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 + a₁x^n-1 + ... + a_n

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

a₀rⁿ + a₁r^n-1 + ... + a_n = 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 + a₁r^n-1 + ... + a_n = 0

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

QED