DP Math AA · HL · Number and Algebra

AHL 1.14—Complex roots of polynomials, conjugate roots, De Moivre’s, powers & roots of complex numbers

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Complex Conjugate Root Theorem

Complex Conjugate: For a complex number z=a+bi, its complex conjugate is zˉ=a−bi. Geometrically, this is the reflection of z across the real axis in the Argand diagram.

One of the most important results in polynomial theory is the Complex Conjugate Root Theorem: if a polynomial has real coefficients and a+bi (where b=0) is a root, then its conjugate a−bi is also necessarily a root.

Why does this happen? Consider a polynomial P(x) with real coefficients. If z=a+bi is a root, then P(z)=0. Taking the conjugate of both sides and using the fact that conjugation distributes over addition and multiplication , and that conjugating a real coefficient leaves it unchanged , gives P(zˉ)=0. So zˉ is also a root.

Note

This theorem applies only when all coefficients are real. A polynomial with complex coefficients does not need to have conjugate pairs of roots , for example, x−i=0 has only the root x=i.

Exam Tip

A direct consequence: a polynomial with real coefficients can only have an even number of non-real complex roots, since they always appear in conjugate pairs.

A simple verification: the polynomial x2+1=0 has roots x=i and x=−i, which are conjugates. We can confirm: (x−i)(x+i)=x2−i2=x2+1. ✓

Using Conjugate Roots to Factorise Polynomials

Because complex roots come in conjugate pairs for real-coefficient polynomials, knowing one complex root immediately gives you another , and their product forms a real quadratic factor.

If z=a+bi is a root, then (x−z)(x−zˉ) is a real quadratic factor:
(x−(a+bi))(x−(a−bi))=(x−a)2+b2=x2−2ax+(a2+b2)

This is always a quadratic with real, positive discriminant-free coefficients.

Example

Finding a cubic with real coefficients given one complex root.

Suppose a cubic polynomial P(x) with real coefficients has roots including z=2+3i.

Step 1: By the conjugate root theorem, z=2−3i is also a root.

Step 2: Form the quadratic factor:
(x−(2+3i))(x−(2−3i))=(x−2)2+9=x2−4x+13

Step 3: Since the cubic has degree 3 and we have a quadratic factor, the remaining root must be real. If told the third root is x=1:
P(x)=(x−1)(x2−4x+13)=x3−5x2+17x−13

Verification: All coefficients are real. ✓

Warning

A common mistake is to forget the conjugate root and try to build a cubic with only two roots. Always account for all n roots of a degree-n polynomial.

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9 more sections in this topic

← Previous topicAHL 1.13—Polar and Euler formNext topic →AHL 1.15—Proof by induction, contradiction, counterexamples
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