Download INTRODUCTION TO COMPLEX NUMBERS

MA 232
Fall d999
Di&erential Equations
Kevin Dempsey
INTRODUCTION TO COMPLEX NUMBERS†
Susanne C. Brenner and D. J. Kaup
Department of Mathematics and Computer Science
Clarkson University
Complex Arithmetic
(Complex conjugation, magnitude of a complex number, division by complex numbers)
Cartesian and Polar Forms
Euler’s Formula
De Moivre’s Formula
Di&erentiation of Complex Functions
One of the most important numbers in complex analysis is i. The number i is a solution of
the equation x2 + 1 = 0 and has the property that i2 = 1. (Of course, another solution of
x2 + 1 = 0 is i.)
We write a complex number z as z = x + iy (or x + yi), where x and y are real numbers.
We call x the real part of z and y the imaginary part of z. We write Re z = x and Im z = y.
Whereas the set of all real numbers is denoted by R, the set of all complex numbers is
denoted by C.
Re(1 + 3i) = 1,
Examples
Im(1 + 3i) = 3
Every real number x can be considered as a complex number x + i0. In other words, a real
number is just a complex number with vanishing imaginary part.
Two complex numbers are said to be equal if they have the same real and imaginary
parts. In other words, the complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are equal if and
only if x1 = x2 and y1 = y2 .
Example
2 3i 6= 2 + 3i
If z = x + iy ; C then x, y ; R and the absolute value of z is |z| = (x2 + y 2 )1/2 and the
conjugate of z is z̄ = x iy.
Addition and multiplication in C are defined by
z1 + z2 = (x1 + iy1 ) + (x2 + iy2 ) = (x1 + x2 ) + i(y1 + y2 )
z1 z2 = (x1 + iy1 )(x2 + iy2 )
= x1 x2 + ix1 y2 + iy1 x2 + i2 y1 y2
= (x1 x2 y1 y2 ) + i(x1 y2 + x2 y1 )
Clearly then
†
zz̄ = (x + iy)(x iy) = x2 + y 2 = |z|2
This handout is an amended version of the original.
Complex Numbers Handout
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Clarkson University
MA 232
Fall d999
Di&erential Equations
Kevin Dempsey
The Rules of Complex Arithmetic
For z1 , z2 , z3 , z = x + iy ; C,
(a) Addition and multiplication are commutative,
z1 + z2 = z2 + z1 , z1 z2 = z2 z1
(b) Addition and multiplication are associative,
(z1 + z2 ) + z3 = z1 + (z2 + z3 ), (z1 z2 )z3 = z1 (z2 z3 )
(c) Multiplication distributes over addition,
z1 (z2 + z3 ) = z1 z2 + z1 z3
(d) Each z has the unique additive identity 0,
z + 0 = (x + iy) + (0 + i0) = x + iy = z
(e) Each z has the unique multiplicative identity 1,
z · 1 = (x + iy) · (1 + i0) = x + iy = z
(f) Each z has the unique additive inverse z,
z + (z) = (x + iy) + (x iy) = 0 + i0 = 0
(g) Each nonzero z has the unique multiplicative inverse z 1 = 1/z = z̄/|z|2 ,
z̄
zz 1 = z 2 = 1
|z|
For example, (a) follows from the definition of addition in C and the commutative law for
real numbers:
z1 + z2 = (x1 + iy1 ) + (x2 + iy2 )
= (x1 + x2 ) + i(y1 + y2 )
= (x2 + x1 ) + i(y2 + y1 )
= (x2 + iy2 ) + (x1 + iy1 )
= z2 + z1
The other properties (b)-(g) may be similarly verified.
Because addition and multiplication in C satisfy (a)-(g), the set C, together with the two
operations of addition and multiplication, is a field. The set R of real numbers is also a field.
Note that for z 6= 0, (g) says that
z 1 =
Example
z̄
1
x iy
x
y
= 2 = 2
= 2
+i 2
2
2
z
|z|
x +y
x +y
x + y2
Find the multiplicative inverse of 1 2i
1
1 + 2i
1 + 2i
1
2
1
=
·
=
= +i
1 2i
1 2i 1 + 2i
1+4
5
5
The operation of taking complex conjugates satisfies two basic algebraic rules:
z1 + z2 = z̄1 + z̄2 ,
z1 z2 = z̄1 z̄2
Note that if z = x + iy then z + z̄ = x + iy + x iy = 2x and z z̄ = x + iy x + iy = 2iy.
From these two equalities we have
z z̄
z + z̄
, Im z = y =
Re z = x =
2
2i
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Clarkson University
MA 232
Fall d999
Di&erential Equations
Kevin Dempsey
We can also find the real and imaginary parts of z1 /z2 using complex conjugates. If z2 6= 0,
1
z̄2
z1 z̄2
z1 z̄2
z1
= z1 ·
= z1 ·
=
=
·
2
z2
z2
|z2 |
z2 z̄2
z2 z̄2
x1 + iy1
x1 + iy1 x2 iy2
(x1 + iy1 )(x2 iy2 )
=
=
·
=
x2 + iy2
x2 + iy2 x2 iy2
x22 + y22
Ã
!
x 1 x 2 + y1 y2
x2 y1 x1 y2
=
+i
2
2
x2 + y2
x22 + y22
Example
Simplify (3 2i)/(1 + i)
5 i
3 2i 1 i
3 3i + 2i + 2i2
5
1
3 2i
=
=
·
=
= i
2
1 + i
1 + i 1 i
1i
2
2
2
Example
Simplify (3i30 i19 )/(2i 1)
3(1)15 (1)9 i
3 + i
3i30 i19 3(i2 )15 (i2 )9 i
=
=
=
2i 1
1 + 2i
1 + 2i
1 + 2i
=
3 + 6i i 2i2
5 + 5i
3 + i 1 2i
·
=
=1+i
=
2
1 + 2i 1 2i
1 4i
5
As mentioned above, we define the
$ modulus or absolute value of a complex number z = x+ iy
as the nonnegative real number x2 + y 2 . We write
|z| =
Example
| 3 + 2i| =
q
x2 + y 2
$
$
9 + 4 = 13
The following modulus properties hold:
|z1 z2 | = |z1 ||z2 |,
¯ ¯
¯ z1 ¯
¯ ¯
¯ ¯
z2
=
|z1 |
(z2 6= 0),
|z2 |
|z|2 = x2 + y 2 = z z̄
Why should we bother with complex numbers?
Theorem (Fundamental Theorem of Algebra)
Any polynomial equation cn z n + cn1 z n1 + · · · + c1 z1 + c0 = 0, with complex coe^cients
c0 , c1 , . . . , cn and cn 6= 0, has n complex roots.
We can therefore solve any polynomial equation completely by using complex numbers. We
can’t say the same thing for reals. Consider x2 + x + 1 = 0. The quadratic formula yields
$
1 ± 1 4
x=
2
Complex Numbers Handout
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Clarkson University
MA 232
Fall d999
Di&erential Equations
Kevin Dempsey
which leads us to complex numbers.
These numbers are also really not “imaginary.” You can visualize z = x + iy as the point
(x, y) ; R2 , which is the (two dimensional) Cartesian plane.
Im 6
z = x + iy = rei
x
O
©
H
© ©
© ©
©
r© © ©
© s
H
y
-
Re
H
H
H
s̄
z = x iy = rei
Polar Coordinates
Let z = x + iy be a non-zero complex number. The point (x, y) has polar coordinates (r, ),
x = r cos ,
y = r sin where r > 0. Therefore
z = x + iy = (r cos + i sin ) = r(cos + i sin )
r(cos + i sin ) is a way of expressing a complex number by using polar coordinates. The
positive number r is just the modulus of z and the angle is called the argument of z. It is
determined up to multiples of 2$.
Examples‡
1+i=
¶
$ µ
$
$
2 cos + i sin
,
4
4
µ
¶
$
2$
2$
+ i sin
1 + i 3 = 2 cos
3
3
Euler’s Formula
We are all quite familiar with the Taylor series of the exponential function ea when the
exponent a is real§
a2 a3 a4 a5
+
+
+
+ ···
ea = 1 + a +
2!
3!
4!
5!
‡
These relations are derived in an example on Page 6.
Here, n! (n factorial) is defined by n! = n(n 1)(n 2) · · · 1 with 0! = 1. Hence, 1! = 1, 2! = 2 · 1 = 2,
3! = 3 · 2 · 1 = 6, 4! = 4 · 3 · 2 · 1 = 24, 5! = 5 · 4 · 3 · 2 · 1 = 120, etc.
§
Complex Numbers Handout
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Clarkson University
MA 232
Fall d999
Di&erential Equations
Kevin Dempsey
However, we may also use the same Taylor series to define the exponential function of an
imaginary exponent (and complex exponents as well). Let be a real number and consider
the exponential series when we replace a by . We then have
ei = 1 + i +
(i)2 (i)3 (i)4 (i)5
+
+
+
+ ···
2!
3!
4!
5!
Now let’s reduce this series by using the relation i2 = 1 and collect the real terms together
and the imaginary terms together. We obtain (after adding some more terms to make the
final result more obvious)
2
3
4
5
6
7
+ i3 + i4 + i5 + i6 + i7 + · · ·
2!
3!
4!
5!
6!
7!
2
4
6
=1
+
+ ···
4!
6!
à 2!
!
3 5 7
+
+ ···
+i 3!
5!
7!
ei = 1 + i + i2
We now have that the real part of ei and the imaginary part of ei are each a familiar Taylor
series. These are
2 4 6
+
+ ···
2!
4!
6!
3 5 7
+
+ ···
sin = 3!
5!
7!
cos = 1 Thus we have obtained the famous formula of Euler which is
ei = cos + i sin A very important property of ei is that it has a modulus of unity:
i
|e | =
q
cos2 + sin2 = 1
for any real . In other words, every complex number of the form ei lies on the unit circle
x2 + y 2 = 1 in the xy-plane.
Examples
$
$
1
$
3
+i
ei$/6 = cos + i sin =
6
6
2
2
$
$
1
1
ei$/4 = cos + i sin = $ + i $
4
4
2
2
ei$ = 1
ein2$ = 1 (n integer)
ei(+n2$) = ei
Complex Numbers Handout
( ; R, n integer)
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Clarkson University
MA 232
Fall d999
Di&erential Equations
Kevin Dempsey
The last result follows from the periodicity of cos and sin ; if n is an integer then cos( +
n2$) = cos and sin( + n2$) = sin .
So now the complex number z written in polar coordinates r(cos + i sin ) can be written
as rei . This latter form will be called the polar form of the complex number z.
Note that if z = rei = r(cos + i sin ), then
z̄ = r(cos i sin ) = r [cos() + i sin()] = rei
When two complex numbers are in polar form, it is very easy to compute their product.
Suppose that z1 = r1 ei1 = r1 (cos 1 + i sin 1 ) and z2 = r2 ei2 = r2 (cos 2 + i sin 2 ) are two
non-zero complex numbers. We first compute their product the hard way as follows:
z1 z2 = (r1 r2 )ei1 ei2
= (r1 r2 )(cos 1 + i sin 1 )(cos 2 + i sin 2 )
= (r1 r2 ) [(cos 1 cos 2 sin 1 sin 2 ) + i(cos 1 sin 2 + sin 1 cos 2 )]
= (r1 r2 ) [cos(1 + 2 ) + i sin(1 + 2 )]
= (r1 r2 )ei(1 +2 )
(We used trigonometric addition formulas to simplify in the second to last step.) So now we
have a very easy formula for multiplying two complex numbers in polar form. We simply
multiply their moduli and add their arguments.
From the rule for multiplication of complex numbers in polar form follows a rule for division.
We can easily see that
1
rei · ei = ei() = e0 = 1
r
Hence
(rei )1 = r1 ei
We now turn our attention to dividing two complex numbers in polar form
1
r1
r1 ei1
= r1 ei1 · ei2 = ei(1 2 )
i
2
r2 e
r2
r2
Therefore, to divide two nonzero complex numbers, divide their moduli and subtract their
arguments.
$
Example Convert z1 = 1 + i and z2 = 1 + i 3 into polar form and compute z1 z2 and z1 /z2
First compute the moduli
r1 = |1 + i| =
$
$
1 + 1 = 2,
Complex Numbers Handout
$
$
r2 = | 1 + i 3| = 1 + 3 = 2
-6-
Clarkson University
MA 232
Fall d999
Di&erential Equations
Kevin Dempsey
Hence
1+i=
$
$ i
2e 1 = 2 (cos 1 + i sin 1 ) ,
from which it is clear that
Ã
(cos 1 , sin 1 ) =
!
1 1
$ ,$ ,
2 2
$
1 + i 3 = 2ei2 = 2 (cos 2 + i sin 2 )
$ !
1 3
(cos 2 , sin 2 ) = ,
2 2
Ã
Either by looking at these points on the unit circle, or by plotting z1 and z2 in the complex
plane, we find that 1 is $/4 (plus any integer multiple of 2$) and 2 is 2$/3 (plus any integer
multiple of 2$). Therefore
¶
$ µ
$
$
$
2 cos + i sin
= 2ei$/4
4
4
µ
¶
$
2$
2$
+ i sin
z2 = 1 + i 3 = 2 cos
= 2ei2$/3
3
3
z1 = 1 + i =
Using the rules for multiplication and division of complex numbers in polar form, we have
z1 z2 = ei11$/12 ,
z1
1
= $ ei5$/12
z2
2
The following formula follows from the rules for multiplication and division of complex
numbers in polar form.
De Moivre’e Formula
³
ei
´n
= ein ,
n = 0, ±1, ±2, . . .
i.e. (cos + i sin )n = cos n + i sin n,
n = 0, ±1, ±2, . . .
Note that this gives us a very easy means for calculating the multiple angle formulas in
trigonometry. For example, let n = 3 in the above. Then, since (a+b)3 = a3 +3a2 b+3ab2 +b3 ,
cos 3 + i sin 3 = (cos + i sin )3
= (cos )3 + 3(cos )2 (i sin ) + 3(cos )(i sin )2 + (i sin )3
= cos3 + 3i cos2 sin 3 cos sin2 i sin3 ³
´
³
= cos3 3 cos sin2 + i 3 cos2 sin sin3 ´
Equating real and imaginary parts gives
cos 3 = cos3 3 cos sin2 ,
sin 3 = 3 cos2 sin sin3 Clearly the real part gives cos 3 in terms of cos and sin while the imaginary part gives
the corresponding expression for sin 3. Thus if you know De Moivre’s formula, and the
binomial expansion (a + b)n , you can always calculate any of the multiple angle formulas.
Complex Numbers Handout
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Clarkson University
MA 232
Fall d999
Di&erential Equations
Kevin Dempsey
Complex Exponential Functions
One can also define a general complex exponential function. The argument of the exponential
can be complex in general in which case
ea+ib := ea eib = ea (cos b + i sin b)
The complex exponential function has the following properties:
³
ea+ib
´n
³
= ea eib
´n
= ean eibn = ean+ibn = e(a+ib)n
ea1 +ib1 · ea2 +ib2 = ea1 eib1 · ea2 eib2
= (ea1 ea2 ) ei(b1 +b2 )
= ea1 +a2 ei(b1 +b2 )
= e(a1 +a2 )+i(b1 +b2 )
= e(a1 +ib1 )+(a2 +ib2 )
ea1 +ib1
= e(a1 +ib1 )(a2 +ib2 )
ea2 +ib2
The details of the last equality are left to the reader. Thus the laws of exponents for real
exponential functions also hold for complex exponential functions.
Example
Compute (1 + i)15
In polar form
¶
$
$ µ
$
$
= 2ei$/4
1 + i = 2 cos + i sin
4
4
and hence by the laws of exponents
¸
³$ ´15
³$ ´15 }
15$
15$
+ i sin
2 ei15$/4 =
2
cos
(1 + i)15 =
4
4
³$ ´15
$
$
Now
2
= 215/2 = 27 2 = 128 2. Also 15$/4 = 4$ $/4. Therefore
Ã
(1 + i)15
Example
µ
¶¸
$ } µ $¶
$
1
$
1
= 128 2 cos + i sin = 128 2 $ i $
4
4
2
2
!
= 128 128i
Solve the equation z 3 = 2i
To solve this equation, start with z = rei , then, expressing 2i in polar form also, gives
z 3 = r3 ei3 = 2i = 2ei$/2 = 2ei($/2+n2$)
Note the n2$ factor in the phase. We don’t know that the phase on the right-hand side
is equal to the phase on the left-hand side. We do know that up to the factor of n2$, the
phases must be the same. Equating moduli and arguments in the above equation gives
$
r3 = 2, 3 = + n2$
2
Complex Numbers Handout
-8-
Clarkson University
MA 232
Fall d999
Di&erential Equations
so
r = 21/3 ,
=
2$
$
4$
$
+n
= +n
6
3
6
6
Plugging in n = 0, 1, 2 we obtain the roots
$
$
3
3
z1 = 2ei$/6 = 2
z2 =
$
3
2ei5$/6
z3 =
$
3
2ei3$/2
Kevin Dempsey
Ã$
!
1
3
+i
2
2
$
Ã
!
$
1
3
3
+i
= 2
2
2
$
3
= i 2
These roots are evenly spaced at 2$/3 = 120H intervals on a circle of radius 21/3 = 1.26.
Note that if we continue and take n = 4, 5, . . . we will not obtain any more or di&erent roots.
(Try it and see.)
Di&erentiating a Complex Function
A function f that depends on a real variable t can take on complex values. Such a complex
function has the form
f(t) = u(t) + iv(t)
where u and v are real functions of t. The derivative of f with respect to t is given by
f 0 (t) = u0 (t) + iv 0 (t)
Example
f (t) = t2 + i cos t,
f 0 (t) = 2t i sin t
Let us see what happens if we di&erentiate the complex exponential function. We proceed
by first decomposing the complex function into real and imaginary parts, di&erentiating, and
then finally reassembling the complex exponential. The individual steps are as follows:
h
i0
³
e(a+ib)t = eat+ibt
´0
h
i0
= eat (cos bt + i sin bt)
³
´0
= eat cos bt + ieat sin bt
³
´
= aeat cos bt + beat ( sin bt) + i aeat sin bt + beat cos bt
= (a + ib)eat (cos bt + i sin bt)
= (a + ib)e(a+ib)t
Therefore, complex exponential functions have the same kind of di&erentiation properties as
real exponential functions.
Complex Numbers Handout
-9-
Clarkson University
MA 232
Fall d999
Di&erential Equations
Kevin Dempsey
Let us now do an example directly related to ODE’s. Let’s take a complex function and see
if it can be a solution of an ODE.
Example
Let L = D4 + 2D2 + 1. Compute L (eit ).
(D4 + 2D2 + 1)(eit ) = D4 (eit ) + 2D2 (eit ) + (eit )
= (i4 + 2i2 + 1)(eit )
=0
4
2
dy
dy
So y(t) = e is a solution of constant-coe^cient ODE L(y) = 4 + 2 2 + y = 0.
dt
dt
it
Answers to Exercises on Page dd
1. (a) 3 + 2i, (b) 2i, (c) 1 + i, (d) 1 + 4i
$
$
2. (a) 5, (b) 5 + 17
3. (a) 1 ± 2i, (b) ±1, ±i
$
4. (a) 2ei$/4 , (b) 2ei5$/6 , (c) 4ei2$/3 , (d) 2e0 = 2
$
5. (a) 32 32i, (b) 1/2 + i 3/2
6. (a) {ei , = 0, 2$/5, 4$, 5, 6$/5, 8$/5}, (b) {ei , = $/2, 7$/6, 11$/6}
8. (a) f 0 (t) = cos t + i4t3
9. (a) ieit , (b) eit + tieit + 2ieit , (c) 0
Complex Numbers Handout
-d0-
Clarkson University
MA 232
Fall d999
Di&erential Equations
Kevin Dempsey
Exercises
1. Express the following in the form a + ib
(a) i(2 3i)
(b) (1 + i)2
(c) (3 + i)/(2 i)
(d) (2 + 3i) + i(1 i)
2. Compute
(a) |3 + 4i|
(b) |i(2 + i)| + |1 + 4i|
3. Solve
(a) z 2 2z + 5 = 0
(b) z 4 1 = 0
4. Express in the polar form rei
(a) 1 $i
i
(b) 3 $
(c) (1 + i 3)2
(d) (1 + i)(1 i)
5. Compute
(a) (1 i)11
(b)
h
1
(1
2
$ i13
+ i 3)
6. Solve completely
(a) z 5 = 1
(b) z 3 = i
7. Show that
(a) ¯z̄ = z
(c) z is real if and only if z = z̄
8. Find f 0 (t) if f (t) = sin t + it4
9. Let L = D2 + D + 1. Compute
(a) L(eit )
it
)
(b) L(te
´#
" ³
(c) L e
12 +i
$
3
2
t
Complex Numbers Handout
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Clarkson University