^ ℓ.^ (3.1.2)
If ℓ < 1 then
n=0 zn^ is absolutely convergent. If^ ℓ >^1 then^
n=0 zn^ diverges.
Remark. Again, if ℓ = 1 in (3.1.2) then we can say nothing: the series may converge absolutely, it may converge but not absolutely converge, or it may diverge.
§3.2 Power series and the radius of convergence
Definition. A series of the form
n=0 an(z^ −^ z^0 )
n (^) where an ∈ C, z ∈ C is called a power
series at z 0.
By changing variables to z′^ = z − z 0 we need only consider power series at 0, i.e. power series of the form (^) ∞ ∑
n=
anzn.
When does a power series converge? Let
R = sup
r ≥ 0 | there exists z ∈ C such that |z| = r and
∑^ ∞
n=
anzn^ converges
(We allow R = ∞ if no finite supremum exists.)
Theorem 3.2. Let
n=0 anz
n (^) be a power series and let R be defined as above. Then
(i)
n=0 anz
n (^) converges absolutely for |z| < R;
(ii)
n=0 anz
n (^) diverges for |z| > R.
Here an = 1/ 2 n. Using Proposition 3.2.2(i) we can calculate the radius of convergence as
1 R
= lim n→∞
|an+1| |an|
= lim n→∞
2 n 2 n+^
= lim n→∞
so that R = 2. Alternatively, we could use Proposition 3.2.2(ii) and see that
1 R
= lim n→∞
|an|^1 /n^ = lim n→∞
2 n
) 1 /n
so that again R = 2.
Proof. We prove (i). Suppose that |an+1/an| converges to a limit, say ℓ, as n → ∞, i.e.
lim n→∞
∣∣^ an+ an
Then
lim n→∞
|an+1zn+1| |anzn|
→ ℓ|z|.
By the ratio test, the power series
n=0 anz
n (^) converges for ℓ|z| < 1 and diverges for
ℓ|z| > 1. Hence the radius of convergence R = 1/ℓ.
∑We prove (ii). Suppose that^ |an|^1 /n^ →^ ℓ^ as^ n^ → ∞. By the root test, the power series ∞ n=0 anz
n (^) converges if limn→∞ |anzn| 1 /n (^) = limn→∞ |an| 1 /n|z| = ℓ|z| < 1 and diverges if
limn→∞ |anzn|^1 /n^ = limn→∞ |an|^1 /n|z| = ℓ|z| > 1. Hence the radius of convergence R = 1/ℓ. 2
Remark. It may happen that neither of the limits in (i) nor (ii) of Proposition 3.2.2 exist. However, there is a formula for the radius of convergence∑ R that works for any power series ∞ n=0 anz
n. Let xn be a sequence of real numbers. For each n, consider supk≥n xk. As n increases, this sequence decreases. Recall that any decreasing sequence of reals converges. Hence
lim n→∞
sup k≥n
xk
exists (although it may be equal to ∞). We denote the limit by lim supn→∞ xn. Thus lim sup xn exists for any sequence xn. (One can show that if limn→∞ xn = ℓ then lim supn→∞ xn = ℓ.) With this definition, it is always the case that 1 R
= lim sup n→∞
|an|^1 /n.
§3.3 Differentiation of power series
We know that for a polynomial
p(z) = a 0 + a 1 z + · · · + anzn
the derivative is given by
p′(z) = a 1 + 2a 2 z + · · · + nanzn−^1.
This suggests that a power series
f (z) =
∑^ ∞
n=
anzn^ (3.3.1)
can be differentiated term by term to give
f ′(z) =
∑^ ∞
n=
nanzn−^1. (3.3.2)
However, because we are dealing with infinite sums, this needs to be proved. There are two steps to this: (i) we have to show that if (3.3.1) converges for |z| < R then (3.3.2) converges for |z| < R, and (ii) that f (z) is differentiable for |z| < R and the derivative is given by (3.3.2).
Lemma 3.3. Let f (z) =
n=0 anz
n (^) have radius of convergence R. Then g(z) = ∑∞ n=1 nanz
n− (^1) converges
for |z| < R.
Proof. Let |z| < R and choose r such that |z| < r < R. Then
n=0 anr
n (^) converges
absolutely. Hence the summands must be bounded, so there exists K > 0 such that |anrn| < K for all n ≥ 0. Let q = |z|/r and note that 0 < q < 1. Then
|nanzn−^1 | = n|an|
∣∣ z r
∣∣n−^1 rn− (^1) < n K r
qn−^1.
But
n=0 nq
n− (^1) converges to (1 − q)− (^2). Hence by the comparison test, ∑∞ n=0 |nanz
n− (^1) |
converges. Hence
n=0 nanz
n− (^1) converges absolutely and so converges. 2
Theorem 3.3. Let f (z) =
n=0 anz
n (^) have radius of convergence R. Then f (z) is holomorphic on the disc
of convergence {z ∈ C | |z| < R} and f ′(z) =
n=1 nanz
n− (^1).
Proof. Let g(z) =
n=1 nanz
n− (^1). By Lemma 3.3.1 we know that this converges for
|z| < R. We have to show that if |z 0 | < R then f (z) is differentiable at z 0 and, moreover, the derivative is equal to g(z 0 ), i.e. we have to show that if |z 0 | < R then
f ′(z 0 ) := lim z→z 0
f (z) − f (z 0 ) z − z 0
= g(z 0 )
or equivalently
zlim→z 0
f (z) − f (z 0 ) z − z 0
− g(z 0 )
For any N ≥ 1 we have the following
f (z) − f (z 0 ) z − z 0
− g(z 0 )
∑^ ∞
n=
an
zn^ − z 0 n z − z 0
− nanz 0 n−^1
Proof. This is a simple induction on k. 2
Instead of using a power series at the origin, by replacing z by z − z 0 we can consider power series at the point z 0. (This will be useful later on when we look at Taylor series.) Suppose that f (z) =
n=0 anz
n (^) has disc of convergence |z| < R. Then, replacing z by
z − z 0 , we have that the power series g(z) =
n=0 an(z^ −^ z^0 )
n (^) has disc of convergence
|z − z 0 | < R. Moreover, inside this disc of convergence all the higher derivatives of g exist and
g(k)(z) =
∑^ ∞
n=k
n! (n − k)!
an(z − z 0 )n−k.
§3.4 Special functions
§3.4.1 The exponential function
You have probably already met the exponential function ex^ =
n=0 x
n/n!, certainly in the
case when x is real. Here we study the (complex) exponential function.
Definition. The exponential function is defined to be the power series
exp z =
∑^ ∞
n=
zn n!
By Proposition 3.2.2(i) we see that the radius of convergence R for exp z is given by
1 R
= lim n→∞
n! (n + 1)!
= lim n→∞
n + 1
so that R = ∞. Hence this series has radius convergence ∞, and so converges absolutely for all z ∈ C. By Theorem 3.3.2 we may differentiate term-by-term to obtain
d dz
exp z =
∑^ ∞
n=
n
zn−^1 n!
∑^ ∞
n=
zn−^1 (n − 1)!
∑^ ∞
n=
zn n!
= exp z,
which we already knew to be true in the real-valued case. In the real case we know that if x, y ∈ R then ex+y^ = exey. This is also true in the complex-valued case, and the proof involves a neat trick. First we need the following fact:
Lemma 3.4. Suppose that f is holomorphic on a domain D and f ′^ = 0 on D. Then f is constant on D.
Remark. This is well-known in the real case: a function with zero derivative must be constant. The proof in the complex case is somewhat more involved and we omit it. (See Stewart and Tall, p.71.)
Proposition 3.4. Let z 1 , z 2 ∈ C. Then exp(z 1 + z 2 ) = exp(z 1 ) exp(z 2 ).
Proof. Let c ∈ C and define the function f (z) = exp(z) exp(c − z). Then
f ′(z) = exp(z) exp(c − z) − exp(z) exp(c − z) = 0
by the product rule. Hence by Lemma 3.4.1 we must have that f (z) is constant; in particular this constant must be f (0) = exp c. Hence exp(z) exp(c − z) = exp(c). Putting c = z 1 + z 2 and z = z 1 gives the result. 2
Remark. In particular, if we take z 1 = z and z 2 = −z in Proposition 3.4.2 then we have that 1 = exp 0 = exp(z − z) = exp(z) exp(−z).
Hence exp z 6 = 0 for any z ∈ C. (We already knew that ex^ = 0 has no real solutions; now we know that it has no complex solutions either.)
Finally, we want to connect the real number e to the complex exponential function. We define e to be the real number e = exp 1. Then, iterating Proposition 3.4.2 inductively, we obtain en^ = exp(1)n^ = exp(1 + · · · + 1) = exp n.
For a rational number m/n (n > 0) we have that
(exp(m/n))n^ = exp(nm/n) = exp(m) = em
so that exp(m/n) = em/n. Thus the notation ez^ = exp z does not conflict with the usual definition of ex^ when z is real. Hence we shall normally write ez^ for exp z. In particular, if we write z = x + iy then Proposition 3.4.2 tells us that
ex+iy^ = exeiy.
We already understand real exponentials ex. Hence to understand complex exponentials we need to understand expressions of the form eiy.
§3.4.2 Trigonometric functions
Define
cos z =
∑^ ∞
n=
(−1)n^
z^2 n (2n)!
, sin z =
∑^ ∞
n=
(−1)n^
z^2 n+ (2n + 1)!
By Proposition 3.2.2(i) it is straightforward to check that these converge absolutely for all z ∈ C. Substituting z = −z we see that cos is an even function and that sin is an odd function, i.e. cos(−z) = cos z, sin(−z) = − sin z.
Moreover, cos(0) = 1, sin(0) = 0. By Theorem 3.3.2 we can differentiate term-by-term to see that
d dz
cos z = − sin z,
d dz
sin z = cos z.
Term-by-term addition of the power series for cos z and sin z shows that
exp iz = cos z + i sin z.
§3.4.5 The logarithmic function
Let z ∈ C and consider the equation
exp w = z (3.4.1)
By §3.4.4, if w 1 is a solution to (3.4.1) then so is w 1 + 2nπi. Each of these values is a called a logarithm of z, and we denote any of these values by log z. Thus the complex logarithm is a multi-valued function. If x ∈ R then exp w = x has a unique real solution (this is just the ordinary definition of log). We write w = log x. In (3.4.1) write w = x + iy. Then
z = exp w = exp(x + iy) = ex(cos y + i sin y).
Hence ex^ = |z| so that x = log |z|, and y is a value of the argument of z. Hence we can make the following definition.
Definition. Let z ∈ C, z 6 = 0. Then a complex logarithm of z is
log z = log |z| + i arg z
where arg z is any argument of z. The principal value of log z is the value of log z when arg z has its principal value, i.e. when arg z ∈ (−π, π]. We denote the principal logarithm by Log z:
Log z = log |z| + i arg z, −π < arg z ≤ π.
Dealing with multivalued functions is tricky. One way is to only consider the logarithm function on a subset of C.
Definition. The complex plane with the negative real-axis (including 0) removed is called the cut plane. See Figure 3.4.1.
Figure 3.4.1: The cut plane: this is the complex plane with the negative real axis removed.
Proposition 3.4. The principal logarithm Log z is continuous on the cut plane.
Proof. This follows from the fact (which we shall not prove, although the proof is easy) that arg z is continuous on the cut-plane. 2
Having seen that the principal logarithm is continuous, we can go on to show that it is differentiable.
Proposition 3.4. The principal logarithm Log z is holomorphic on the cut plane and
d dz
Log z =
z
Proof. Let w = Log z. Then z = exp w. Let Log(z+h) = w+k. Then by Proposition 3.4. Log is continuous on the cut plane so we have that k → 0 as h → 0. Then
d dz
Log z = lim h→ 0
Log(z + h) − Log(z) h
= lim k→ 0
(w + k) − w exp(w + k) − exp(w)
= lim k→ 0
exp(w + k) − exp(w) k
d dw
exp(w)
z
§3.5 Exercises
Exercise 3. Let zn ∈ C. Show that
n=0 zn^ is convergent if, and only if, both^
∑∞^ n=0^ Re(zn) and n=0 Im(zn) are convergent.
Exercise 3. Find the radii of convergence of the following power series:
(i)
∑^ ∞
n=
2 nzn n
, (ii)
∑^ ∞
n=
zn n!
, (iii)
∑^ ∞
n=
n!zn, (iv)
∑^ ∞
n=
npzn^ (p ∈ N).
Exercise 3. By multiplying two series together, show using Proposition 3.1.2 that for |z| < 1, we have
∑^ ∞
n=
nzn−^1 =
(1 − z)^2
Exercise 3. Suppose that
anzn^ has radius of convergence R. Use the formula 1/R = limn→∞ |an|^1 /n to find the radii of convergence of the following power series in terms of R:
(i)
∑^ ∞
n=
n^3 anzn, (ii)
∑^ ∞
n=
anz^3 n, (iii)
∑^ ∞
n=
a^3 nzn.
