Learn9709Paper 3 · Pure Mathematics 3Complex numbersPolynomial equations and square roots

Complex numbers · Subtopic 4 of 5

Polynomial Equations and Square Roots

Every quadratic now has two roots. For polynomials with real coefficients, complex roots come in conjugate pairs, which lets you find all the roots from one. Square roots of $a+bi$ come from equating real and imaginary parts.

Cambridge 9709 Paper 3Edexcel YFM01 WFM01Edexcel 9FM0 Paper 1Edexcel 9FM0 Paper 2AQA 7367 Paper 1AQA 7367 Paper 2≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 15

Learning objectives

Solve quadratic equations with a negative discriminant.
Use conjugate pairs to find the remaining roots of a cubic or quartic from one given complex root.
Build a quadratic factor from a pair of conjugate roots.
Find the two square roots of a complex number in the form $x+iy$ .

Pause and think

Solve $z^2-4z+13=0$ with the formula. What do you notice about the two answers?

Section 2 of 15

Core idea

Start hereEvery polynomial has its full set of roots

$x^2-4x+13=0$ has no real roots: its graph floats above the axis. With complex numbers, it has two: $2\pm3i$ . In fact (the Fundamental Theorem of Algebra), every polynomial of degree $n$ has exactly $n$ complex roots, counted with repeats.

The big idea

Quadratics with $\Delta<0$ have roots $\frac{-b\pm i\sqrt{-\Delta}}{2a}$ . If a polynomial has real coefficients, non-real roots come in conjugate pairs: $z$ root ⇒ $z^*$ root.

1. Quadratics

y = x² − 4x + 13 never touches the x-axis

No real roots, because the graph stays above the axis. The two roots are complex:

2\pm3i

. The real part

2

is the line of symmetry.

Worked example $x^2-4x+13=0$

Solve.

Goal: Both roots.
Strategy: Formula; $\sqrt{-36}=6i$ .

$x=\frac{4\pm\sqrt{16-52}}2=\frac{4\pm\sqrt{-36}}2$
Formula.
$=\frac{4\pm6i}2=2\pm3i$
$\sqrt{-36}=6i$ .

Answer

2\pm3i

✓ CheckSum

4=-b

, product

4+9=13=c

✓.

2. Conjugate pairs

Why non-real roots pair up

$p(z)=a_nz^n+\cdots+a_0=0$
Real coefficients $a_k$ .
$\left(p(z)\right)^*=a_n(z^*)^n+\cdots+a_0=p(z^*)$
Conjugate of a sum/product is the sum/product of conjugates; $a_k^*=a_k$ .
$p(z^*)=0^*=0$
So $z^*$ is also a root.

Worked example $1+2i$ is a root of $z^3-5z^2+11z-15=0$

Find the other roots.

Goal: All roots.
Strategy: Conjugate root; quadratic factor; divide.

$1-2i\text{ is a root}$
Real coefficients.
$(z-1-2i)(z-1+2i)=(z-1)^2+4=z^2-2z+5$
Difference of squares.
$z^3-5z^2+11z-15=(z^2-2z+5)(z-3)$
Divide.

Answer

1\pm2i

and

3

✓ Check

3^3-45+33-15=0

✓.

3. Square roots of complex numbers

Worked example $\sqrt{3+4i}$

Find both square roots.

Goal: $a+ib$ .
Strategy: $(a+ib)^2=3+4i$ ; equate parts.

$a^2-b^2=3,\quad2ab=4$
Equate parts.
$b=\tfrac2a\Rightarrow a^2-\tfrac4{a^2}=3\Rightarrow a^4-3a^2-4=0$
Substitute.
$(a^2-4)(a^2+1)=0\Rightarrow a=\pm2,\ b=\pm1$
$a$ real so $a^2=4$ .

Answer

\pm(2+i)

✓ Check

(2+i)^2=4+4i-1=3+4i

✓.

The three cube roots of 8 sit on a circle

z^3=8

has one real root and a conjugate pair, spaced

120^\circ

apart round a circle of radius

2

Pause and think

Quadratic with roots $3\pm i$ ?

Summary

Degree $n$ → $n$ roots.
Conjugate pairs (real coefficients).
Square roots by equating parts.

Conjugate pair: $a\pm bi$ .

HistoryGauss and the Fundamental Theorem of Algebra. In his doctoral thesis of 1799, the 22-year-old Carl Friedrich Gauss gave a proof that every polynomial equation has a root among the complex numbers. It follows that a polynomial of degree

n

has exactly

n

roots (counting repeats). Gauss later gave three more proofs, and he also popularised the name "complex number".

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