number an is real or complex. Such a formal sum is also denoted by an. n=1 + "
Sometimes it is appropriate to consider infinite series an of the form n=m am + am+1 + am+2 + · · ·, where m " Z. Clearly results for such sequences may be deduced immediately from corresponding results in the case m = 1. + "
Definition An infinite series an is said to converge to some complex n=1 number s if and only if, given any µ > 0, there exists some natural number N
m
such that an - s < µ for all natural numbers m satisfying m e" N. If
n=1 + " + "
the infinite series an converges to s then we write an = s. An infinite n=1 n=1 series is said to be divergent if it is not convergent. For each natural number m, the mth partial sum sm of the infinite se- + " + "
ries an is given by sm = a1 + a2 + · · · + am. Note that an converges n=1 n=1 to some complex number s if and only if sm s as m + ". The fol- lowing proposition therefore follows immediately on applying the results of Proposition 1.2. + " + "
Proposition 2.1 Let an and bn be convergent infinite series. Then n=1 n=1 + " + " + " + " + "
(an+bn) is convergent, and (an+bn) = an+ bn. Also (an) = n=1 n=1 n=1 n=1 n=1 + "
an for any complex number . n=1 + "
If an is convergent then an 0 as n + ". Indeed n=1 lim an = lim (sn - sn-1) = lim sn - lim sn-1 = s - s = 0, n+ " n+ " n+ " n+ " m + "
where sm = an and s = an = limm+ " sm. However the condition n=1 n=1 that an 0 as n + " is not in itself sufficient to ensure convergence. For + "
example, the series 1/n is divergent. n=1 25 Proposition 2.2 Let a1, a2, a3, a4, . . . be an infinite sequence of real num- + "
bers. Suppose that an e" 0 for all n. Then an is convergent if and only if n=1 there exists some real number C such that a1 + a2 + · · · + an d" C for all n. + "
Proof The sequence s1, s2, s3, . . . of partial sums of the series an is non- n=1 decreasing, since an e" 0 for all n. The result therefore is a consequence of the fact that a non-decreasing sequence of real numbers is convergent if and only if it is bounded above (see Theorem 1.3). Example Let z be a complex number. The infinite series 1+z +z2 +z3 +· · · is referred to as the geometric series. This series is clearly divergent whenever |z| e" 1, since zn does not converge to 0 as n + ". We claim that the series converges to 1/(1 - z) whenever |z| < 1. Now 1 - zm+1 1 + z + z2 + . . . zm = . 1 - z (To see this, multiply both sides of the equation by 1 - z.) Thus if |z| < 1 then 1 lim (1 + z + z2 + . . . zm) = . m+ " 1 - z 2.1 The Comparison Test and Ratio Test + "
Proposition 2.3 An infinite series an of real or complex numbers is con- n=1 vergent if and only if, given any µ > 0, there exists some natural number N with the property that |am + am+1 + · · · + am+k| < µ for all m and k satisfying m e" N and k e" 0. Proof The stated criterion is equivalent to the condition that the sequence of partial sums of the series be a Cauchy sequence. The required result thus follows immediately from Cauchy s Criterion for convergence (Theo- rem 1.9). Proposition 2.4 (Comparison Test) Suppose that 0 d" |an| d" bn for all n, + " + "
where an is complex, bn is real, and bn is convergent. Then an is n=1 n=1 convergent. 26 Proof Let µ > 0 be given. Then there exists some natural number N such that bm + bm+1 + · · · + bm+k < µ for all m and k satisfying m e" N and k e" 0. But then |am + am+1 + · · · + am+k| d" |am| + |am+1| + · · · + |am+k| d" bm + bm+1 + · · · + bm+k < µ + "
when m e" N and k e" 0. Thus an is convergent, by Proposition 2.3. n=1 Let us apply the Comparison Test in the case when an and bn are non- + "
negative real numbers satisfying 0 d" an d" bn for all n. If bn is convergent, n=1 + " + " + "
then so is an. Thus if an is divergent then so is bn. These results n=1 n=1 n=1 also follow directly from Proposition 2.2. Example Comparison with the geometric series shows that the infinite series + "
zn/n is convergent whenever |z| < 1. n=1 Proposition 2.5 (Ratio Test) Let a1, a2, a3 . . . be complex numbers. Sup- + "
an+1 pose that r = lim exists and satisfies |r| < 1. Then an is conver- n+ " an n=1 gent. Proof Choose Á satisfying |r| < Á < 1. Then there exists some natural number N such that |an+1/an| < Á for all n e" N. Let
|a1| |a2| |a3| |aN| K = maximum , , , . . . . Á Á2 Á3 ÁN Now |an+1| d" Á|an| whenever n e" N. Therefore |an| d" Án-N|aN| d" KÁn whenever n e" N. But the choice of K also ensures that |an| d" KÁn when n < + "
N. Moreover KÁn converges, since Á < 1. The desired result therefore n=1 follows on applying the Comparison Test (Proposition 2.4). + "
zn Example Let z be a complex number. Then converges for all values n! n=0 of z. For if an = zn/n! then an+1/an = z/(n + 1), and hence an+1/an 0 as n + ". The result therefore follows on applying the Ratio Test. 27 + "
Let an be an infinite series for which r = lim an+1/an is well-defined. n+ " n=1 The series clearly diverges if |r| > 1, since |an| increases without limit as n + ". If however |r| = 1 then the Ratio Test is of no help in deciding whether or not the series converges, and one must try other more sensitive tests. 2.2 Absolute Convergence + " Definition An infinite series an is said to be absolutely convergent if n=1 + " the infinite series |an| is convergent. A convergent series which is not n=1 absolutely convergent is said to be conditionally convergent. An absolutely convergent infinite series is convergent, and the sum of any two absolutely convergent series is itself absolutely convergent. These results follow on applying the Comparison Test (Proposition 2.4). Moreover the following criterion for absolute convergence follows directly from Propo- sition 2.3. + "
Proposition 2.6 An infinite series an is absolutely convergent if and n=1 only if, given any µ > 0, there exists some natural number N such that |am| + |am+1| + · · · + |am+k| < µ for all m and k satisfying m e" N and k e" 0. Many of the tests for convergence described above do in fact test for absolute convergence; these include the Comparison Test and the Ratio Test. 2.3 The Cauchy Product of Infinite Series + " + "
The Cauchy product of two infinite series an and bn is defined to be n=0 n=0 + "
The convergence of an and bn is not in itself sufficient to ensure the n=0 n=0 convergence of the Cauchy product of these series. Convergence is however + " + "
assured provided that the series an and bn are absolutely convergent. n=0 n=0 28 + "
Theorem 2.7 The Cauchy product cn of two absolutely convergent infi- n=0 + " + "
nite series an and bn is absolutely convergent, and n=0 n=0 + " + " + "
cn = an bn . n=0 n=0 n=0 Proof For each non-negative integer m, let Sm = {(j, k) " Z × Z : 0 d" j d" m, 0 d" k d" m}, Tm = {(j, k) " Z × Z : j e" 0, k e" 0, 0 d" j + k d" m}.
m m m
Now cn = ajbk and an bn = ajbk. Also n=0 n=0 n=0 (j,k)"Tm (j,k)"Sm + " + " m
cn = lim cn = an bn , m+ " n=0 n=0 n=0 n=0 as required. 2.4 Uniform Convergence for Infinite Series Let f1, f2, f2, . . . be complex-valued functions defined over a subset D of C. + "
The infinite series series fn(z) is said to converge uniformly on D to some n=1 function s if, given any µ > 0, there exists some natural number N (which
n
s(z) does not depend on the value of z) such that - fm(z) < µ whenever
m=0 z " D and n e" N. + "
Note that an infinite series fn(z) of functions converges uniformly if n=0 and only if the partial sums of this series converge uniformly. It follows immediately from Theorem 1.20 that if the functions fn are continuous on + "
D, and if the series fn(z) converges uniformly on D to some function, n=0 then that function is also continuous on D. Proposition 2.8 (The Weierstrass M-Test) Let D be a subset of C and let f1, f2, f3, . . . be a sequence of functions from D to C, let M1, M2, M3, . . . be non-negative real numbers satisfying |fn(z)| d" Mn for all natural numbers n + " + "
and z " D. Suppose that Mn converges. Then fn(z) converges abso- n=1 n=1 lutely and uniformly on D. Proof It follows immediately from the Comparison Test (Proposition 2.4) + "
that the series fn(z) is absolutely convergent for all z " D. We must n=1 show that the convergence is uniform. Let µ > 0 be given. Then there exists some natural number N such + " + " 1 that Mn < µ, since Mn converges. Now if m and k are integers n=N 2 n=1 30 satisfying m e" N and k e" 0 then m+k m m+k m+k + "
1 fn(z) - fn(z) = fn(z) d" Mn d" Mn < µ
2 n=1 n=m+1 n=1 n=m+1 n=N for any z " D. On taking the limit as k + ", we see that + " m
1 fn(z) - fn(z) d" µ < µ
2 n=1 n=1 for all z " D and m e" N. However N has been chosen independently of z. Thus the infinite series converges uniformly on D, as required. 2.5 Power Series + "
A power series is an infinite series of the form an(z - z0)n, where the n=0 coefficients a0, a1, a2, . . . are complex numbers. + "
Definition Let an(z - z0)n be a power series centred on some complex n=0 number z0. Suppose that the set of complex numbers z for which the power series converges is bounded. Then the radius of convergence R0 of the power series is defined to be the smallest non-negative real number with the property that every complex number z for which the power series converges satisfies |z - z0| d" R0. The circle {z " C : |z - z0| = R0} is then referred to as the circle of convergence of the power series. We set R0 = + " if the set of complex numbers z for which the power series converges is unbounded. + "
Theorem 2.9 Let an(z - z0)n be a power series with radius of conver- n=0 gence R0, and let s(z) denote the sum of the power series at those complex numbers z at which the series converges. (i) If R0 = + " then s(z) is a continuous function of z defined over the entire complex plane C. (ii) If R0 < + " then s(z) is a continuous function of z defined over the whole of the disk {z " C : |z - z0| < R0} bounded by the circle of convergence of the power series. 31 Proof Let z1 be any complex number satisfying |z1 -z0| < R0. Then we can choose R such that |z1 -z0| < R < R0 and R < + ". Now it follows from the definition of the radius of convergence that there exists some complex num- + "
ber w such that R < |w| < R0 and anwn converges. Choose some positive n=0 real number A with the property that |anwn| d" A for all n, and set Á = R/|w| and Mn = AÁn. |z - z0| < R then |an(z - z0)n| d" |an|Rn d" AÁn = Mn for + "If all n. Also Mn converges to A/(1 - Á). Thus we can apply the Weier- n=0 + "
strass M-Test (Proposition 2.8) to deduce that the power series an(z-z0)n n=0 converges uniformly on the disk {z " C : |z - z0| < R} of radius R about z0. It then follows from Theorem 1.20 that the restriction of the function s to this disk is continuous on the disk, and, in particular, is continuous around z1. We deduce that the function s is continuous throughout the complex plane when R0 = + ", and is continuous inside the circle of convergence when R0 < + ", as required. A power series with finite radius of convergence will converge everywhere within its circle of convergence, and will diverge everywhere outside this cir- cle. However Theorem 2.9 provides no information concerning the behaviour of the power series on the circle of convergence itself. 2.6 The Exponential Function Definition The exponential function exp: C C is defined for all complex numbers z by the formula + "
zn exp(z) = . n! n=0 A straightforward application of the Ratio Test shows that radius of con- vergence of the power series defining exp(z) is infinite, and it therefore follows from Theorem 2.9 that the exponential function is continuous on C. Lemma 2.10 The exponential function has the property that exp(z + w) = exp(z) exp(w). for all complex numbers z and w. Proof Let z and w be complex numbers. The infinite series defining exp(z) and exp(w) are absolutely convergent. It therefore follows from Theorem 2.7 32 that the value of exp(z) exp(w) is the sum of the Cauchy product of the infinite series defining exp(z) and exp(w), and therefore + "
exp(z) exp(w) = cn n=0 where n
zj wn-j cn = j! (n - j)! j=0 On applying the Binomial Theorem, we see that
n n
1 n! 1 n 1 cn = zjwn-j = zjwn-j = (z + w)n, n! j!(n - j)! n! j n! j=0 j=0 and thus exp(z) exp(w) = exp(z + w), as required. Let z be a complex number. Then exp(z) exp(-z) = exp(z - z) = exp(0) = 1. It follows that exp(z) = 0 and exp(-z) = 1/ exp(z) for all
complex numbers z. Let sin: C C and cos: C C be the functions defined as sums of power series according to the formulae " "
(-1)nz2n+1 (-1)nz2n sin z = , cos z = . (2n + 1)! (2n)! n=0 n=0 If z is real then the power series appearing on the right hand side of these identities correspond to the Taylor expansions of the standard sine and cosine functions. A routine application of the Ratio Test shows that these power se- ries both have infinite radius of convergence, and therefore define continuous functions defined over the whole complex plane. These continuous functions therefore constitute a natural extension to the complex plane of the standard sine and cosine functions of basic trigonometry. On comparing power series, we see that 1 1 cos z = (exp(iz) + exp(-iz)), sin z = (exp(iz) - exp(-iz)), 2 2i cos z + i sin z = exp(iz), cos z - i sin z = exp(-iz) for all complex numbers z. Thus if z = x + iy, where x and y are real numbers, then exp(z) = exp(x + iy) = exp(x) exp(iy) = ex(cos y + i sin y). 33 The exponential map exp: C C \ {0} is surjective. For, given any complex number w, there exist real numbers r and ¸ such that r > 0 and w = r(cos ¸ + i sin ¸). Moreover there exists a real number x such that r = ex. (This real number x is the natural logarithm loge r of r.) Then w = exp(x + i¸). Lemma 2.11 Let z and w be complex numbers. Then exp(z) = exp(w) if and only if w = z + 2Ä„in for some integer n. Proof If w = z + 2Ä„in for some integer n then exp(w) = exp(z) exp(2Ä„in) = exp(z)(cos 2Ä„n + i sin 2Ä„n) = exp(z). Conversely suppose that exp(w) = exp(z). Let w-z = u+iv, where u, v " R. Then eu(cos v + i sin v) = exp(w - z) = exp(w) exp(z)-1 = 1. Taking the modulus of both sides, we see that eu = 1, and thus u = 0. Also cos v = 1 and sin v = 0, and therefore v = 2Ä„n for some integer n. The result follows. Proposition 2.12 Let D0 = C \ {x " R : x d" 0}. Then there exists a unique continuous function log: D0 C characterized by the properties that log(1) = 0, | Im log(z)| < Ä„ and exp(log(z)) = z for all z " D0. This function has the property that log(rei¸) = loge r + i¸ for all real numbers r and ¸ satisfying r > 0 and -Ä„ < ¸ < Ä„, where loge r denotes the natural logarithm of the real number r. Proof Given any complex number w belonging to the open set D0 there exist unique real numbers r and ¸ for which r > 0, -Ä„ < ¸ < Ä„ and rei¸ = w. It follows that there is a well-defined function log: D0 C such that log(rei¸) = loge r+i¸ for all real numbers real numbers r and ¸ satisfying r > 0 and -Ä„ < ¸ < Ä„, where loge r, the natural logarithm of r, is the unique real number t satisfying et = r. Moreover exp(log(rei¸)) = exp(loge r) exp(i¸) = rei¸ for all real numbers r and ¸ with r > 0. It follows that exp(log(z)) = z for all z " D0. 34 Let x and y be real numbers. Then
Å„Å‚ ôÅ‚ x ôÅ‚ 1 ôÅ‚ loge(x2 + y2) + i arccos if y > 0, ôÅ‚ 2 ôÅ‚ ôÅ‚ x2 + y2 ôÅ‚ ôÅ‚ ôÅ‚ òÅ‚ x 1 log(x + iy) = loge(x2 + y2) - i arccos if y < 0, 2 ôÅ‚ x2 + y2 ôÅ‚ ôÅ‚
ôÅ‚ ôÅ‚ ôÅ‚ ôÅ‚ y 1 ôÅ‚ ôÅ‚ loge(x2 + y2) + i arcsin if x > 0, ół 2 x2 + y2 where arcsin: [-1, 1] [-Ä„/2, Ä„/2] and arccos: [-1, 1] [0, Ä„] are the in- verses of the restrictions of the sine and cosine functions to the intervals [-Ä„/2, Ä„/2] and [0, Ä„] respectively. Now the functions loge, arcsin and arccos are continuous, since the inverse of any strictly increasing or strictly de- creasing function defined over some interval in R is itself continuous. (see Corollary 1.19). We conclude that the restrictions of log: D0 C to each of the open sets {z " C : Im z > 0}, {z " C : Im z < 0}, {z " C : Re z < 0} is continuous, and the open set D0 is the union of these three open sets. It follows that log: D0 C is continuous throughout D0. A straightforward application of Lemma 2.11 then shows that this function log is the unique function from D0 to C with the required properties. The function log: D0 C characterized by the properties set out in the statement of Proposition 2.12 is referred to as the principal branch of the logarithm function. It is impossible to define a continuous logarithm function L: C \ {0} C with the property that exp(L(z)) = z for all z " C \ {0}. The best that one can achieve is to define continuous inverses of the exponential map over appropriately chosen open subsets of C \ {0}. Such functions are referred to as branches of the logarithm function . Corollary 2.13 Let w be a non-zero complex number, and let Dw,|w| = {z " C : |z - w| < |w|}. Then there exists a continuous function Fw: Dw,|w| C with the property that exp(Fw(z)) = z for all z " Dw,|w|. Proof Let D0 = C \ {x " R : x d" 0}, let log: D0 C be the principal branch of the logarithm function, and let Å› be a complex number satisfying exp Å› = w. Then z/w " D0 for all z " Dw,|w|. A function Fw: Dw,|w| C with the required properties may therefore be obtained on defining Fw(z) = Å› + log(z/w) for all z " Dw,|w|. 35