## 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
• H.A. Thurston, , Dover Publications, 1967
• Bruce E. Meserve, , Dover Books, 1981.

Jimmy T. said...

I think there is small typo in the proof of Theorem 14: i^2 = (-1,0)

It should be:
(0,1)*(0,1) = (0*0 - 1*1,0*1 - 1*0) = (-1,0)

Thanks

Larry Freeman said...

Hi Jimmy T.,

Thanks for noticing that! I fixed the typo.

Cheers,

-Larry