Download Episode 3 Slides - Department of Mathematical Sciences

MATH 57091 - Algebra for High School Teachers
Division of Complex Numbers
Professor Donald L. White
Department of Mathematical Sciences
Kent State University
D.L. White (Kent State University)
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Complex Conjugates
The computation of the multiplicative inverse of a complex number
will require use of the “complex conjugate” of the number.
√
This will
√ be analogous to, for example, using the 1“conjugate” 3 − 5
of 3 + 5 to “rationalize” the denominator of 3+√5 :
√
√
√
1
1
3− 5
3− 5
3− 5
3 1√
√ =
√ ·
√ =
=
= −
5.
9−5
4
4 4
3+ 5
3+ 5 3− 5
This method is based on the “difference of squares” factorization,
(a + b)(a − b) = a2 − b 2 .
With this in mind, we make the following definition:
Definition
The (complex) conjugate of the complex number z = a + bi
(where a, b ∈ R) is z = a + bi = a − bi.
D.L. White (Kent State University)
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Complex Conjugates
We have the following properties of the complex conjugate.
Theorem 1
If z and w are complex numbers, then
i
z + w = z + w;
ii
z · w = z · w.
Proof: Exercise.
(Let z = a + bi and w = c + di, so z = a − bi and w = c − di,
and compute the right hand sides of the equations.)
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Complex Conjugates
Theorem 2
If z = a + bi is a complex number, then
i
z + z = 2a = 2Re(z);
ii
z − z = 2bi = 2Im(z) · i;
iii
z · z = a2 + b 2 , a non-negative real number,
and z · z = 0 if and only if z = 0 + 0i = 0.
Proof: We leave (i) and (ii) as exercises and prove (iii). We have
z · z = (a + bi)(a − bi)
= a2 − abi + bai − (bi)2
= a2 − abi + abi − b 2 (−1)
= a2 + b 2 .
Since a and b are real numbers, so is a2 + b 2 .
Also, a2 > 0 and b 2 > 0, and so a2 + b 2 > 0,
and a2 + b 2 = 0 if and only if a = b = 0; i.e., z = 0.
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Multiplicative Inverses
1
We want to write the multiplicative inverse of a + bi, that is, a+bi
,
a+bi
or the quotient c+di , in the standard form A + Bi, where A, B ∈ R.
Of course, if 0 6= r ∈ R, then r has a multiplicative inverse, r −1 or 1r .
It is easy to verify, using the definition of multiplication in C,
that for a complex number a + bi, we have
r ( ar + br i) = (r + 0i)( ar + br i) = a + bi.
That is,
a + bi
=
r
1
r
· (a + bi) =
a
r
+ br i,
and so we can divide a complex number by a real number.
It follows that in order to write any quotient of complex numbers
in the standard form, we only need to “real-ize” the denominator.
D.L. White (Kent State University)
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Multiplicative Inverses
By Theorem 2(iii),
if z = a + bi 6= 0, then zz = a2 + b 2 is a positive real number, and so
a − bi
a
b
z
= 2
= 2
− 2
i
2
2
zz
a +b
a +b
a + b2
as above. But notice also that
z·
z
zz
=
= 1,
zz
zz
z
hence zz
is the multiplicative inverse of z.
In particular, every non-zero complex number has a multiplicative inverse.
Therefore, we have the following theorem.
Theorem
The set of complex numbers is a field under the operations defined above.
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Multiplicative Inverses
If z = a + bi 6= 0, then we have
(a + bi)−1 =
1
a
b
= 2
− 2
i.
2
a + bi
a +b
a + b2
Despite the fact that the formula is red, and is inside a red box,
it is not worth memorizing.
Just learn the method:
1
“real-ize” the denominator of a+bi
just as you would
“rationalize” the denominator of 3+1√5 .
Multiply numerator and denominator by the conjugate of the denominator:
1
1
a − bi
a − bi
a
b
=
·
= 2
= 2
− 2
i.
2
2
a + bi
a + bi a − bi
a +b
a +b
a + b2
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Examples
1
Find the multiplicative inverse of 2 + 3i.
(2 + 3i)−1 =
=
2 − 3i
2 − 3i
1
·
=
2 + 3i 2 − 3i
(2 + 3i)(2 − 3i)
2 − 3i
2
3
2 − 3i
=
=
− i.
2
2
2 +3
13
13 13
In standard form, (2 + 3i)−1 =
2
2
13
−
3
13 i
(and not
2−3i
13 ).
Find the multiplicative inverse of 4 − 7i.
(4 − 7i)−1 =
=
D.L. White (Kent State University)
4 + 7i
4 + 7i
1
·
=
4 − 7i 4 + 7i
(4 − 7i)(4 + 7i)
42
4 + 7i
4 + 7i
4
7
=
=
+ i.
2
+ (−7)
65
65 65
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Division
a + bi
, by the same method as an inverse:
c + di
a + bi
(a + bi)(c − di)
(ac + bd) + (−ad + bc)i
=
=
,
c + di
(c + di)(c − di)
c2 + d2
We compute a quotient,
and so
ac + bd
a + bi
ad − bc
= 2
− 2
i
2
c + di
c +d
c + d2
is another formula that is not worth memorizing. For example,
3 + 2i
−4 + 5i
=
(3 + 2i)(−4 − 5i)
(−4 + 5i)(−4 − 5i)
=
[3(−4) − 2(−5)] + [3(−5) + 2(−4)]i
(−4)2 + 52
=
(−12 + 10) + (−15 − 8)i
2
23
= − − i.
16 + 25
41 41
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