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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(...
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 Spring '08
 Snyder
 Complex Numbers, Complex number

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