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)


(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)


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.


(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.


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


(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)


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)


(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.



1 comment :

Anonymous said...

Thanks for your practical math info.
Do you have any idea, WHY
"In general, if \phi\, is a holomorphic function whose restriction to the real numbers is real-valued, and \phi(z)\, is defined, then

\phi(\overline{z}) = \overline{\phi(z)}.\,\! "

Please if you know any reason, mail me.
Thanks in advance