Download MATH 117 The Development of Complex Numbers

MATH 117
The Development of
Complex Numbers
In the real number system, a square root of a negative number is undefined. The real
line has an ordering so that for any two distinct real numbers a and b , we can determine
2
if a < b or if a > b . In particular for any a ≠ 0, we know that a > 0. Thus if we were
2
to define a = −1 , then a = −1 < 0, which is a contradiction in the real number
system.
We most often encounter square roots of negative numbers when using the
2
quadratic formula. For example if 6x + 7x + 8 = 0 , then
x=
−7 ± 7 2 − 4 × 6 × 8 −7 ± −143
.
=
2×6
12
Usually, we simply ignore the square roots of negative numbers in these cases and state
that there are no real solutions.
The First Encounter
Gerolamo Cardano (1501 – 1576) was the first to observe that square roots of negative
numbers could be studied as a new entity. However he made this observation not in
the study of quadratic equations, but with the study of cubic equations.
In 1545, Cardano published Ars Magna which was the first work to describe the
algebraic solutions of cubic and quartic equations. He initially gave a solution to the
3
“depressed” cubic equation of the form x + m x = n . Every such equation has at least
one real solution given by
x
n
+
= 3
2
n2 m3
+
4
27
−
3−n +
2
n2 m3
+
.
4
27
3
For example if x + 6x = 20 , then m = 6, n = 20, and
x = 3 10 +
400 216
+
4
27
− 3 −10 +
400 216
+
= 3 10 + 108 − 3 −10 + 108 .
4
27
Although not easily simplified, the result actually
reduces to x = 2 as shown via calculator. In any case,
108 is defined and can be approximated without a
calculator, and then the two cube roots also can be
estimated to give a difference of 2.
When creating such depressed cubic equations, one often determines the solution in
3
advance. For example, begin with x − 15x . Then suppose we want x = 4 to be a
3
solution. Substitute x = 4 and we obtain 4 − 15 × 4 = 4 ; thus, the equation must be
x 3 − 15x = 4 . To recover the solution, we apply Cardano’s formula with m = –15 and n
= 4 to obtain
16 −3375
x = 3 2+
+
4
27
− 3 −2 +
16 −3375
+
= 3 2 + −121 − 3 −2 + −121 .
4
27
However we now have −121 in the expression for the solution. Cardano realized
that this square root, and then the difference of the cube roots, should be able to be
manipulated algebraically to obtain the real number x = 4. He was not able to go any
further though.
Imaginary Numbers
In 1572, Rafael Bombelli (1526 – 1573) wrote the treatise Algebra which contained the
first mathematical treatment of the square root of a negative number. He called such
values imaginary numbers, which is a typical deragatory name like “negative” numbers
or “irrational” numbers. Bombelli was the first to define the arithmetic operations for
these numbers. In particular,
(
2
−1) = −1
and
(
3
−1) =
2
(
−1) × −1 = − −1 .
Also for a positive real number a , we let −a2 = a −1 . For instance −121 =
11 −1 . These “imaginary” numbers then could be treated like “real” numbers. For
example using the formula (a + b) 3 = a 3 + 3a 2 b + 3a b2 + b3 , we have
(2 +
3
2
−1) = 8 + 12 −1 + 6 ( −1 ) + ( −1 )
3
= 8 + 12 −1 − 6 − −1
= 2 + 11 −1 .
Similarly
(−2 +
3
−1 ) = −2 + 11 −1 .
Thus by taking cube roots, 2 + −1 =
3
2 + 11 −1 and −2 + −1 = 3 −2 + 11 −1 . So now the solution to the above equation
3
x − 15x = 4 becomes
x = 3 2 + −121 − 3 −2 + −121 = 3 2 + 11 −1 − 3 −2 + 11 −1
= 2 + −1 − (−2 + −1) = 4 .
Unfortunately, the algebra most often does not work out so nicely so as to give the
desired solution. In fact, how did we know in advance to choose 2 + −1 so that
(2 +
3
−1) = 2 + 11 −1 ? What we normally would have to do to solve for x in this cubic
equation would be to compute the cube roots 3 2 + 11 −1 and 3 −2 + 11 −1 without
knowing in advance what they are. To do so, we will need further development of the
imaginary numbers.
The Formal Arithmetic
In 1777 Euler provided the symbol i for
expression of the form
−1 . A complex number is then defined to be an
z = a + bi ,
where a and b are real numbers. The number a is called the real part of z denoted by
Re( z ), and the number b is called the imaginary part of z denoted by Im( z ). It was Carl
Friedrich Gauss who introduced the name “complex number” in 1831.
Addition: We add two complex numbers by adding the respective real parts and
imaginary parts:
( a + bi ) + ( c + d i ) = ( a + c ) + ( b + d ) i .
Multiplication: We multiply two complex numbers by product distribution or “foil”
2
method. Using the fact that i = –1, we have
(a + bi) × (c + di) = ac + a d i + bc i + bd i 2
= (ac − b d) + (a d + bc)i .
Conjugate: (Defined by Cauchy in 1821) If z = a + bi , then the conjugate of z is
z = a − bi .
2
2
Note: The product z × z always gives the real number a + b :
(a + bi) × (a − b i) = a2 − a bi + ba i − b2 i 2
= a2 − b 2 (−1)
= a2 + b 2 .
Division: If z and w are two complex numbers, then we define
z
z× w
.
=
w w×w
z
by
w
Example. Let z = 4 + 6i and w = 8 − 2i . Compute z + w and z × w and z / w .
Solution. Adding componentwise, z + w = 12 + 4 i . Multiplying, we have
2
z × w = ( 4 + 6i ) × ( 8 − 2i ) = 32 – 8 i + 48 i – 12 i = 32 – 12 (–1) + 40 i = 44 + 40 i .
For division, we multiply numerator and denominator by the conjugate:
z 4 + 6i 8 + 2i 32 + 8i + 48i + 12i 2
=
×
=
w 8 − 2 i 8 + 2i
64 + 4
=
20 + 56i 20 56
5 14
=
+ i=
+ i.
68
68 68 17 17
These calculations are now trivial to
do with calculators that allow “complex”
arithmetic. The TI-84 has an i symbol
(2nd . ). Use the ˛Frac command when
dividing on the TI-84 to obtain the exact
form:
The Cyclic Powers
i1 = i
5
i =i
i 2 = –1
6
i = –1
i3 = – i
7
i = –i
i4 = 1
8
i =1
The pattern continues for higher powers of i . What happens for negative powers of i ?
Euler’s Formula
Euler discovered several key results regarding complex numbers. One of these is now
known as Euler’s Formula:
For any real number x ,
e
ix
= cos x + i sin x .
To derive Euler’s formula, we need the Taylor series expansions for cos x , sin x , and
e that are usually derived in second-semester calculus:
x
x2 x4 x6 x8
(−1)n x 2n
= 1−
+
−
+
−...
2! 4! 6! 8!
(2n)!
n =0
∞
cos x = ∑
x3 x 5 x7 x 9
(−1)n x 2n +1
= x−
+
−
+
−...
3! 5! 7! 9!
n = 0 (2n + 1)!
∞
sin x = ∑
x2 x3 x 4 x5
xn
= 1+ x +
+
+
+
+...
2! 3! 4! 5!
n = 0 n!
∞
ex = ∑
x
Now replace x by i x in the series for e and simplify the powers i down to i , –1, –
i , and 1. Then collect the even powers and the odd powers:
e i x = 1+ i x +
(i x) 2 (i x)3 (i x)4 (i x)5 (i x)6 (i x)7
+
+
+
+
+
+...
2!
3!
4!
5!
6!
7!
x2
x3 x 4
x5
x6
x7
−i
+
+i
−
−i
+ .. .
2!
3!
4!
5!
6!
7!


x2
x4
x6
x 3 x5
= 1−
+
−
+ . . . + ix −
+
− . . .
2!
4!
6!
3! 5!


= 1+ i x −
= cos x + i sin x .
Letting x = π, we obtain e
famous Euler Identity
iπ
= cos π + isin π = –1 + 0 i = –1. Thus we obtain the
e
iπ
+1= 0
Note that by replacing x with n x in Euler’s Formula, we obtain
( )
cos(nx ) + isin(n x) = e i n x = ei x
n
= (cos x + isin x )n .
This result is called De Moivre’s Formula after Abraham de Moivre (1667 - 1754).
Exercises
1. Let z = 4 − 3i and w = 6 + 4i . Compute z + w and z × w and z / w .
2. The length of a complex number z = a + bi is defined by z =
a2 + b2 .
(a) For the complex numbers in Exercise 1, compute z , w , and z × w .
(b) Verify that z × w = z × w .
−1
−2
−3
3. Display the pattern that occurs for negative powers of i : i , i , i , etc. (That is,
1 1 1
consider these as fractions , 2 , 3 , then simplify with long-division to get a pattern.)
i i i
4. Use Euler’s Formula to write the following numbers in the form a + bi :
(a) e
i (π / 6)
(b) e
i (2π /3)
(c) e
i (3π /2)