Complex Conjugates and Polynomial Roots

This lesson explores the complex conjugate and the powerful conjugate root theorem, which tells us how complex roots of polynomials with real coefficients must come in conjugate pairs. Understanding conjugates is essential for solving polynomial equations and for simplifying complex expressions.

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The Complex Conjugate

If $z = a + bi$z=a+bi, the complex conjugate (or simply conjugate) of $z$z is:

$\bar{z} = a - bi$zˉ=a−bi

The conjugate is obtained by changing the sign of the imaginary part.

$z$z	$\bar{z}$zˉ
$3 + 2i$3+2i	$3 - 2i$3−2i
$-1 - 4i$−1−4i	$-1 + 4i$−1+4i
$5$5	$5$5
$7i$7i	$-7i$−7i

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Properties of the Conjugate

For any complex numbers $z$z and $w$w:

Property	Formula
Double conjugate	$\overline{\bar{z}} = z$zˉ=z
Sum	$\overline{z + w} = \bar{z} + \bar{w}$z+w​=zˉ+wˉ
Difference	$\overline{z - w} = \bar{z} - \bar{w}$z−w​=zˉ−wˉ
Product	$\overline{zw} = \bar{z}\bar{w}$zw=zˉwˉ
Quotient	$\overline{\left(\frac{z}{w}\right)} = \frac{\bar{z}}{\bar{w}}$(wz​)​=wˉzˉ​
Sum with conjugate	$z + \bar{z} = 2\,\text{Re}(z)$z+zˉ=2Re(z)
Difference with conjugate	$z - \bar{z} = 2i\,\text{Im}(z)$z−zˉ=2iIm(z)
Product with conjugate	$z\bar{z} =

Key Point: $z\bar{z}$zzˉ is always a non-negative real number. This is why multiplying by the conjugate is used to realise the denominator.

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Worked Example: Using Conjugate Properties

Question: Given $z = 4 - 3i$z=4−3i, find $z + \bar{z}$z+zˉ, $z - \bar{z}$z−zˉ, and $z\bar{z}$zzˉ.

$\bar{z} = 4 + 3i$zˉ=4+3i

$z + \bar{z} = (4 - 3i) + (4 + 3i) = 8 = 2\,\text{Re}(z)$z+zˉ=(4−3i)+(4+3i)=8=2Re(z)

$z - \bar{z} = (4 - 3i) - (4 + 3i) = -6i = 2i\,\text{Im}(z)$z−zˉ=(4−3i)−(4+3i)=−6i=2iIm(z)

$z\bar{z} = (4 - 3i)(4 + 3i) = 16 + 9 = 25 = |z|^2$zzˉ=(4−3i)(4+3i)=16+9=25=∣z∣2

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The Conjugate Root Theorem

Theorem: If a polynomial with real coefficients has a complex root $z = a + bi$z=a+bi (where $b \neq 0$b=0), then its conjugate $\bar{z} = a - bi$zˉ=a−bi is also a root.

This means complex roots of real-coefficient polynomials always come in conjugate pairs.

Consequence: A polynomial of degree $n$n with real coefficients has:

• For odd $n$n: at least one real root (since complex roots come in pairs)
• For even $n$n: possibly all complex roots, but they come in conjugate pairs

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Factorising Polynomials with Complex Roots

If $z = a + bi$z=a+bi and $\bar{z} = a - bi$zˉ=a−bi are roots, the corresponding quadratic factor is:

$(x - z)(x - \bar{z}) = x^2 - (z + \bar{z})x + z\bar{z} = x^2 - 2ax + (a^2 + b^2)$(x−z)(x−zˉ)=x2−(z+zˉ)x+zzˉ=x2−2ax+(a2+b2)

This quadratic factor has real coefficients — exactly as expected.

Worked Example: Find a quadratic with real coefficients that has $z = 2 + 3i$z=2+3i as a root.

The other root is $\bar{z} = 2 - 3i$zˉ=2−3i.

Sum of roots: $z + \bar{z} = 4$z+zˉ=4

Product of roots: $z\bar{z} = 4 + 9 = 13$zzˉ=4+9=13

Quadratic: $x^2 - 4x + 13 = 0$x2−4x+13=0

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Solving Cubics and Quartics

Worked Example (Cubic): The polynomial $p(x) = x^3 - 5x^2 + 11x - 15$p(x)=x3−5x2+11x−15 has a root $x = 1 + 2i$x=1+2i. Find the other roots.

Since the coefficients are real, $x = 1 - 2i$x=1−2i is also a root.

The quadratic factor from these two roots is: