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Unformatted text preview: CHAPTER 0 Preliminaries 1. Introduction In this preliminary chapter we consider brieﬂy some important concepts
from calculus and algebra which we shall require for our study of differen
tial equations. Many of these concepts may be familiar to the student, in
which case this chapter can serve as a review. First the elementary proper—
ties of complex numbers are outlined. This is followed by a discussion of
functions which assume complex values, in particular polynomials and
power series. Some consequences of the Fundamental Theorem of Algebra
are given. The exponential function is deﬁned using power series; it is of
central importance for linear differential equations with constant coeﬂi
cients. The role that determinants play in the solution of systems of linear
equations is discussed. Lastly we make a few remarks concerning principles
of discovery, and methods of proof, of mathematical results. 2. Complex numbers It is a fundamental fact about real numbers that the square of any such
number is never negative. Thus there is no real x which satisﬁes the equa
tion 902 + 1 = 0. We shall use the real numbers to deﬁne new numbers which include numbers
which satisfy such equations.
A complex number 2 is an ordered pair of real numbers (cc, y), and we
write
2 = (96,21).
If 21 = ($11111), 22 = (x2) y2)) 4 Preliminaries Chap. 0 agrees with our earlier identiﬁcations 0 = (0, 0) , 1 = (1, 0) .] In this
sense, the complex numbers contain the real numbers. The properties
(i)—(ix), which hold for complex numbers, are also valid for real numbers,
and thus we see that we have succeeded in enlarging the set of real numbers
without losing any of these algebraic properties. We have gained something
also, since there are complex numbers 2 which satisfy the equation z2+1=0. One such number the imaginary unit 1' = (0, 1), as can be easily checked,
and this provides one justiﬁcation for our deﬁnition of multiplication. If 2 = (x, y) is a complex number, the real number x is called the real
part of z, and we write Re 2 = x; whereas y is called the imaginary part
of z, and we write Im z = y. Thus 2 = (x,y) =x(1,0) +y(0,1) =x+iy = Rez+i(Imz). Hereafter it will be convenient to denote a complex number (x, y) as
:c + z'y. It is clear that the complex numbers are in a onetoone correspondence
with the points of the (re, y)plane, the complex number 2 = x + iy corre
sponding to the point with coordinates (x, y). Then thought of in this way
the x—axis is of ten called the real axis, the y—axis is called the imaginary axis,
and the plane is called the complex plane. If 2 = x + iy, its mirror image in the real axis is the point x — iy. This
number is called the complex conjugate of z, and is denoted by 2. Thus
2 = x — iy if z = x + iy. We see immediately that Z = Z, 21 + 22 = 21 + 52, 2122 = 5152, 21 = (2)_l, for any complex numbers 2, 21, 22.
Introducing polar coordinates (r, 0) in the complex plane via x=rcos(9, y=rsin6, (r;0,0§0<21r),
we see that we may write
2 = x+iy = r(cos0+isin0).
The magnitude of z = x + iy, denoted by l2 l , is deﬁned to be 7‘. Thus
lzl = (x2 + ye)”2 = (25)“, where the positive square root is understood. Clearly = Suppose
z is real (that is, Inn 2 = 0). Then 2 = x + £0, for some real 2:, and Izl = (222)1’2, Sec. 2 Preliminaries ' 5 which is the magnitude of :1: considered. as a real number. In addition the
magnitude of a complex number obeys the same rules as the magnitude of a
real number, namely: lzl 20:
[2 =0 ifandonlyif 2:0,
lzl, 21+22l § W + [22L 1
33.
ll lzlllz2l lzlzzl
We show that I21 +22] g lzll + I22], for example. First we note that
Rez g [2
for any complex number 2. Then
I21+222 (21+22)m = 212+ I222+zlé2+2122
I212+ 222+2Re (2152)
l212+ lz22+22152l \
21l2+222+2l21l l22l
(l21l+lzzl)2, from which it follows that zl + all g [21] + [22 I.
From the above rules one can deduce further that II "A ll ll llzll “IZle §l31+z2 é lzll‘l‘lzzb
ﬁg =lzll
22 lzzl Geometrically we see that [21 — 22I represents the distance between the
two points 21 and 22 in the complex plane. EXERCISES 1. Compute the following complex numbers, and express in the form x + iy,
Where x, y are real: (a) (2 — i3) + (1+ 756) . (b) (4 i2) — (6 — 2'3) (c) (6— iﬁ)(2+i4) (d) 1+5
' _ 7:
(e) I4—i5l (f) Re(4—i5) (g) Im (6 + 2'2) 6 Preliminaries Chap. 0 2. Express the following complex numbers in the form r (cos 0 + t sin 0) with
r20and0§0<21rz _
,. (a) 1+ N37 0)) (1+ 02 (c) I“ (d) (1+i)<1— 2') l—t ,3. Indicate graphically the set of all complex numbers 2 satisfying: (a)z—2l=1 .(b)[z+2<2
(c)Rez[§3 ‘ (d)[Imz[>1
.(e) z—1[+Iz+2l=8. .4. Prove that:
(a)z+é=2Rez (b)z—2=2iImz
(c)lRezI§z ‘(d)z§Rez!+IImz 5. If r is a real number, and 2 complex, show that
Re (72) = 7‘ (Re 2), Im (r2) = 7* (Im z). 6. Provethat
l21l*lz2ll§l21+22. (Hint: 21 = 21 + 22 + (Z2), and 22 = 21 + 22 + (—21).) 7. Provethat
I21+22l2+ I21—22l2=2l21l2+2l22l2, for all complex 21, Z2. _ ISI? 8. If I a I < 1, what complex 2 satisfy _ _
l 1 — azl , 9. If n is any positive integer, prove that
7"" (cos n0 + i sin n0) = [r (cos 0 + i sin (9)1".
(H int: Use induction.) 10. Use the result of Ex. 9 to ﬁnd
(a) two complex numbers satisfying 22 = 2,
(b) three complex numbers satisfying 23 = 1. 3. Functions Suppose D is a set whose elements are denoted by P, Q,    , which are
called the points of the set. Let R be another set. A function on D to R is a
law f which associates with each point P in D exactly one point in R, which
we denote by f(P). The set D is called the domain of f. The point f(P)
is called the value of f at P. We can Visualize the concept of a function as See. 3 Preliminaries 7 Figure l in Fig. 1, where each P in D is connected to a unique f ( P) in R by a string
according to some rule. This rule, or what amounts to the same thing, the
collection of all these strings, is the function f on D to R. We say that two functions f and g are equal, f = 9, if they have the
same domain D, and f(P) = g(P) for all P in D. The idea of a function is very general, and is a fundamental one in
mathematics. We shall consider some examples which are of importance
for our study of differential equations. (a) Complexvalued functions. If the set R which contains the values of
f is the set of all complex numbers, we say that f is a complexvalued func— tion. If f and g are two complexvalued functions with the same domain D,
we can deﬁne their sum f + g and product fg by (f+g)(P) =f(P) +9(P),
(fg)(P) =f(P)g(P), for each P in D. Thus f + g and fg are also functions with domain D. If a
is any complex number the function which assigns to each P in a domain D
the number a is called a constant function, and is also denoted by (2. Thus
if f is any complex—valued function on D we have (af)(P) = 04(1))
for all P in D.
A real—valued function f deﬁned on D is one whose values are real num
bers. Such a function is a special case of a complexvalued function. Clearly
the sum and product of two real—valued functions on D are real—valued functions. Realvalued functions are usually the principal object of study
in ﬁrst courses in calculus. Every complex—valued function f deﬁned on a domain D gives rise to
two real—valued functions Re f, Im f deﬁned by (Ref)(P) = Re [f(P)],
(Imf)(P) = ImEf(P)l 8 Preliminaries Chap. 0 for all P in D. Re f and Im f are called the real and imaginary parts of f
respectively and we have f=Ref+iImf. Thus the study of complexvalued functions can be reduced to the study of
pairs of realvalued functions. To obtain examples of complex—valued func
tions we must specify their domains. (b) Complex—valued functions with real domains. Many of the functions
we consider in this book have a domain D which is an interval I of the real
axis. Recall that an interval is a set of real as satisfying one of the nine
inequalities "A
II/\ a a b, a§x<b, a<x§b, a<x<b, IIA
/\ a x 00, —OO<1’§I), a<x<°°, —°°<x<b) —oo<x<00, where a, b are distinct real numbers. The calculus of complexvalued func
tions deﬁned on real intervals is entirely analogous to the calculus of real Valued functions deﬁned on intervals. We sketch the main ideas.
Suppose f is a complexvalued function deﬁned on a real interval I. Then f is said to have the complex number L as a limit at so in I, and we
write limf(x) = L, or ﬂat) —> L, (cc—>960), 192:0
if
f(x)—L—>0, as 0<Ix—xol—)0. This means that given any 6 > 0 there is a 6 > 0 such that
f(:v) — LI < e, whenever 0 <13: — 230! < 6, xinI. Note that here we are using the magnitude of complex numbers. Formally
our deﬁnition is the same as that for real limits of realvalued functions.
Because of this the usual rules for limits, and their proofs, are valid. In
particular, if f and g are cOmplexvalued functions deﬁned on I such that
for some so in I (x6530):
then ' ' (f+g)(x)—>L+M, (f9)(x)+LM, (ac—>960) Sec. 3 Preliminaries 9 Suppose I has a limit L
since ll L; + iLg at so, where L1, L2 are real. Then [(Rech)  Lll lReU(x)  LJI é f(x) — Li, and
l(Imf)(x) — Lzl = IImEﬂx)  L]! é We) — Ll,
it follows that (Ref)($)—*L1, (Imf)(x)>L2, (as—>330). Conversely, if Re f and Im f have limits L1, L2 respectively at me, then f
will have the limit L = L1 + iLz at me. We say that a complexvalued function f deﬁned on an interval I is
continuous at are in I if f has the limit f (are) at so, that is, f(a:) —f(xo)I——>0, as 0<Ix—:co[—>0. Equivalently, f is continuous at $0 if both Re f and Im f are continuous at
are. We say f is continuous on I if it is continuous at each point of I. The sum and product of two functions which are continuous at so are continuous
there. The complexvalued function f deﬁned on an interval I is said to be
diﬁerentiable at we in I if the ratio f(:v)  f(xo) x__x0 1 (x;£x0)7 has a limit at so. If f is differentiable at so we deﬁne its derivative at 220,
f’ (we), to be this limit. Thus, iff’(xo) exists, f(z)  f(xo) x_xo —f’(xo) —>0, as 0<fx—x0[—+0. An equivalent deﬁnition is: f is differentiable at are if both Ref and Im f
are differentiable at :50. The derivative of f at me is given by f’(xo) = (Ref)’(xo)+i(1mf)’(xo) Using these deﬁnitions one can show that the usual rules for differentiating
real—valued functions are valid for complex—valued functions. For example,
if f, g are differentiable at so in I , then so are f + g and fg, and (f + g)’(xo) = f’(xo) + g’(xo),
(fg)’(xo) =f’(xo)g(xo) +f(xo)g’(xo). If f is differentiable at every x in an interval I , then f gives rise to a new
function f’ on I whose value at each x on I is f’ (x) . 10 Preliminaries Chap. 0 A complex—valued function f with domain'the interval a g :1: § b is
said to be integrable there if both Re f and Im f are, and in this case we deﬁne its integral by
b b b
/ f(x) d3; =/ (Re f) (x) dx +if (Im f) (x) dx.
Every function f which is continuous on a § 00 g b is integrable there. This deﬁnition implies the usual integration rules. In particular, if f and
g are integrable on a g x g b, and a, B are two complex numbers, b b b
/ W + new dz = a [ f(x) dx + a / gm) dz. An important inequality connected with the integral of a continuous
complex—valued function f deﬁned on (1 § 5c é b is free) dx s flies) Idrc* This inequality is valid if f is realvalued, and the proof for the case when
f is complex—valued can be based on this fact. Let F = [7 f(x) dx. If F = 0 the inequality is obvious. If F 75 0, let
F=Fu, u=cos0+isin0, (0§0<27r). Then ml = 1, and we have [fee dx =iI/abf(x) dx = Re[a£bf(x) dx] = beeEﬁﬂxﬂdx g [him ldx, *By b
f lf(x)ldx is meant the integral of the function If  given by If I (x) = lf(x) for a g :3 § b. Thus
a more appropriate notation would be b
/ f(x)dx. We shall use the former notation since it is commonly used, and there will be no chance
of confusion, Sec. 3 Preliminaries 11 since Re [mm s We)! = We) I. As particular examples of complexvalued functions let
f(x) = x + (1 — 17762,
9(x) = (1 +0962, for all real 95. Then (Re f) (x) = a: + 962, (1m f) (x) = x2, (f+g)(x) =x+2x2,
(EH18) = (1+i)$3+2$‘,
f'($) =1+ (2*2i)$y
[0f(x)dx=Axdx+(1—i)£x2dx=g_g (c) Complex—valued functions with complex domains. We shall need to
know a little about complexvalued functions whose domains consist of
complex numbers. An example is the function f given by f(3) = z": for all complex z, where n is a positive integer.
Let f be a complex—valued function which is deﬁned on some disk D: Iz—al<r with center at the complex number a and radius r > 0. Much of the calculus
for such functions can be patterned directly after the calculus of complex valued functions deﬁned on a real interval I. We say that f has the com
plex number L as a limit at 20 in D if f(z) — LI —>0, as 0 < Iz—zol —)0, and we write limf(z) =L, or f(z)—>L, (2—920). 2—»:
0 If f and g are two complexvalued functions deﬁned on D such that for some
20 in D I‘M—>11, 9(2)—>M, (zMo), 12 Preliminaries Chap. 0 then
(f+ g)(2) —>L + M, (fg)(z) —>LM, (2—wa The proofs are identical to those for functions deﬁned on real intervals.
The function f, deﬁned on the disk D, is said to be continuous at 20
in D if
f(z) —f(zo) l——>0, as 0 < [2 — all—>0. It is said to be continuous on D if it is continuous at each point of D. The
sum and product of two functions which are continuous at 20 are continuous
there. Examples of continuous functions on the whole complex plane are f0?) = lzt 9(2) = 23. Let g be deﬁned on some disk D1 containing 20, and let its values be in
some disk D2, where a function f is deﬁned. If g is continuous at 20, and f
is continuous at g(zo), then “the function of a function” F given by 17(2) = (z in D1); is continuous at 20. The proof follows the same lines as in calculus for real
valued functions deﬁned for real 2:.
If f is deﬁned on a disk D containing .20 we say that f is diﬁferentiable at 20 f(z) — f(zo) 2’20 ’ (z ¢ 20)! has a limit at 20. If f is differentiable at 20 its derivative at 20, f’ (zo) , is deﬁned
to be this limit. Thus 2"‘20 ——f’(zo) ——>0, asO<z—zo—>0. Formally our deﬁnition is the same as that for the derivative of a complex
valued function deﬁned on a real interval. For this reason if f and g are
functions which have derivatives at 20 in D then f + g, fg have derivatives
there, and (f + g)'(Zo) = f’(Zu) + g'(zo),
(3.2) (fg)’(zo) = f’(Zo)g(Zo) + f(20)g'(20)~ Also, suppose f and g are two functions as given in (3.1), and that g is
differentiable at 20, whereas f is differentiable at g(20). Then F is differen
tiable at zo, with 'F' (20) = f'(9(zo) )9' (20} Sec. 3 Preliminaries 13 It is clear from the deﬁnition of a derivative that the function (1 deﬁned
by (1(2) = c, where c is a complex constant, has a derivative which is zero
everywhere, that is, q’(z) = 0. Also, if 121(2) = z for all 2, then p{(z) = 1.
Combining these results with the rules (3.2) we obtain the fact that every polynomial has a derivative for all z. A polynomial is a function 1) whose
domain is the set of all complex numbers and which has the form 72(2) = aoz" + an“ +   + (In—12 + an, where a0, a1,  ~ , a" are complex constants. The rules (3.2) imply that for
such a p p’(z) = %7zz”“1 + a,1(n —— Dan—2 +  + and. Thus 12’ is also a polynomial. It is a rather strong restriction on a function deﬁned on a disk D to
demand that it be differentiable at a point 20 in D. To illustrate this we note
that the real—valued function f given by f(x) = [$1, for all real .1, is differentiable at all cc 75 0. Indeed f ’( x) is +1 or ——1 accord
ing as a: is positive or negative. However the continuous complexvalued
function g given by 9(2) = Izl, for all complex 2, is not differentiable for any 2. Suppose 20 = x0 + 1101' sf 0,
for example, and let 2 = x + yi. Then for z ¢ 20 V!  leol = (962 + 7J2)”2  (x3 + 2/3)“2
2—20 (x—xo)+i(yyo)
= (x2 + W  (xi + 2/3)
[(96 — me) + My * yoljﬂwz + 3/2)“2 + (x3 + 313W] If we let Iz — col 90 using 2 of the form 2 = $0 + yz‘ (that is y—> go)
we see that izl_izoi ) 1/0
z—a' un+nww (3.3) Whereas if we let I2 — 20[ —>0 using 2 of the form 2 = x + W (that is
x —+ me) we obtain
l2!  IZOI we a . .4.
2—a "Wa+aw2 (3) 14 Preliminaries Chap. 0 The two limits (3.3) and (3.4) are different. However, in order that g be
differentiable at 20 we must obtain the same limit no matter how I2 —— zol —> 0. This shows that g is not differentiable at 20.
(d) Other functions. Other types of functions which are important for our study of differential equations are usually combinations of the types
discussed in (b), (c) above. Typical is a complex—valued function f which
is deﬁned for real as on some interval Ix — xol g a (mo real, (1 > 0), and for
complex 2 on some disk lz —— zol § b (20 complex, b > 0). Thus the do
main D of f is given by
D: lx~xol ga, Iz—zol éb, and the value of f at (x, z) is denoted by f (x, 2). Such a function f is said to
be continuous at (E, n) in D if lf(x,z) “f(£,n)l—*0, as 0 < Ix— 21 + 12— nl >0 There are two important facts which we shall need in Chap. 5 concern
ing such continuous functions. The ﬁrst is that a continuous f on the D
given above (with the equality signs included) is bounded, that is, there is a
positive constant M such that lf(x,z)l é M, for all (x, z) in D. This result is usually proved in advanced calculus
courses. The second result relates to “plugging in” a complex—valued func
tion 4) into f. Suppose 45 is a complex—valued function deﬁned on lit—550' gay which is continuous there, and has values in la — zol g b. Then if f is
continuous on D, the function F given by FCC) =f(x,¢(x)), for all x such that [as —— :50] § (1, is continuous for such as.
A slightly more complicated type of complex—valued function f is one
which is deﬁned for real 2' and complex 21,    , 2,. on a domain
D: [1— fuel é a, In —' zlol + + lzn— znol é b. Here are is real, 2m,   , 2,0 are complex, and a, b are positive. The value of
fat cc, 21,  ~  , 2,. is denoted byf(x, zl,    , 2,.) . Continuity offis deﬁned just Sec. 3 as in the case of one 2. Thqu is continuous at E, m,   o, 17,. in D if 0< latEl + IZi—ml +~+ lzn_nn—)0. Such an f is bounded on D, and if (351, ~ 
valued functions deﬁned on Ix — xol _S_ (1, having the property that Preliminaries “(37,211 "'5zn) fQNTI: "3771!” *0; 15 , qsn are n continuous complex ¢1(x) "Ziol + "' + ¢n(x) — Znol é b for all such 25, then the function F given by Ft”) =f(xi¢l(x)) ""¢n(x)) for x — x0] é a is continuous there. EXERCISES ,l. Leta: 2+i3,b= 1— i. Ifforallreala:
f(2) = ax+ (bx)? compute : (a) (Ref)(rc)
(c) f’(x) ‘ 2. If for all real 1: compute : f(rv) = x+ ix”. (b) (Imoe)
l d dz < > [0 f(x) x2 9(90) = —, (a) The function F given by F (x) = f (g(:c)) 2 (b) F’(x) 3. If a is a real—valued function deﬁned on an interval I, and f is a complex valued function deﬁned there, show that Re (W) = a(Re D, 4. Let f(z) = 22 for all complex 2, and let 24(33, y) = (313 fXx + W);
(a) Compute u(x, y) and 0(23, y). (b) Show that du 82) 62: w ﬂy, Im (af) = a(Im 0(1', y) = (1m f)(x + iy). 16 Preliminaries Chap. 0 (0) Show that
6% 62a 62v 620 6x2 + 6y2 ’ 6x2 ﬂy2 5. Let f be a complexvalued function deﬁned on a disk
D: Iz<r (r>0),
which is differentiable there. Let
"(95, y) = (Re f)($ + 1'31), 9(3), y) = (1m no + W) Show that
6a 80 Ba _ 6v ace—6y, 53—; 6:15, forallz = x+ iyinD. (Hintlfzo = x0+ iyoisinD,let0 < [2 — eel—>0,
in the deﬁnition of f’(2o), through 2 of the form 2 = as + iyo, and then of the
form 2 = x0 + lg, to obtain 6 6
rec) = is... 2/0) + rice, yo) =—( )—“( l
x ’t x, .
0,110 a 03/0 The equations (*) are called the CauchyRiemann equations.) . 6. Let f be the complexwalued function deﬁned on DI lxlél, HMS—2,
(a: real, 2 complex) by
f(x, 2) = 32;2 + x2 + 22,
and let d) be the function deﬁned on I a:  g l by
We) = x + i.
(a) Compute the function F given by
F03) = f(x, ¢(x)), (l xl é 1). (b) Compute F'(:v).
(0) Compute [01 F(x) dz. — — (*)4 Sec. 4 Preliminaries 17 , 7. If r is a complex number, and
29(2) = (2 — r)", Where n is a positive integer, show that W) = p’(r) = = p‘"“’(r) = 0, p‘"’(r) = n! 4. Polynomials We have deﬁned a polynomial as a complexvalued function 20 whose
domain is the set of all complex numbers and which has the form 17(2) = 04:2" + (112"‘1 +    + an_1z + an, where n is a nonnegative integer, and a0, a1    , an are complex constants.
The highest power of z with nonzero coefﬁcient which appears in the
expression deﬁning a polynomial p is called the degree of p, and written
deg 17. A root of a polynomial p is a complex number r such that p(r) = 0. A root of p is sometimes called a zero of 12. We shall require, and assume, the
following important result.* Flmdamental theorem of algebra. If p is a polynomial such that
deg p g 1, then p has at least one root. This is a rather remarkable result, and justiﬁes our introduction of the
complex numbers. We have seen that not every polynomial with real
coefﬁcients (for example 22 + 1) has a real root, but polynomials of degree
greater than zero with complex coefﬁcients always have a complex root.
The remarkable fact is that we do not need to invent new numbers, which
include the complex numbers, to guarantee a complex root. We derive some consequences of this fundamental theorem. Corollary 1. Let p be a polynomial of degree n g 1, with leading coeﬁ‘i—
cient 1 (the coeﬁlcient of z"), and let r be a root of p. Then p(z) = (z — r)q(2)
where q is a polynomial of degree n — 1, with leading coeﬁicient 1.
Proof. Let 12(2) have the form 22(2) = z" + aiz'”l +    + tin—12 + an, * A proof can be found in G. Birkhoﬂ' and S. MacLane, A survey of modern algebra, New York, rev. ed., 1953, p. 107, and also in K. Knopp, Theory of functions, New York,
1945, p. 114. 18 Preliminaries ‘ Chap. 0 and let 6 be any complex number. Then
22(2) — 10(6) = (2" — c") + (1102"—1 — c“) + + an1(z — c)
= (z  c)q(z),
where q is the polynomial given by
q(z) = 2"‘1 + cam—2 + czz"‘3 +  ' ' + 0"”1
+ a1(z"‘2 + 62"‘3 +    + c"") +    :I am. Clearly deg q = n — 1 and g has leading coefﬁcient 1. In particular if
c = r, a root of p, then we have p(z) = (z  104(2), as desired.
If n —— 1 _Z_ 1, the polynomial q has a root, and this root is also a root of p by Corollary 1. Thus applying the Fundamental Theorem of Algebra
n times, together with Corollary 1, we obtain Corollary 2. If p is a polynomial, deg p = n g 1, with leading coeﬁ‘l—
cz'ent a0 ;é 0, then p has exactly 11 roots. If r1, r2,   ~ , rn are these roots, then pt?) = ao(z — 1'1) (z — r2) (z — r"). (4.1) Note that aglp is a polynomial which has leading coefﬁcient 1. We re—
mark that the roots need not all be distinct. If r is a root of p; the number
of times 2 — r appears as a factor in (4.1) is called the multiplicity of 7'. Theorem 1. If r is, a root of multiplicity In of a polynomial p, deg p g 1,
then
W”) = IN“) = = p‘“‘”(r) = 0,
and
p('"’(r) # 0. Proof. Let p have leading coefﬁcient on ¢ 0, and degree n g m. It
follows from Corollary 2 that 22(2) = aoe — We, (42) where q is a polynomial of degree n —— m, and q(r) ;é 0. Clearly p0“) = 0
by the deﬁnition of a root. Also p’Cz) = agm(z — r)'"“q(z) + (10(3 — r)"‘g’(z), Sec. 4 Preliminaries 19
and this implies that, if m — 1 > 0, p’(r) = 0. If m = 1 we have
P’(z) = aoq(2) + ao(z  r)q’(z), and thus p’(r) = a0q(r) 75 0.
The general argument can be based on (4.2) and the formula
Ic(k —— 1) (fg)"°’ = My + (CW“9’ + 2. far—mg” + + fgoc) (4.3) I for the kth derivative of the product fg of two functions having 19 deriva
tlves. Formula (4.3) can be established by induction. Applying (4.3) to
the funct10nsf(z) = (z — 1')“, 9(2) = q(z) in (4.2), we obtain p‘Wz) = ao[m(m — 1) (m — k + 1) (z — r)"'”°q(z)
+ (terms with higher powers of (z — r) as a factor) ]. It is now clear that p0.) ___ pi“) = ... = p(m—1)(r) = 0’
and Mo) = aom! go) :75 0, which is the desired result. EXERCISES 1. Compute the roots, with multiplicities, of the following polynomials: (a)zz+z—6 (b)22+2+,1
(c) 23 — 322+ 4 (d) 23 — (2+ 7322+ (1+ 72).? i
(8)24—3 ’ 2. If r is such that r3 = 1, and 1' ¢ 1, prove that 1+ 7' + r2 = 0. 3. Let p be the polynomial given by
'. M2) = aoz"+ W“ + + am
with a0, a1,   , an all real. Show that
17?) = W).
As a consequence show that if r is a root of p, then so is 7". . 4. Prove that every polynomial of degree 3 with real coefﬁcients has at least
one real root. ' 20 Preliminaries Chap. 0 5. Prove that if p is a polynomial, deg 12 > 1, and r is a complex number such that
11(r) = 11’0") = = p‘m‘1)(r) = 0. p‘“’(r) 75 0,
then r is a root of p with multiplicity m. This is the converse of Theorem 1. 6. (a) Use the result of Ex. 5 to show that i is a root of the polynomial 1)
given by 73(2) = 25+ (2 — 3024+ (fl — 6023+ (—6  51322 + (6 + 2i)z + 2i, and compute the multiplicity of i.
(b) Find the other roots of the polynomial p in (a). 7. Prove the formula (4.3). This can be written in the form k k
_ ~ I (lg—2) II
(for) — ﬂk’g + (1)“ “g + (2)f g
+ + (Dr's—“gm + + fg‘k’, (k) _ ls!
l _ l!(k— l)! is a binomial coefﬁcient. H int: Use induction, and show that
k + 1)  ( k ) (k)
< l ' z— 1 + z ' 5. Complex series and the exponential function where If a: is a real number, and e is the base for the natural logarithms, the
number 6" exists, and
m xiv vr = —, 0! = 1 , ° a. k! ( )
where theseries converges for all real Indeed, this series may be taken
as the deﬁnition of e”. We shall need to know what (22 is for complex 2. One
way is to deﬁne e3 by a) 2k 6‘ = . (5.1)
k=0 kl
Now we have to prove that this series converges for all complex 2, and in
fact there is the problem of deﬁning what we mean by a convergent series
with complex terms. The method is the sameas that used to deﬁne con—
vergcnt series with real terms. ,V “sageswwhw. . , u. “WWWWWY‘. inﬂnmn, . > «vm kwrr" «may»? mmm “ml”... was“ New“, We See. 5 Preliminaries 21 A series I; Ck, V (5.2) where all ck are complex numbers, is said to be convergent if the sequence of
partial sums sn=zck1 (nz0r192yn')!
k—O tends to a limit 3, as n —> 00. That is, s is a complex number such that lsn  SI —>0, (n—+ 00), Where the magnitude is the magnitude for complex numbers. If the series
(5.2) is convergent, and 3,. —> s, we call 8 the sum of the series, and write an
S = Z 6),.
kﬁl If the series is not convergent we say that it is divergent.
The series (5.2) with 00mplex terms or gives rise to two series with
real terms, namely ‘ f: Re ck, k=0 go: Im ca (5.3) and it is not difﬁcult to see that the series (5.2) is convergent with sum
8 = Re 3 + iIm 8 if, and only if, the two real series in (5.3) are con
vergent with sums Re s and Im s respectively. In principle, therefore, the
study of series with complex terms is the study of pairs of real series. The series (5.2) is said to be absolutely convergent if the series k;  a.  (5.4) is convergent. It can be shown that every absolutely convergent series is
convergent. Since the series (5.4) has terms which are real and nonnegative,
any condition which implies the convergence of such series can be applied
to guarantee the convergence of the series (5.2). One of the most important
testsfor convergence is the ratio test. One version of this is the following. Ratio test. Consider the series a: Zlckl, 22 Preliminaries Chap. 0 where the 0,. are complex. If Ickl > 0 for all k beyond a certain positive
integer, and ick+1i
I Ch I —> L. (k —* w), (55) then the series is convergent if L < 1, and divergent for L > 1. Thus the series (5.2) is convergent if (5.5) is valid for an L < 1.
An immediate application of this result is to the series co 3k
En
Here ck = z"/Ic!and
6k+1 2"“ k! lzl
= ———— = k .
Ickl (Ic+1)!z'= Ic+l—)0’ (Tm) Thus this series converges for every z such that < 00, that is, for all
complex 2. Hence our deﬁnition (5.1) of e’ as the sum of this series makes
sense. The function which associates with each 2 the complex number 6’
is called the exponential function. The series deﬁning e‘ is an example of a power series 0:) Z ak(z '— 20),“ kO about some point 20, the a]. being complex. Many of the properties of a
power series of the type (D
Z ak(x _ 1:0)";
i=0 where the ca, x, :30 are real, remain true for series of the form (5.6), and
the proofs are identical. In particular, if a series (5.6) is convergent on a
disk Dzlz — 201 < r (r > 0), then the functionf deﬁned by on f(2) = Z M2 — Zo)’°, k—O (z in D), has all derivatives in D, and these may be computed by differentiating term
by term. Thus f’(z) = 5: me — 20),”! = 5': me — sow,
luO i=1 ? .
~’ .. g,
if,
t}
X5"
é‘, '
«i i
i ., Sec. 5 Preliminaries 23 where the last series converges in D. Applying this result to (5.1) we ﬁnd
that zk—l a, zk—l a, 2k (6‘) =Ekﬁ=§(k—1)!=.§m Another important property of the exponential function is that = e‘. ez1+22 = 621622 for every complex 21, 22. This can be proved by justifying the following
steps I: k
€21,622 = _ __ = .
(2s, this, k!) 2 c" k Ic—O
Here Thus formally we have the product of the series deﬁning e'1 and e'2 is the
series deﬁning e'h‘”, and these steps can be justiﬁed to give a proof of the
equality (5.17) . A consequence of (5.7) is that (ez)n = en: for every integer n. In particular l/e' = e“.
Another property of the exponential function is that for all real 0, e“ = cos 0 + isin 0, (5.8)
and the proof results from adding the series involved. Indeed, i2 = —1,
i3 = —i, i4 = 1, etc., and thus
62 94
0089:]. "‘
(M2 (179)“
._ 1 + 2' 4! _ 03 65
sm0—0—§+ﬁ ",
‘ . 3 ' 5
ismﬂ = w+(1)+(w) 3! 5! 24 Preliminaries Chap. 0
Hence . . I “(9)2 (my
cosﬁ+zsm0=1+w+ + +
2! 3!
= 20.: (my: = e“.
k=0 k!
A consequence of (5.8) is that
r” = cos 0 — isin 0, (5.9)
since cos (—0) = cos 0, and sin (—0) = — sin 0. Using (5.8) and (5.9)
we can solve for cos 0 and sin 6, obtaining
eiO + 6—129
cos 0 = ———2— ,
eiB _ e—i9
sin 0 = ——————
22‘ If 2 is a complex number with polar coordinates (r, 0) , then
z=r(cos0+isin0), (r;0,0§0<27r),
and we have, using (5.8) , z = re”. (5.10) Note that [2] = r, Ieial = 1 for every real 6. The relation (5.10) can be
employed to ﬁnd the roots of polynomials p of the form 10(2) = z" — c, (5.11) where c is a complex constant. Suppose c = [cl 6"“, where a is real, 0 g a <
21r, and re“ is a root. Then rneino = iclem’
and taking magnitudes of both sides we see that
r" = Icl, or r =cl1/",
where the positive n—th root is understood. Further em = eia, or ei(n9—a) = 1, Sec. 5 Preliminaries 25 There are exactly n distinct values of 0 satisfying this relation and 0 g 0 <
21r, namely, those for which n0 — a = 21rk,
or 2k
o=9~+7£, (lo=0,1,u,n—l). Thus the roots zl,   , 2,. of the polynomial p in (5.11) are given by zk+l = I c ll/n6i(a+21rlc)/n = lcll’" [cos(%12—dc)+ isin(c—¥—j—;L~2Lk)], (k =0,1, n,n — 1). Geometrically we can describe the roots of p as follows. All roots lie on a
circle about the origin with radius Icl 1’". One root has an angle 01/11 with
the real axis, if c has angle a with the real axis. The remainder of the roots
are located by cutting the circle into n even parts, with the ﬁrst cut being
at the root at angle a/n. As a particular example let us ﬁnd the three cube roots of 41'. Thus we
want the roots of 23 — 41'. Here 0 = 41', and hence the cube roots will all
have a magnitude of [41' [1’3 = 4‘“. If we write 6 = Icle‘“, we see that
a = 7r/2 in this case. Thus the three cube roots of 42’ are given by 21 = 41/361316, 22 = 41/361151”, as = 41/3ei9rI6’ Imaginary
axis Figure 2. Three cube roots of 413 26 Preliminaries Chap. 0
or since 7r/6 represents 30°, 2.», = — 41/321
These roots are sketched in Fig. 2.
E X E R C I S E S
1. Find the three cube roots of l.
2. Find the two square roots of i.
3. Find all roots of the polynomials:
(a) 23 + 24 (h) z4 + 2'64
(c)z‘+4z2+4 x(d)2‘°°—1
: 4. If 2 = x + iy, where x, y are real, show that I e‘ I = e”. As a consequence / show that there is no complex 2 such that e‘ = 0. _ 5. If a, b, a; are real show that: (a) Re [e(“+"’)’] = e” cos b2: (b) Im [e(“+"’)’] = 9‘” sin bx 6. (a) If r = a + 2'6 75 0, where a, b are real, show that (e”)’ = re".
(b) Using (a) compute: (i) [01 e" da: 1
(ii) / e” cos b2; d2:
0 1
(iii) / c“ sin bx d1:
0 , 7. (a) If ¢(:v) = e”, where r is a complex constant, and :1: is real, show that
¢’(x) — WW) = 0
(b) If (Mac) = em, where a is a real constant, show that:
(i) ¢'(x) — ia¢(x) = 0
(ii) 45"(35) + a2¢(x) = 0 8. For What values of the constant r will the function 4) given by ¢(x) = e”
satisfy ¢"(x) + 3¢’(x) — 24:05) = 0 for all real 1;? regime 22:, Sec. 6 Preliminaries 27 9. Let as = k! + (i/kl). For what real 1:, are the following series convergent? (a) f, (Re ak)x" (b) 2“: (Im mack
ko k—O (0) 2 am"
[50
I}, 10. Consider the series
. E 2k, (*) where z is complex.
(a) Show that the partial sum 8,.(2) = Z": 2" =
tan 1 _ zn+l l—z ’ ifz¢ l. (b) Show that the series (*) converges absolutely for I z I < l.
(0) Compute the sum 8(2) of the series (*) for I zI < l. 6. Determinants We shall need to know the connection between determinants and the
solution of systems of linear equations. Suppose we have such a system of n
equations (11121 + (11232 + ' ' ' + alnzn 01 02 (6.1) can + (12222 +    + aznzn «1.121 + c.222 + '   + annzn = on, Where the a5, and c; are given complex constants. The problem is to ﬁnd
complex numbers 21, , 2,. satisfying these equations. Such a set of n
numbers is called a solution of (6.1). We say that two solutions 21,   o, 2,.
andzi, ,z,:of (6.1) areequal ile = 21’, ,z,. = 2;. Ifcl = 02 = =
6,. = O we say that the system is a homogeneous system of n linear equations, otherwise we say (6.1) is a nonhomogeneous system. The determinant A
of the coefﬁcients in (6.1) is denoted by an (112 "' aln 1121 (122 "‘ Gen 28 Preliminaries Chap. 0
and is shorthand for the number A given by
A = 2(i)ali1a2i2 ' '. ailin) where the sum is over all indices i1,   , in such that i1,   , in is a permuta
tion of 1,   , n and each term occurs with a + or — sign according as
i1,   , in is an even or odd permutation of 1,   , n. Thus an 012
=a114122 " (1121121,
€121 (122 and ’ an 012 (113
= auazzaaa ‘ (11111231132 + (112023031 “21 €122 023
“ alzamaaz + (1130211132 — a13a22aal
031 (132 (133 The principal results we require concerning determinants are contained in
the following theorems. They are usually proved in elementary texts on linear algebra. Theorem 2. If the determinant A of the coeﬁicients in (6.1) is not zero
there is a unique solution of the system for 21, ~  , 2". It is given by 216:2" (k=1)"’7n)a
where Ah is the determinant obtained from A by replacing its kth column
a”, ,a,.kbycl, ,c,.. Proof for the case n = 2. In this case suppose 21, 22 satisfy (11121 + (11252 = 01 (6.2)
02131 + 02222 = 02 Multiply the ﬁrst equation by (122, the second equation by —a12, and add.
There results 01 612
21A 2 (12261 —' (21262 = = A1
02 (122 Sec. 6 Preliminaries 29 Multiply the ﬁrst equation by —a21, and the second by an, and add, ob
taining
an 01
22A = —a2101 + (11162 = = A2.
021 62 Thus if A ¢ 0, zk must be Ak/ A (k = 1, 2), and it is readily veriﬁed that
these values satisfy (6.2). We note that for a homogeneous system (cl = 02 =   = 6,. = 0 in
(6.1)) there is always the solution zl=z2=luu=zn=0. This solution is called the trivial solution. Theorem 3. If cl = (32 = = 0,. = 0 in (6.1), and the determinant
of the coeﬁicients A = 0, there is a solution of (6.1) such that not all the zk are 0. Proof for the case n = 2. We are dealing with the case
01121 + (11222 = 0 azlzi + a2222 = 0,
where (111022 — 021012 = 0.
If an 75 0,
_a12 21 = , 2'2 = 1,
an is a solution. If an = 0, and a2; 75 0, _a22 is a solution. If an = 0, and a2; = 0, is a solution. ‘
Combining Theorem 3 with Theorem 2 we obtain Theorem 4. The system of equations (6.1) has a unique solution if, and
only if, the determinant A of the coeﬁicients is not zero. Proof. If A ¢ 0 Theorem 2 says that there is a unique solution. Con
versely, suppose there is a unique solution 21,   , 2,. of (6.1). If A = 0, by 30 Preliminaries Chap. 0 Theorem 3 there is a solution {1,   , {n of the corresponding homogeneous
system, which is not the trivial solution. Then it is easy to check that
z; + {1,   , 2.. + £3. is a solution of (6.1) distinct from 21, , z”, and
forces us to conclude that A ;é 0. EXERCISES 1. Consider the system of equations
izl + 22 = 1 + 'I;
221 + (2 — 1:)22 = 1. (a) Compute the determinant of the coefﬁcients.
(b) Solve the system for 21 and 22. 2. Solve the following system for 21, 22 and 23:
321 + 22 — 23 = 0
221 — 23 = 1 32+2Z3=2 3. Does the following system of equations have any solution other than
21 = 22 = 23 = 0? If so find one. 421+222+223=0
321+ 722+ 223: 0
221+ 22+ 23:0 4. Consider the homogeneous system corresponding to (6.1) (the case 01 ——
62 =  °  = cn = O). Show that if the determinant of the coefficients A = 0, there are an inﬁnite number of solutions. (Hint: If 21,   , 2,. is a nontrivial
solution, show that 0121,    , azn is also a solution for any complex number a.) 5. Prove that if the determinant A of the coefficients in (6.1) is zero then
either there is no solution of (6.1), or there are an inﬁnite number of solutions. (Hint: Use Ex. 4.) 7. Remarks on methods of discovery and proof Often a student studying mathematics has difﬁculty in understanding
why or how a particular result, or method of proof, was ever conceived in
the ﬁrst place. Sometimes ideas seem to appear from nowhere. Now it is
true that mathematical geniuses do invent radically new results, and meth
ods for proving old results, which often appear quite strange. The most
that ordinary people can do is to accept these brilliant ideas for what mmwm —. We..,...m...w,m..w mwsmwmwww... , Sec. 7 Preliminaries 31 they are, try to understand their consequences, and build on them to
obtain further information. However, there are a few general principles
which, if followed, can lead to a better understanding of mathematical dis
covery and proof. Concerning discovery, we mention two principles: (a) use simple examples as a basis for conjecturing general results,
(b) argue in reverse. Both of these principles are illustrated in the proof we gave of Theorem 2
for the case n = 2. We were faced with trying to find out whether the
system (6.1) of n linear equations has a solution or not, and what condition,
or conditions, would guarantee a unique solution. We looked at the simplest
example, which occurs for n = 2 (using ((1)). Then we assumed that we
had a solution (principle (b)), and found out what must be true for a
solution, namely, that 21A = A1, 22A = A2. We immediately saw that if A ¢ 0, then A1 A2
Zl=‘_ Zz=~.i A . A (7.1) Note that at this point we have not yet shown that there is a solution. All
we have shown is that if 21, 22 is a solution, and A 75 0, it must be given by
(7.1). We can now guess that if A 75 0, then 21, 22 given by (7.1) is a solu
tion. This can be readily veriﬁed by substituting ( 7 .1) into the given
equations. An alternate procedure is to check that the steps leading to
(7 .1) can be reversed, if A # 0. Once we have discovered the right condi
tion for the case n = 2, it is natural to conjecture that a similar condition
will work for a general n.
Three important methods of proving mathematical results are: (i) a constructive method,
(ii) method of contradiction,
(iii) method of induction. A typical example of aconstructive method appears in the proof of Theorem 3
for the case n = 2. We wanted to show that nontrivial solutions of the
two homogeneous equations exist if A = 0. To do this we constructed
solutions explicitly. An example of the method of contradiction appears in
the proof of Theorem 4. We supposed that the system (6.1) had a unique
solution. We assumed that A = 0, and, using logical arguments, we arrived
at the fact that ( 6.1) does not have a unique solution. This is a contradic
tion, and the only thing that can be wrong is our assumption that A — 0. 32 Preliminaries Chap. 0 The only other alternative is that A 75 0, which is the conclusion we de—
sired. ' The method of induction is concerned with proving an inﬁnite number
of statements 31, 82,   , one for each positive integer n. If 81 is true, and
if for any positive integer k the statement 3,, implies the statement 8H1,
then all the statements 81, 82,   , are true. An example of a result which
can be proved using induction is the formula k Ic k k!
(k) = (In—z) a) = “—
(fg) Half ‘7 ’ (l) l!(Ic—l)!’
for the kth derivative of the product of two complexvalued functions
f, g which have It derivatives; see (4.3). The proof is the same as the induc
tion used to prove the binomial formula I:
(a + w = 120 aklbl, (Is =1,2,),
for the poweis of the sum of two complex numbers a, b. The method of
induction is equivalent to a property of the positive integers, and conse
quently we assume that this method is a valid method of proof. The principles of discovery (a), (b), and the methods of proof (i),
(ii), (iii), will be used many times throughout this book. It will be instruc—
tive for the student to identify which principles and methods are being
used in any particular situation. CHAPTER 1 Introduction—Linear Equations of the First Order 1. Introduction In Sec. 2 we discuss what is meant by an ordinary differential equation
and its solutions. Various problems which arise in connection with differ
ential equations are considered in Sec. 3, notably initial value problems,
boundary value problems, and the qualitative behavior of solutions. In a
succession of easy steps we solve the linear equation of the ﬁrst order in
Secs. 4—7. 2. Diﬂ'erential equations Suppose f is a complexvalued function deﬁned for all real :6 in an
interval I , and for complex y in some set S. The value of f at (x, y) is
denoted by f (x, 1/). An important problem associated with f is to ﬁnd a
(complexvalued) function d; on I , which is differentiable there, such that
for all x on I, (i) ¢>(x) is in S,
(ii) ¢’(x) =f(x, ¢(x)). This problem is called an ordinary diﬁerential equation of the ﬁrst order,
and is denoted by y’ = f(x, fl) (21) The ordinary refers to the fact that only ordinary derivatives enter into
the problem, and not partial derivatives. If such a function ¢ exists on I
satisfying (i) and (ii) there, then qs is called a solution of (2.1) on I. 33 ...
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This note was uploaded on 05/12/2008 for the course MATH 364 taught by Professor Snyder during the Spring '08 term at Simons Rock.
 Spring '08
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