Reference

Conjugation

Concept

Conjugate

The conjugate of an irrational binomial is the same number, but the sign of the irrational term is switched. Similarly, the conjugate of a complex number is the same complex number where the sign of the complex part is opposite to its original sign.

	Binomial	Conjugate
Irrational conjugate	$a + b c$	$a - b c$
Irrational conjugate	$a - b c$	$a + b c$
Complex conjugate	$a + b i$	$a - b i$
Complex conjugate	$a - b i$	$a + b i$

Concept

Complex Conjugate

The complex conjugate of a complex number has the same real part, but the imaginary part is the opposite of its original sign. Therefore, changing the sign of the imaginary part of a complex number creates its complex conjugate. It is denoted by a line drawn above the complex number.

\overline{a + b i} = a - b i or \overline{a - b i} = a + b i

For example, the complex conjugate of

z = 3 - 5 i

\overset{z}{ˉ} = 3 + 5 i .

It is worth noting that the product of a complex number and its conjugate is a real number.

z \cdot \overset{z}{ˉ}

Substitute

a + b i

for

z

and simplify

Substitute

$z = a + b i$

(a + b i) (\overline{a + b i})

Substitute

$\overline{a + b i} = a - b i$

(a + b i) (a - b i)

Multiply

a^{2} - a b i + a b i - b^{2} i^{2}

AddTerms

Add terms

a^{2} - b^{2} i^{2}

IDefPow

$i^{2} = - 1$

a^{2} - b^{2} (- 1)

NegNeg

$- (- a) = a$

a^{2} + b^{2}

This fact is very useful when simplifying quotients whose denominators are complex numbers.

Concept

Irrational Conjugate

Let

a,

b,

and

c

be rational numbers, with

c

irrational. The irrational conjugate of

a + b c

is obtained by switching the sign of the irrational term.

Number a + b c a - b c switch sign switch sign Conjugate a - b c a + b c

For example, the conjugate of

2 - 53

2 + 53 .

Rule

Sum of Conjugates

For two numbers $z_{1}$ and $z_{2},$ the conjugate of the sum is equal to the sum of the conjugates.

$\overline{z_{1} + z_{2}} = \overline{z_{1}} + \overline{z_{2}}$

This is true for both the complex conjugates of complex numbers in the form of

a + b i

and for irrational conjugates of numbers in the form of

a + b r .

Proof

Complex Conjugates

Let

z_{1} = a + b i

and

z_{2} = c + d i

be complex numbers. Consider the left- and right-hand sides of the equality separately. Then, beginning with the left, substitute the complex numbers and simplify each side.

\overline{z_{1} + z_{2}}

SubstituteII

$z_{1} = a + b i$ , $z_{2} = c + d i$

\overline{(a + b i) + (c + d i)}

Simplify

RemovePar

Remove parentheses

\overline{a + b i + c + d i}

CommutativePropAdd

Commutative Property of Addition

\overline{a + c + b i + d i}

FactorOut

Factor out $i$

\overline{(a + c) + (b + d) i}

Definition of complex conjugate

(a + c) - (b + d) i

Similar operations can be used for the right-hand side.

\overline{z_{1}} + \overline{z_{2}}

SubstituteII

$z_{1} = a + b i$ , $z_{2} = c + d i$

\overline{a + b i} + \overline{c + d i}

Simplify

Definition of complex conjugate

(a - b i) + (c - d i)

RemovePar

Remove parentheses

a - b i + c - d i

CommutativePropAdd

Commutative Property of Addition

a + c - b i - d i

FactorOut

Factor out $- i$

(a + c) - (b + d) i

Since both sides simplify to the same expression, they are equal.

\overline{z_{1} + z_{2}} = \overline{z_{1}} + \overline{z_{2}}

Proof

Irrational Conjugates

Let

z_{1} = a + b r

and

z_{2} = c + d r

be two irrational numbers, where

a,

b,

c,

d,

and

r

are all rational. Consider the left- and right-hand sides of the equality separately. Then, beginning with the left, substitute the complex numbers and simplify each side.

\overline{z_{1} + z_{2}}

SubstituteII

$z_{1} = a + b r$ , $z_{2} = c + d r$

\overline{(a + b r) + (c + d r)}

Simplify

RemovePar

Remove parentheses

\overline{a + b r + c + d r}

CommutativePropAdd

Commutative Property of Addition

\overline{a + c + b r + d r}

FactorOut

Factor out $r$

\overline{(a + c) + (b + d) r}

Definition of irrational conjugate

(a + c) - (b + d) r

Similar operations can be used for the right-hand side.

\overline{z_{1}} + \overline{z_{2}}

SubstituteII

$z_{1} = a + b r$ , $z_{2} = c + d r$

\overline{a + b r} + \overline{c + d r}

Simplify

Definition of irrational conjugate

(a - b r) + (c - d r)

RemovePar

Remove parentheses

a - b r + c - d r

CommutativePropAdd

Commutative Property of Addition

a + c - b r - d r

FactorOut

Factor out $- r$

(a + c) - (b + d) r

Since both sides simplify to the same expression, they are equal.

\overline{z_{1} + z_{2}} = \overline{z_{1}} + \overline{z_{2}}

Rule

Product of Conjugates

For two numbers $z_{1}$ and $z_{2},$ the conjugate of the product is equal to the product of the conjugates.

$\overline{z_{1} \cdot z_{2}} = \overline{z_{1}} \cdot \overline{z_{2}}$

This is true both for the complex conjugates of complex numbers of the form

a + b i

and for irrational conjugates of numbers of the form

a + b r .

Proof

Complex Conjugates

Let

z_{1} = a + b i

and

z_{2} = c + d i

be two complex numbers. Consider the left and right hand side of the equality separately, starting with the left-hand side.

\overline{z_{1} \cdot z_{2}}

SubstituteII

$z_{1} = a + b i$ , $z_{2} = c + d i$

\overline{(a + b i) \cdot (c + d i)}

Simplify

MultPar

Multiply parentheses

\overline{a c + a d i + b c i + b d i^{2}}

IDefPow

$i^{2} = - 1$

\overline{a c + a d i + b c i - b d}

CommutativePropAdd

Commutative Property of Addition

\overline{a c - b d + a d i + b c i}

FactorOut

Factor out $i$

\overline{(a c - b d) + (a d + b c) i}

Definition of complex conjugate

(a c - b d) - (a d + b c) i

Similar operations can be used for the right hand-side.

\overline{z_{1}} \cdot \overline{z_{2}}

SubstituteII

$z_{1} = a + b i$ , $z_{2} = c + d i$

(\overline{a + b i}) \cdot (\overline{c + d i})

Simplify

Definition of complex conjugate

(a - b i) \cdot (c - d i)

MultPar

Multiply parentheses

a c - a d i - b c i + b d i^{2}

IDefPow

$i^{2} = - 1$

a c - a d i - b c i - b d

CommutativePropAdd

Commutative Property of Addition

a c - b d - a d i - b c i

FactorOut

Factor out $- i$

(a c - b d) - (a d + b c) i

Since both sides simplify to the same expression, they must be equal.

\overline{z_{1} \cdot z_{2}} = \overline{z_{1}} \cdot \overline{z_{2}}

Proof

Irrational Conjugates

Let

z_{1} = a + b r

and

z_{2} = c + d r

be two irrational numbers, where

a,

b,

c,

d,

and

r,

are all rational. Consider the left and right hand side of the equality separately, starting with the left-hand side.

\overline{z_{1} \cdot z_{2}}

SubstituteII

$z_{1} = a + b r$ , $z_{2} = c + d r$

\overline{(a + b r) \cdot (c + d r)}

Simplify

MultPar

Multiply parentheses

\overline{a c + a d r + b c r + b d (r)^{2}}

PowSqrt

$(a)^{2} = a$

\overline{a c + a d r + b c r + b d r}

CommutativePropAdd

Commutative Property of Addition

\overline{a c + b d r + a d r + b c r}

FactorOut

Factor out $r$

\overline{(a c + b d r) + (a d + b c) r}

Definition of irrational conjugate

(a c + b d r) - (a d + b c) r

Similar operations can be used for the right-hand side.

\overline{z_{1}} \cdot \overline{z_{2}}

SubstituteII

$z_{1} = a + b r$ , $z_{2} = c + d r$

(\overline{a + b r}) \cdot (\overline{c + d r})

Simplify

Definition of irrational conjugate

(a - b r) \cdot (c - d r)

MultPar

Multiply parentheses

a c - a d r - b c r + b d (r)^{2}

PowSqrt

$(a)^{2} = a$

a c - a d r - b c r + b d r

CommutativePropAdd

Commutative Property of Addition

a c + b d r - a d r - b c r

FactorOut

Factor out $- r$

(a c + b d r) - (a d + b c) r

Since both sides simplify to the same expression, they must be equal.

\overline{z_{1} \cdot z_{2}} = \overline{z_{1}} \cdot \overline{z_{2}}

Rule

Power of Conjugates

For a number $z$ and a positive integer $n,$ the conjugate of $z^{n}$ is equal to the $n th$ power of the conjugate of $z .$

$\overline{z^{n}} = (\overset{z}{ˉ})^{n}$

This is true for both the complex conjugates of complex numbers in the form of $a + b i$ and for irrational conjugates of numbers in the form of $a + b r .$

Proof

This can be proved by mathematical induction, using the Product of Conjugates Property.

Show That the Statement Is True for

n = 1

The statement is true for $n = 1,$ since $\overline{z^{1}} = \overset{z}{ˉ}$ and $(\overset{z}{ˉ})^{1} = \overset{z}{ˉ} .$

Assume That the Statement Is True for Some Natural Number

k

Assume that $\overline{z^{k}} = (\overset{z}{ˉ})^{k}$ for some positive integer $k .$

Show That the Statement Is True for the Next Number

k + 1

Multiply the assumption by

\overset{z}{ˉ}

and simplify both sides.

\overline{z^{k}} = \overset{z}{ˉ}^{k}

MultEqn

$LHS \cdot \overset{z}{ˉ} = RHS \cdot \overset{z}{ˉ}$

\overline{z^{k}} \cdot \overset{z}{ˉ} = \overset{z}{ˉ}^{k} \cdot \overset{z}{ˉ}

$\overline{z_{1} \cdot z_{2}} = \overline{z_{1}} \cdot \overline{z_{2}}$

\overline{z^{k} \cdot z} = \overset{z}{ˉ}^{k} \cdot \overset{z}{ˉ}

MultBasePow

$a \cdot a^{m} = a^{1 + m}$

\overline{z^{k + 1}} = \overset{z}{ˉ}^{k + 1}

The last line is the statement for

n = k + 1 .

Draw a Conclusion

The statement is true for $n = 1,$ and if it is true for $n = k,$ then it is also true for $n = k + 1 .$ Therefore, by the principle of mathematical induction, the statement holds true for all natural numbers $n .$

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Proof

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Proof

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