Thursday, September 24, 2009

The Conjugate of a Complex Number

Definition 1: conjugate of complex number

Let z=(a,b) be a complex number. Then z = (a,-b)

For a review of complex numbers, see here.


Theorem 1: x is a real number if and only x = x

Proof:

(1) Assume that x is a real number such that x = (x,0)

(2) x = (x,-0) = (x,0) = x

(3) Assume that x = x

(4) Let x = (a,b) so that we have:

(a,b) = (a,-b)

(5) But by the definition of equality for complex numbers (see definition 4, here), this is only true if a=a and b=-b

(6) But b=-b only if b+b=2b=0 so b = 0.

(7) So x must be a real number.

QED

Theorem 2:

Let ηi be an nth root of unity that is not real.

Then:

ηi = η-i

Proof:

(1) Any root of unity has the following form (see Corollary 1.1, here):



(2) So, there exists an integer k such that:

ηi = cos [(2kπ)/n] + isin[(2kπ)/n]

(3) The conjugate of this value is (see Definition 1 above):

cos[(2kπ)/n] - isin[(2kπ)/n]
(4) We note that:



(5) Using the well known cos2(x) + sin2(x) = 1 [see Corollary 2, here], we get:

1/(cos[(2kπ)/n] - isin[(2kπ)/n]) = cos [(2kπ)/n] + isin[(2kπ)/n]

(6) Which shows that:

cos[(2kπ)/n] - isin[(2kπ)/n] = 1/ηi = η-i

QED


Lemma 3: (a + b) = a + b

Proof:

(1) Let a = s + ti

(2) Let b = u + vi

(3) a + b = (s + u) + (t+v)i

(4) a + b = (s + u) - (t+v)i

(5) a + b = s - ti + u - vi = (s + u) - (t+v)i

QED


Lemma 4: (ab) = a * b

Proof:

(1) Let a = s + ti

(2) Let b = u + vi

(3) a * b = (s*u - t*v) + (s*v + u*t)i

(4) a * b = (s*u - t*v) - (s*v+u*t)i

(5) a * b = (s - ti)*(u - vi) = (s*u - t*v) - (s*v + u*t)i

QED


Theorem 5:

Let f(x) be a polynomial with real coefficients.

if r is a root of f(x), then r is also a root

Proof:

(1) Let f(x) = a0xn + a1xn-1 + ... + an

(2) Since r is a root, we have:

a0rn + a1rn-1 + ... + an = 0

(3) But then taking the complex conjugate of both sides, we get (using Theorem 1 above as well as Lemma 3 and Lemma 4 above):

a0rn + a1rn-1 + ... + an = 0

(4) Which shows that f(r) = 0.

QED

Tuesday, September 22, 2009

Complex Conjugates and Properties of Complex Numbers

The content in today's blog is taken from Bruce E. Meserve's Fundamental Concepts of Algebra.

The presentation builds on a previous blog I did on the definitions needed to build the complex numbers from the real numbers.

Definition 1: Complex Conjugate

For any complex number a+bi (see Definition 6, here), the complex conjugate is the form a-bi.


In other words, for the complex number (a,b), its complex conjugate is (a,-b). The complex conjugate of (a,-b) is likewise (a,b).


Definition 2: Norm of complex number: n(z)

The norm of a complex number which is represented as n(z) is the product of a complex number with its conjugate.

This definition is demonstrated in the following lemma:

Lemma 1: For any complex number (a,b), its norm is a real number: (a2 + b2,0)

Proof:

(1) Let (a,b) be any complex number.

(2) It's complex conjugate is (a,-b) [see Definition 1 above]

(3) Using the definition for multiplication of complex numbers (see Definition 3, here) and the definition for norms (see Definition 2 above):

It's norm is (a,b)*(a,-b) = (a*a - (b*-b),a*(-b) + (b*a)) = (a2 + b2,0)

(4) We can see that this is a real number (see Theorem 15, here)

QED


Definition 3: Absolute value of a complex number (the modulus)

The absolute value of a complex number is the nonnegative square root of the norm so that:

if z = (a,b), then abs(z) = √a2 + b2


Definition 4: Division of complex numbers

(a,b) / (c,d) = (p,q) if and only if (a,b) = (c,d)*(p,q)


That this definition is well-defined is established in the next theorem.

Theorem 2: For the set of complex numbers, division by an nonzero number is well-defined.

Proof:

(1) Let (a,b), (c,d) be two complex numbers.

(2) Using the Definition of Multiplication (see Definition 3, here):

(c,d)*(p,q) = (cp - dq,cq + dp)

(3) So, if (a,b) = (c,d)*(p,q), then we have (see Definition 4, here):

a = cp - dq b = cq + dp

(4) Solving the first equation in terms of p and the second equation in terms of q gives us:

p = (a + dq)/c q = (b - dp)/c

(5) Combining the equations for p, we have:



(6) Now, rearranging the equation gives us:



And solving for p gives us:




which results in:




(7) Combining the equations for q, we have:



(8) Now, rearranging the equation gives us:



And solving for q gives us:



which results in:



(9) By assumption (c,d) is not nonzero so c2 + d2 is nonzero

(8) Since a,b,c,d are real numbers and the operations on real numbers are well-defined, it follows that p,q are well-defined and therefore division of (a,b) by (c,d) is well-defined.

QED


Theorem 3: The norm of a product is equal to the product of the norms.

Proof:

(1) Using Lemma 1 above, the norm for (a,b) is (a2 +b2,0)

(2) Likewise, the norm for (c,d) is (c2 + d2,0)

(3) Using the definition of multiplication for complex numbers (see Definition 3, here),

(a,b)*(c,d) = (a*c -b*d,a*d + b*c)

(4) Norm(a*c - b*d,a*d + b*c) is ([ac-bd]2 + [ad+bc]2,0)

(5) Finally:

(ac - bd)2 + (ad + bc)2 = a2c2 - 2abcd + b2d2 + a2d2 + 2abcd + b2c2 =

= a2c2 + b2d2 + a2d2 + b2c2 = (a2 + b2)(c2 + d2)

which shows that:

Norm[(a,b)*(c,d)] = Norm(a,b)*Norm(c,d)

QED


Theorem 4:

The absolute value of a sum of complex numbers is less than or equal to the sum of the absolute values:

abs(z1 + z2) ≤ abs(z1) + abs(z2)

Proof:

(1) Assume that abs(z1 + z2) is greater than abs(z1) + abs(z2)

(2) Let:

z1 = a + bi

z2 = c + di

(3) Using the definition of absolute values for complex numbers (see Definition 3 above):

(a + c)2 + (b + d)2 is greater than a2 + b2 + √c2 + d2

(4) Squaring both sides gives us:

(a + c)2 + (b+d)2 is greater than a2 + b2 + 2√(a2 + b2)(c2 + d2) + c2 + d2

So that:

a2 + c2 + b2 + d2 + 2ac + 2bd is greater than a2 + b2 + c2 + d2 + 2√(a2 + b2)(c2 + d2)

So that:

ac + bd is greater than (a2 + b2)(c2 + d2)

(5) Squaring both sides gives us:

a2c2 + b2d2 + 2abcd is greater than a2c2 + b2d2 + a2d2 + b2c2

So that:

0 is greater than a2d2 + b2c2 +-2abcd

So that:

0 is greater than (bc - ad)2

(6) But this is impossible since (bc -ad)2 ≥ 0.

(7) So, we reject our assumption in step #1.

QED


References

Monday, September 21, 2009

The Set of Complex Numbers

In today's blog I go over the definition of complex numbers and show how this definition can be used to prove that the set of complex numbers forms a field.

Definition 1: Complex Number


A complex number is an ordered pair of real numbers: (x,y) where x,y are real numbers.


Definition 2: Addition of Complex Numbers

(a,b) + (c,d) = (a+c,b+d)


Definition 3: Multiplication of Complex Numbers

(a,b)*(c,d) = (a*c - b*d,a*d + b*c)


Definition 4: Equality

(a,b) = (c,d) if and only if a = c and b = d


Lemma 1: The set of complex numbers is closed on addition

Proof:

This follows directly from the fact that the real numbers are closed on addition [see Lemma 1, here] and Definition 2 above.

QED


Lemma 2: The set of complex numbers is closed on multiplication.

Proof:

This follows directly form the fact that the real numbers are closed on multiplication [see Lemma 2, here] and Definition 3 above.

QED


Lemma 3: The set of complex numbers supports the commutative rule for addition

Proof:

(1) By Definition 2 above:

(a,b) + (c,d) = (a+c,b+d)

(2) Since the real numbers support the commutative rule for addition [see Lemma 3, here]:

(a+c,b+d) = (c+a,d+b) = (c,d) + (a,b)

QED


Lemma 4: The set of complex numbers supports the associative rule for addition

Proof:

(1) By Definition 2 above:

[(a,b) + (c,d)] + (e,f) = (a+c,b+d) + (e,f) = ([a+c]+e,[b+d]+f)

(2) Since the real numbers support the associative rule for addition [see Lemma 4, here]:

([a+c]+e,[b+d]+f) = (a+[c+e],b+[d+f]) = (a,b) + [(c,d) + (e,f)]

QED


Lemma 5: The set of complex numbers support the commutative rule for multiplication

Proof:

(1) By definition 3 above, we have:

(a,b)*(c,d) = (a*c - b*d,a*d + b*c)

(2) Since the real numbers support the commutative rule for multiplication (see Lemma 5, here):

(a*c - b*d,a*d + b*c) = (c*a - d*b,d*a + c*b)

(3) Since the real numbers support the commutative rule for addition (see Lemma 3, here):

(c*a - d*b,d*a + c*b) = (c*a - d*b,c*b + d*a)

(4) Using Definition 3 above again:

(c,d)*(a,b) = (c*a - d*b,c*b + d*a)

QED


Lemma 6: The set of complex numbers support the associative rule for multiplication

Proof:

By definition 3 above, we have:

[(a,b)*(c,d)]*(e,f) = ([a*c - b*d],[a*d + b*c])*(e,f) =

= ([a*c-b*d]*e - [a*d+b*c]*f,[a*c - b*d]*f + [a*d+b*c]*e) =

= (a*c*e - b*d*e -a*d*f + b*c*f, a*c*f - b*d*f + a*d*e + b*c*e) =

= (a*[c*e - d*f] - b*[c*f + d*e],a*[c*f + d*e] + b*[c*e - d*f]) =

= (a,b)*([c*e - d*f],[c*f + d*e]) = (a,b)*[(c,d)*(e,f)]

QED


Lemma 7: The set of complex numbers support the distributive rule

Proof:

(1) By Definition 2 above:

(a,b)*[(c,d) + (e,f)] = (a,b)*(c+e,d+f)

(2) By Definition 3 above:

(a,b)*(c+e,d+f) = (a*(c+e) - b*(d+f),a*(d+f) + b*(c+e)) =

= (a*c+a*e - b*d -b*f, a*d+a*f +b*c+b*e) =

= ([a*c - b*d] + [a*e - b*f],[a*d + b*c] + [a*f + b*e]) =

= (a*c-b*d,a*d + b*c) + (a*e-b*f,a*f + b*e) =

= (a,b)*(c,d) + (a,b)*(e,f)

QED


Lemma 8: The set of complex numbers have an additive identity (0,0)

Proof:

(a,b) + (0,0) = (a+0,b+0) = (a,b)

QED


Lemma 9: Every element of the set of complex numbers has an additive inverse (-a,-b)

Proof:

(a,b) + (-a,-b) = (a+-a,b+-b) = (0,0)

QED


Lemma 10: The set of complex numbers has a multiplicative identity (1,0)

Proof:

(a,b)*(1,0) = (a*1 - b*0, a*0 + b*1) = (a,b)

QED


Lemma 11: If (a,b) ≠ (0,0) then a2 + b2 ≠ 0

Proof:

(1) From Definition 4 above:

(a,b) ≠ (0,0) implies either a ≠ 0 or b ≠ 0.

(2) a≠ 0 → a2 is greater than 0

(3) b≠ 0 → b2 is greater than 0

(4) So a2 is 0 or positive and b2 is 0 or positive

(5) In all cases, a2 + b2 is positive since:

position + 0 = positive

0 + positive = positive

positive + positive = positive

QED


Lemma 12: Every nonzero element of the set of complex numbers has a multiplicative inverse

Proof:

(1) Let (a,b) be any nonzero element of the set of complex numbers

(2) Let c =1/(a2 + b2)

We know that a2 + b2 is nonzero from Lemma 11 above.

(3) The multiplicative inverse is (a*c,-b*c)

(4) And we see that:

(a,b)*(a*c,-b*c) = (a*a*c - b*(-b*c),a*(-b*c) + b*(a*c)) =

= (a*a*c +b*b*c,-a*b*c + a*b*c) = (a*a*c + b*b*c,0)

(5) a*a*c + b*b*c = a*a/(a*a + b*b) + b*b/(a*a + b*b) =

= (a*a + b*b)/(a*a + b*b) = 1

QED


Theorem 13: The complex numbers form a field

Proof:

(1) The complex numbers are closed on addition [see Lemma 1 above] and multiplication [see Lemma 2 above].

(2) The complex numbers support the commutative property of addition [see Lemma 3 above], the associative property of addition [see Lemma 4 above], the commutative property of multiplication [see Lemma 5 above], an associative property of multiplication [see Lemma 6 above], and a distributive property [see Lemma 7 above].

(3) The set of complex numbers has an additive identity property [see Lemma 8 above], an additive inverse property [see Lemma 9 above], a multiplicative identity property [see Lemma 10 above], and a multiplicative inverse property [see Lemma 12 above].

(4) From all these properties, the complex numbers form a field. [see Definition 3, here]

QED


Definition 5: i

i = (0,1)


Theorem 14: i2 = (-1,0)

Proof:

The result follows directly from Definition 3 above:

(0,1)*(0,1) = (0*0 - 1*1,0*1 - 1*0) = (-1,0)

QED


Theorem 15: All real numbers can be represented as complex numbers

(1) Any real number x can be represented as (x,0) [see definition 1 above]

(2) This correspondence holds over addition

x + y = z if and only if (x,0) + (y,0) = (x+y,0) = (z,0)

(3) This correspondence holds over multiplication

x*y = z if and only if (x,0)*(y,0) = (x*y - 0*0,x*0 + 0*y) = (x*y,0) = (z,0)

QED


Definition 6: a+bi

a+bi = (a,b)

References

The Set of Real Numbers

In a previous blog, I showed how the Dedekind cut could be used to define the real numbers.

In today's blog, I will show that the real numbers form a field.

Lemma 1: The real numbers are closed on addition.

Proof:

(1) We can define the real numbers based on a Dedekind cut. [see Definition 2, here]

(2) Let x,y be the real numbers.

(3) From the definition of the Dedekind cut, x is the set of rational numbers that are less than x and y is the set of rational numbers that are less than y.

(4) x+y is defined as the set of rational numbers in x added to the set of rational numbers in y so that x+y is the set of all of possible sums.

(5) Since the rational numbers are closed on addition [see Lemma 2, here], it follows that x+y is also closed on addition.

QED

Lemma 2: The real numbers are closed on multiplication

Proof:

(1) We can define the real numbers based on a Dedekind cut. [see Definition 2, here]

(2) Let x,y be the real numbers.

(3) From the definition of the Dedekind cut, x is the set of rational numbers that are less than x and y is the set of rational numbers that are less than y.

(4) xy is defined as the set of rational numbers in x multiplied to the set of rational numbers in y so that xy is the set of all of possible products.

(5) Since the rational numbers are closed on multiplication [see Lemma 3, here], it follows that xy is also closed on multiplication.

QED

Lemma 3: The set of real numbers support the commutative rule for addition

Proof:

(1) By the definition of addition for real numbers, addition follows the properties of the set of rationals. [see Definition 5, here]

(2) So, the conclusion follows from the fact that the rational numbers support the commutative rule for addition. [see Lemma 4, here]

QED

Lemma 4: The set of real numbers support the associative rule for addition

Proof:

(1) By the definition of addition for real numbers, addition follows the properties of the set of rationals. [see Definition 5, here]

(2) So, the conclusion follows from the fact that the rational numbers support the associative rule for addition. [see Lemma 5, here]

QED

Lemma 5: The set of real numbers support the commutative rule for multiplication.

Proof:

(1) By the definition of multiplication for real numbers, multiplication follows the properties of the set of rationals. [see Definition 7, here]

(2) So, the conclusion follows from the fact that the rational numbers support the commutative rule for multiplication. [see Lemma 11, here]

QED

Lemma 6: The set of real numbers support the associative rule for multiplication

Proof:

(1) By the definition of multiplication for real numbers, multiplication follows the properties of the set of rationals. [see Definition 7, here]

(2) So, the conclusion follows from the fact that the rational numbers support the associative rule for multiplication. [see Lemma 8, here]

QED

Lemma 7: The set of real numbers support the distributive rule

Proof:

(1) The properties of multiplication of reals is based on the properties of rational numbers [see Definition 7, here] and the properties of addition of reals is based on the properties of rational numbers [see Definition 5, here].

(2) So, the conclusion follows from the fact that the rational numbers support the associative rule for multiplication. [see Lemma 10, here]

QED

Lemma 8: The set of real numbers have an additive identity

Proof:

(1) The additive identity is the set of all rational numbers less than 0.

(2) Let x be any real number.

(3) The x+0 be the set of all rational numbers less than x added to all rational numbers less than 0.

(4) Let a be any rational number less than x.

(5) Let b be any rational number less than 0 so that be must be negative.

(6) a + b is thus less than a which is less than x.

(7) Since b can be as close to 0 as we want, a+b can be as close to x as we want.

QED

Lemma 9: The real numbers support an additive inverse property

Proof:

(1) Let x be any real number

(2) Let -x be the set of rational numbers that are less than -x.

(3) Using the definition for addition (see Definition 5, here):

x+-x is the set of all rational numbers less than 0.

QED

Lemma 10: The set of real numbers supports a multiplicative identity property

Proof:

(1) Let x be any real number

(2) The multiplicative inverse is 1 which is the set of rational numbers less than 1.

(3) It is clear that x*1 = {the set of rational numbers less than x } = x.

QED

Lemma 11: The set of real numbers supports a multiplicative inverse property

Proof:

(1) Let x = be any nonzero real number

(2) The multiplicative inverse is 1/x

(3) This is clear since the set defined by x*1/x is the set of all rational numbers less than 1.

QED

Theorem 12: The real numbers form a field

(1) The real numbers are closed on addition [see Lemma 1 above] and multiplication [see Lemma 2 above].

(2) The real numbers support the commutative property of addition [see Lemma 3 above], the associative property of addition [see Lemma 4 above], the commutative property of multiplication [see Lemma 5 above], an associative property of multiplication [see Lemma 6 above], and a distributive property [see Lemma 7 above].

(3) The set of real numbers has an additive identity property [see Lemma 8 above], an additive inverse property [see Lemma 9 above], a multiplicative identity property [see Lemma 10 above], and a multiplicative inverse property [see Lemma 11 above].

(4) From all these properties, the real numbers form a field. [see Definition 3, here]

QED

Sunday, September 20, 2009

The Set of Rational Numbers

In today's blog, I show how to formally construct the set of rational numbers from the set of integers.

I will then give a proof that the set of rational numbers forms a field. If you need a review of fields, check out here.

Definition 1: Set of rational numbers

We can define the set of rational numbers as the ordered pair of integers (a,b) where a,b are integers and b ≠ 0.


Definition 2: Addition of rationals

(a,b) + (c,d) = (ad + bc, bd)


Definition 3: Multiplication of rationals

(a,b) * (c,d) = (ac,bd)


Definition 4: Equality of rationals

Two rational numbers (a,b) and (c,d) are equal if and only if ad=bc.


Definition 5: Comparison of rationals

(a,b) is less than (c,d) if and only if abd2 is less than b2cd.


Lemma 1: All integers can be represented as rational numbers

(1) Any integer x can be represented as (x,1) [see definition 1 above]

(2) This correspondence holds over addition

x + y = z if and only if (x,1) + (y,1) = (x*1+y*1,1*1) = (x+y,1) = (z,1)

(3) This correspondence holds over multiplication

x*y = z if and only if (x,1)*(y,1) = (xy,1*1) = (xy,1) = (z,1)

QED


Lemma 2: The set of rational numbers is closed on addition

Proof:

(1) The integers are closed on addition [see Lemma 1, here] and multiplication [see Lemma 2, here].

(2) So, it follows from Definition 2 above that the rational numbers are closed on addition.

QED


Lemma 3: The set of rational numbers is closed on multiplication

Proof:

The follows directly from Definition 3 and the fact that the integers are closed on multiplication [see Lemma 2, here].

QED


Lemma 4: The set of rational numbers satisfy the Commutative Rule for Addition

Proof:

(1) From Definition 2 above:

(a,b) + (c,d) = (ad + bc, bd)

(2) Since integers are commutative by addition (see Lemma 7, here):

(ad + bc, bd) = (bc + ad, bd)

(3) From Definition 2 again, we get:

(bc + ad,bd) = (c,d) + (a,b)

QED


Lemma 5: The set of rational numbers satisfy the Associative Rule for Addition

Proof:

From Definition 2 above:

[(a,b) + (c,d)] + (e,f) = (ad+bc,bd) + (e,f) = (adf+bcf + bde,bdf) = (a,b) + (cf + de,df) =

= (a,b) + [(cf + de,df)] = (a,b) + [(c,d) + (e,f)]

QED


Lemma 6: The set of rational numbers has an Additive Identity for all elements.

Proof:

(1) Using Definition 2 above, we have:

(a,b) + (0,c) = (a*c + 0*b,b*c) = (ac,bc)

(2) Using Definition 4 above, we note that:

(ac,bc) = (a,b) since acb = bca [using the commutative property multiplication for integers, see Lemma 8, here]

QED


Lemma 7: The set of rational numbers has an Additive Inverse for all elements.

Proof:

(1) Let (a,b) be a rational number.

(2) Then, it's additive inverse is (-a,b) since:

(a,b) + (-a,b) = (a*b + -a*b,b*b) = (ab-ab,b*b) = (0,b*b)

QED


Lemma 8: The set of rational numbers supports the Associative Rule for Multiplication

Proof:

Using Definition 3 above, we have:

[(a,b)*(c,d)]*(e,f) = [(ac,bd)]*(e,f) = (ace,bdf) = (a,b)*[(ce,df)] = (a,b)*[(c,d)*(e,f)]

QED


Lemma 9: (c,c) = (1,1)

Proof:

The follows directly from definition 4 above since:

c*1 = 1*c

QED


Lemma 10: The set of rational numbers supports the Distributive Rule

Proof:

(1) Using Definition 2 and Definition 3 above, we have:

(a,b)[(c,d) + (e,f)] = (a,b)*(cf+de,df) = (acf + ade,bdf)

(2) Using Definition 3 above and Lemma 8 above, we have:

(acbf + aebd,bdbf) = (b,b)*(acf + aed,bdf) = (1,1)*(acf + aed,bdf) = (acf + aed,bdf)

(3) Using Definition 2 above, we have:

(ac,bd) + (ae,bf) = (acbf + aebd,bdbf)

QED


Lemma 11: The set of rational numbers supports the Commutative Rule for Multiplication

Proof:

(1) Using Definition 3 above, we have:

(a,b)*(c,d) = (ac,bd)

(2) Using the Commutative Property of Multiplication for Integers (see Lemma 8, here):

(ac,bd) = (ca,db)

(3) Using Definition 3 above, we have:

(ca,db) = (c,d)*(a,b)

QED


Lemma 12: The set of rational numbers has a Multiplicative Identity

Proof:

For any rational number (a,b), we have (see Definition 3 above):

(a,b)*(1,1) = (a*1,b*1) = (a,b)

QED


Lemma 13: For every nonzero element, the set of rational numbers has a Multiplicative Inverse

Proof:

(1) Let (a,b) be any rational number.

(2) Then (b,a) will be its multiplicative inverse since:

(a,b)*(b,a) = (ab,ba)

(3) Using the Commutative Property of Multiplication of Integers (see Lemma 8, here), we have:

(ab,ba) = (ab,ab)

(4) Using Lemma 8 above, we have (ab,ab)=1.

QED


Theorem 14: The set of rational numbers forms a field

Proof:

This follows directly from Lemma 2 through Lemma 13 above and from Definition 3, here.

QED

References