Complex Roots of Polynomials (DP IB Analysis & Approaches (AA)): Revision Note

Written by: Amber

Reviewed by: Mark Curtis

Updated on 15 July 2025

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Complex roots of quadratics

When does a quadratic have complex roots?

The quadratic equation $a z^{2} + b z + c = 0$ where $a, b, c \in ℝ$ has complex roots if
- $b^{2} - 4 a c < 0$
  - the discriminant is negative

How do I solve a quadratic with complex roots?

To solve a quadratic equation with complex roots, $a z^{2} + b z + c = 0$
- either use the quadratic formula
  - e.g. $\pm \sqrt{- 9} = \pm 3 i$
- or complete the square

Examiner Tips and Tricks

You can can check your answer by substituting your complex roots back into the equation $a z^{2} + b z + c = 0$ (which should given $0 + 0 i$ if correct).

The two complex roots to the quadratic equation $a z^{2} + b z + c = 0$ where $a, b, c \in ℝ$ are complex conjugate pairs
- i.e. if $z = p + q i$ is a root, then $z^{*} = p - q i$ is the other root
This is not true if any of the coefficients $a$ , $b$ or $c$ are complex

How do I factorise a quadratic expression using complex roots?

If a quadratic expression $a z^{2} + b z + c$ has a negative discriminant ( $b^{2} - 4 a c < 0$ ), then it can factorised as follows:
- set the expression equal to zero
  - $a z^{2} + b z + c = 0$
- solve this equation to find the complex roots
  - $z_{1} = p + q i$ and $z_{2} = {z_{1}}^{*} = p - q i$
- rewrite $a z^{2} + b z + c$ in the factorised form $a (z - z_{1}) (z - z_{2})$
  - You could expand inside each bracket
  - $a (z - p - q i) (z - p + q i)$

How do I find a quadratic equation when given its root?

If you are given that $z_{1} = p + q i$ is a root of a quadratic equation
- then $z_{2} = {z_{1}}^{*} = p - q i$ is the other root
- and $(z - z_{1}) (z - z_{2}) = 0$ is the factorised equation
To find the quadratic equation $z^{2} + b z + c = 0$
- expand $(z - z_{1}) (z - z_{2}) = 0$
  - e.g. expanding $(z - 5 - 6 i) (z - 5 + 6 i)$

Examiner Tips and Tricks

A trick to reduce the algebra is to expand $(z - z_{1}) (z - z_{2})$ in $z_{1}$ and $z_{2}$ first (before substituting them in) to get $z^{2} - (z_{1} + z_{2}) z + z_{1} z_{2}$ , then substitute in $z_{1}$ and $z_{2}$ .

Worked Example

(a) Factorise $z^{2} - 2 z + 5$ into the form $(z + a + b i) (z + c + d i)$ where $a, b, c, d \in ℝ$ .

1-9-3-ib-aa-hl-complex-roots-we-solution-1-a

(b) Given that $2 - 3 i$ is a root of a different quadratic equation, $z^{2} + B z + C = 0$ where $B, C \in ℝ$ , find the values of $B$ and $C$ .

1-9-3-ib-aa-hl-complex-roots-we-solution-1-b

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