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A-Level Further Mathematics – Chapter 1: Complex Numbers

Edexcel · AQA · OCR A-Level Further Mathematics · Updated March 2026

Complex numbers are the gateway to Further Mathematics. They extend the real number line into the complex plane, unlocking solutions to equations that have no real roots and providing the language for Fourier analysis, quantum mechanics, and control theory. Every topic in this chapter — from arithmetic in Cartesian form through to de Moivre's theorem — is assessed directly in the Further Pure papers and underpins later work on series, differential equations, and transforms.

Specification Note

All content in this chapter is Further level and forms part of the compulsory Further Pure 1 component for Edexcel, AQA, and OCR. Loci in the Argand diagram are introduced here and developed fully in Chapter 2.

Contents

  1. 1.1 Introduction to Complex Numbers
  2. 1.2 Arithmetic of Complex Numbers
  3. 1.3 Modulus and Argument
  4. 1.4 Exponential (Euler) Form
  5. 1.5 Applications of de Moivre's Theorem
  6. 1.6 Loci in the Complex Plane (Preview)
  7. Practice Problems

1.1 Introduction to Complex Numbers Further

The equation $x^2 + 1 = 0$ has no solution in the real numbers because $x^2 \geq 0$ for all $x \in \mathbb{R}$. To resolve this, mathematicians introduce the imaginary unit $\mathrm{i}$, defined by $\mathrm{i}^2 = -1$.

Definition 1.1 — Complex Number (Cartesian Form)

A complex number is an expression of the form

$$z = a + b\mathrm{i}, \quad a,b \in \mathbb{R},$$

where $a = \operatorname{Re}(z)$ is the real part and $b = \operatorname{Im}(z)$ is the imaginary part. The set of all complex numbers is denoted $\mathbb{C}$.

Note that $\mathbb{R} \subset \mathbb{C}$: every real number $a$ is complex with imaginary part zero. A complex number with $a = 0$ and $b \neq 0$ is called purely imaginary.

Equality of Complex Numbers

Two complex numbers $z_1 = a + b\mathrm{i}$ and $z_2 = c + d\mathrm{i}$ are equal if and only if $a = c$ and $b = d$. This seemingly simple fact is the basis of the method of equating real and imaginary parts, used throughout the course.

Example 1.1 — Solving by Equating Parts

Find real $p$ and $q$ such that $(p + q) + (2p - q)\mathrm{i} = 5 + \mathrm{i}$.

Step 1 Equate real parts: $p + q = 5$.

Step 2 Equate imaginary parts: $2p - q = 1$.

Step 3 Add the two equations: $3p = 6$, so $p = 2$.

Step 4 Substitute back: $q = 5 - 2 = 3$.

Therefore $p = 2$, $q = 3$.

Powers of $\mathrm{i}$

Since $\mathrm{i}^2 = -1$, the powers of $\mathrm{i}$ cycle with period 4:

$$\mathrm{i}^1 = \mathrm{i}, \quad \mathrm{i}^2 = -1, \quad \mathrm{i}^3 = -\mathrm{i}, \quad \mathrm{i}^4 = 1, \quad \mathrm{i}^5 = \mathrm{i}, \ldots$$

Example 1.2 — Simplifying Powers of i

Simplify $\mathrm{i}^{23}$.

Write $23 = 4 \times 5 + 3$, so $\mathrm{i}^{23} = (\mathrm{i}^4)^5 \cdot \mathrm{i}^3 = 1^5 \cdot (-\mathrm{i}) = -\mathrm{i}$.

1.2 Arithmetic of Complex Numbers Further

Addition, subtraction, and multiplication follow from the ordinary rules of algebra, with $\mathrm{i}^2$ replaced by $-1$ wherever it appears.

Definition 1.2 — Complex Conjugate

The complex conjugate of $z = a + b\mathrm{i}$ is $\bar{z} = a - b\mathrm{i}$. Key properties:

Multiplication

$$( a + b\mathrm{i})(c + d\mathrm{i}) = ac + ad\mathrm{i} + bc\mathrm{i} + bd\mathrm{i}^2 = (ac - bd) + (ad + bc)\mathrm{i}.$$

Example 1.3 — Multiplication

Expand $(3 + 2\mathrm{i})(1 - 5\mathrm{i})$.

$= 3(1) + 3(-5\mathrm{i}) + 2\mathrm{i}(1) + 2\mathrm{i}(-5\mathrm{i})$

$= 3 - 15\mathrm{i} + 2\mathrm{i} - 10\mathrm{i}^2$

$= 3 - 13\mathrm{i} + 10 = 13 - 13\mathrm{i}$.

Division

To divide $z_1$ by $z_2$, multiply numerator and denominator by $\bar{z}_2$:

$$\frac{z_1}{z_2} = \frac{z_1 \bar{z}_2}{z_2 \bar{z}_2} = \frac{z_1 \bar{z}_2}{|z_2|^2}.$$

Example 1.4 — Division

Write $\dfrac{3 + \mathrm{i}}{2 - 3\mathrm{i}}$ in the form $a + b\mathrm{i}$.

Multiply by $\dfrac{2 + 3\mathrm{i}}{2 + 3\mathrm{i}}$:

$$\frac{(3 + \mathrm{i})(2 + 3\mathrm{i})}{(2 - 3\mathrm{i})(2 + 3\mathrm{i})} = \frac{6 + 9\mathrm{i} + 2\mathrm{i} + 3\mathrm{i}^2}{4 + 9} = \frac{6 + 11\mathrm{i} - 3}{13} = \frac{3 + 11\mathrm{i}}{13} = \frac{3}{13} + \frac{11}{13}\mathrm{i}.$$

Complex Roots of Real Polynomials

Theorem 1.1 — Complex Conjugate Root Theorem

If $P(z)$ is a polynomial with real coefficients and $\alpha \in \mathbb{C}$ is a root, then $\bar{\alpha}$ is also a root. In particular, non-real complex roots of real polynomials always occur in conjugate pairs.

Example 1.5 — Finding a Cubic Given One Complex Root

A cubic $P(z)$ with real coefficients has roots $2$ and $1 + 3\mathrm{i}$. Find $P(z)$ in the form $z^3 + bz^2 + cz + d$.

Step 1 By the conjugate root theorem, $1 - 3\mathrm{i}$ is also a root.

Step 2 The quadratic factor from the complex pair is $(z - (1+3\mathrm{i}))(z - (1-3\mathrm{i})) = (z-1)^2 + 9 = z^2 - 2z + 10$.

Step 3 The linear factor from root $2$ is $(z - 2)$.

Step 4 $P(z) = (z-2)(z^2 - 2z + 10) = z^3 - 2z^2 + 10z - 2z^2 + 4z - 20 = z^3 - 4z^2 + 14z - 20$.

Example 1.6 — Solving a Quartic with Complex Roots

Solve $z^4 - 2z^3 + 3z^2 - 2z + 2 = 0$ given that $z = \mathrm{i}$ is a root.

Step 1 Since coefficients are real, $z = -\mathrm{i}$ is also a root. Quadratic factor: $(z - \mathrm{i})(z + \mathrm{i}) = z^2 + 1$.

Step 2 Divide: $z^4 - 2z^3 + 3z^2 - 2z + 2 \div (z^2 + 1) = z^2 - 2z + 2$.

Step 3 Solve $z^2 - 2z + 2 = 0$: $z = \dfrac{2 \pm \sqrt{4-8}}{2} = 1 \pm \mathrm{i}$.

The four roots are $z = \mathrm{i},\; -\mathrm{i},\; 1+\mathrm{i},\; 1-\mathrm{i}$.

1.3 Modulus and Argument Further

Definition 1.3 — Modulus and Argument

For $z = a + b\mathrm{i}$:

The modulus-argument form (polar form) is $z = r(\cos\theta + \mathrm{i}\sin\theta)$ where $r = |z|$ and $\theta = \arg(z)$.

To determine $\arg(z)$ correctly, always sketch the point in the Argand diagram to identify the correct quadrant before applying the arctangent formula.

Example 1.7 — Finding Modulus and Argument

Express $z = -1 + \sqrt{3}\,\mathrm{i}$ in modulus-argument form.

Step 1 Modulus: $|z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$.

Step 2 The point $(-1, \sqrt{3})$ lies in the second quadrant. The reference angle is $\arctan\!\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$, so $\arg(z) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

Step 3 $z = 2\!\left(\cos\frac{2\pi}{3} + \mathrm{i}\sin\frac{2\pi}{3}\right)$.

Properties of Modulus and Argument

Theorem 1.2 — Multiplication and Division Rules

For $z_1 = r_1 e^{\mathrm{i}\theta_1}$ and $z_2 = r_2 e^{\mathrm{i}\theta_2}$:

Geometric interpretation: multiplying by $z_2$ scales by $|z_2|$ and rotates by $\arg(z_2)$.

Figure 1.1 — The point $z = 3 + 4\mathrm{i}$ (blue), its conjugate $\bar{z} = 3 - 4\mathrm{i}$ (red), and the modulus $|z| = 5$ illustrated on an Argand diagram.

1.4 Exponential (Euler) Form Further

Theorem 1.3 — Euler's Formula

For all $\theta \in \mathbb{R}$:

$$e^{\mathrm{i}\theta} = \cos\theta + \mathrm{i}\sin\theta.$$

Consequently, any complex number with modulus $r$ and argument $\theta$ may be written in exponential form as $z = r e^{\mathrm{i}\theta}$.

Euler's formula may be established by comparing the Maclaurin series for $e^x$, $\cos x$, and $\sin x$:

$$e^{\mathrm{i}\theta} = \sum_{n=0}^{\infty}\frac{(\mathrm{i}\theta)^n}{n!} = \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots\right) + \mathrm{i}\!\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right) = \cos\theta + \mathrm{i}\sin\theta.$$

The special case $\theta = \pi$ gives Euler's identity: $e^{\mathrm{i}\pi} + 1 = 0$, often cited as one of the most beautiful results in mathematics.

Example 1.8 — Converting Between Forms

Write $z = 4e^{\mathrm{i}\pi/3}$ in Cartesian form.

$z = 4\!\left(\cos\frac{\pi}{3} + \mathrm{i}\sin\frac{\pi}{3}\right) = 4\!\left(\frac{1}{2} + \frac{\sqrt{3}}{2}\mathrm{i}\right) = 2 + 2\sqrt{3}\,\mathrm{i}.$

Example 1.9 — Product in Exponential Form

Let $z_1 = 3e^{\mathrm{i}\pi/4}$ and $z_2 = 2e^{\mathrm{i}\pi/6}$. Find $z_1 z_2$ and $z_1/z_2$.

$z_1 z_2 = 6\,e^{\mathrm{i}(\pi/4 + \pi/6)} = 6\,e^{\mathrm{i}5\pi/12}$.

$z_1/z_2 = \tfrac{3}{2}\,e^{\mathrm{i}(\pi/4 - \pi/6)} = \tfrac{3}{2}\,e^{\mathrm{i}\pi/12}$.

Figure 1.2 — The unit circle $|z| = 1$ with selected points at standard arguments $\theta = 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \pi, -\frac{\pi}{2}$ shown in Euler form $e^{\mathrm{i}\theta}$.

1.5 Applications of de Moivre's Theorem Further

Theorem 1.4 — de Moivre's Theorem

For any integer $n$ and $\theta \in \mathbb{R}$:

$$(\cos\theta + \mathrm{i}\sin\theta)^n = \cos n\theta + \mathrm{i}\sin n\theta,$$

equivalently, $(e^{\mathrm{i}\theta})^n = e^{\mathrm{i}n\theta}$.

Proof sketch: For $n \geq 1$, use induction together with the addition formulae. For negative $n$, use $z^{-1} = \bar{z}/|z|^2$. The result for rational $n$ extends to roots; see below.

Expressing Multiple-Angle Identities

De Moivre's theorem combined with the binomial theorem enables the derivation of identities for $\cos n\theta$ and $\sin n\theta$ in terms of powers of $\cos\theta$ and $\sin\theta$.

Example 1.10 — Triple Angle Formulae

Derive expressions for $\cos 3\theta$ and $\sin 3\theta$.

Step 1 By de Moivre: $\cos 3\theta + \mathrm{i}\sin 3\theta = (\cos\theta + \mathrm{i}\sin\theta)^3$.

Step 2 Expand using the binomial theorem (let $c = \cos\theta$, $s = \sin\theta$): $$= c^3 + 3c^2(\mathrm{i}s) + 3c(\mathrm{i}s)^2 + (\mathrm{i}s)^3 = c^3 + 3c^2 s\,\mathrm{i} - 3cs^2 - s^3\mathrm{i}.$$

Step 3 Equate real parts: $\cos 3\theta = c^3 - 3cs^2 = \cos^3\!\theta - 3\cos\theta\sin^2\!\theta$.

Using $\sin^2\theta = 1 - \cos^2\theta$: $\cos 3\theta = 4\cos^3\!\theta - 3\cos\theta$.

Step 4 Equate imaginary parts: $\sin 3\theta = 3c^2 s - s^3 = 3\sin\theta\cos^2\!\theta - \sin^3\!\theta = 3\sin\theta - 4\sin^3\!\theta$.

Expressing Powers in Terms of Multiple Angles

Let $z = e^{\mathrm{i}\theta}$ so $z + z^{-1} = 2\cos\theta$ and $z - z^{-1} = 2\mathrm{i}\sin\theta$. Then:

$$(2\cos\theta)^n = (z + z^{-1})^n, \qquad (2\mathrm{i}\sin\theta)^n = (z - z^{-1})^n.$$

Expanding the right-hand side using the binomial theorem and applying de Moivre yields identities expressing $\cos^n\theta$ or $\sin^n\theta$ in terms of $\cos k\theta$ or $\sin k\theta$.

Example 1.11 — Expressing $\cos^4\theta$

Show that $\cos^4\!\theta = \tfrac{1}{8}\cos 4\theta + \tfrac{1}{2}\cos 2\theta + \tfrac{3}{8}$.

$(2\cos\theta)^4 = (z + z^{-1})^4 = z^4 + 4z^2 + 6 + 4z^{-2} + z^{-4}$.

$= (z^4 + z^{-4}) + 4(z^2 + z^{-2}) + 6 = 2\cos 4\theta + 8\cos 2\theta + 6$.

So $16\cos^4\!\theta = 2\cos 4\theta + 8\cos 2\theta + 6$, giving $\cos^4\!\theta = \tfrac{1}{8}\cos 4\theta + \tfrac{1}{2}\cos 2\theta + \tfrac{3}{8}$.

nth Roots of Complex Numbers

The equation $z^n = w$ has exactly $n$ solutions in $\mathbb{C}$. If $w = Re^{\mathrm{i}\Phi}$, the $n$ roots are:

$$z_k = R^{1/n}\exp\!\left(\mathrm{i}\,\frac{\Phi + 2k\pi}{n}\right), \quad k = 0, 1, \ldots, n-1.$$

These roots are equally spaced around a circle of radius $R^{1/n}$, separated by angles of $\frac{2\pi}{n}$.

Example 1.12 — Cube Roots of Unity

Find all solutions to $z^3 = 1$ and plot them on an Argand diagram.

Write $1 = 1 \cdot e^{\mathrm{i} \cdot 0}$, so $R = 1$, $\Phi = 0$.

The three roots are $z_k = e^{\mathrm{i} \cdot 2k\pi/3}$ for $k = 0, 1, 2$:

The roots form an equilateral triangle inscribed in the unit circle. Denoting $\omega = e^{2\pi\mathrm{i}/3}$, note that $1 + \omega + \omega^2 = 0$, a result that recurs throughout Further Mathematics.

Example 1.13 — Fourth Roots of $-16$

Solve $z^4 = -16$.

Write $-16 = 16e^{\mathrm{i}\pi}$, so $R = 16$, $\Phi = \pi$.

$z_k = 16^{1/4} e^{\mathrm{i}(\pi + 2k\pi)/4} = 2\,e^{\mathrm{i}(1+2k)\pi/4}$ for $k = 0,1,2,3$.

Exam Tip — Arguments and Principal Values

Always express the principal argument in the range $(-\pi, \pi]$. When computing $\arg(z_1 z_2)$, add the arguments then adjust by $\pm 2\pi$ if necessary to return to $(-\pi, \pi]$. Forgetting this step is one of the most common sources of mark loss on Further Pure papers.

1.6 Loci in the Complex Plane (Preview) Further

A locus is a set of complex numbers satisfying a given condition. For example, the set $\{z \in \mathbb{C} : |z - (2 + \mathrm{i})| = 3\}$ is a circle in the Argand diagram centred at $2 + \mathrm{i}$ with radius 3. This material is treated in full generality in Chapter 2, where Cartesian equations, half-lines, perpendicular bisectors, and regions are all covered systematically.

Chapter Link

Locus problems featuring $|z - a| = r$, $\arg(z - a) = \theta$, and $|z - a| = |z - b|$ are examined together with Argand diagram sketches. Practise converting between complex locus notation and Cartesian equations — this skill is assessed on every Further Pure paper.

Practice Problems

Problem 1 — Cartesian Arithmetic

Given $z_1 = 2 + 3\mathrm{i}$ and $z_2 = 1 - 4\mathrm{i}$, find $z_1 z_2$ and $\dfrac{z_1}{z_2}$, expressing each answer in the form $a + b\mathrm{i}$.

Show solution

Product: $(2+3\mathrm{i})(1-4\mathrm{i}) = 2 - 8\mathrm{i} + 3\mathrm{i} - 12\mathrm{i}^2 = 2 - 5\mathrm{i} + 12 = 14 - 5\mathrm{i}$.

Quotient: $\dfrac{(2+3\mathrm{i})(1+4\mathrm{i})}{(1-4\mathrm{i})(1+4\mathrm{i})} = \dfrac{2+8\mathrm{i}+3\mathrm{i}+12\mathrm{i}^2}{1+16} = \dfrac{2+11\mathrm{i}-12}{17} = \dfrac{-10+11\mathrm{i}}{17} = -\dfrac{10}{17} + \dfrac{11}{17}\mathrm{i}$.

Problem 2 — Modulus and Argument

Find the modulus and principal argument of $z = -3 - 3\mathrm{i}$.

Show solution

$|z| = \sqrt{9 + 9} = 3\sqrt{2}$.

The point $(-3, -3)$ lies in the third quadrant. Reference angle $= \arctan(3/3) = \pi/4$, so $\arg(z) = -\pi + \pi/4 = -3\pi/4$.

Problem 3 — Euler Form

Write $z = \sqrt{2}\,e^{-\mathrm{i}\pi/4}$ in Cartesian form and find $|z|$ and $\arg(z)$.

Show solution

$z = \sqrt{2}\!\left(\cos(-\pi/4) + \mathrm{i}\sin(-\pi/4)\right) = \sqrt{2}\!\left(\dfrac{\sqrt{2}}{2} - \dfrac{\sqrt{2}}{2}\mathrm{i}\right) = 1 - \mathrm{i}$.

$|z| = \sqrt{2}$, $\arg(z) = -\pi/4$.

Problem 4 — de Moivre's Theorem

Use de Moivre's theorem to find $(\cos(\pi/12) + \mathrm{i}\sin(\pi/12))^{12}$.

Show solution

By de Moivre: $(\cos(\pi/12) + \mathrm{i}\sin(\pi/12))^{12} = \cos(12 \cdot \pi/12) + \mathrm{i}\sin(12 \cdot \pi/12) = \cos\pi + \mathrm{i}\sin\pi = -1$.

Problem 5 — Double-Angle Identity via de Moivre

Use de Moivre's theorem to show that $\sin 2\theta = 2\sin\theta\cos\theta$.

Show solution

$(\cos\theta + \mathrm{i}\sin\theta)^2 = \cos^2\theta + 2\mathrm{i}\sin\theta\cos\theta + \mathrm{i}^2\sin^2\theta = (\cos^2\theta - \sin^2\theta) + 2\mathrm{i}\sin\theta\cos\theta$.

By de Moivre this equals $\cos 2\theta + \mathrm{i}\sin 2\theta$. Equating imaginary parts gives $\sin 2\theta = 2\sin\theta\cos\theta$.

Problem 6 — Powers Identity

Show that $\sin^3\theta = \tfrac{3}{4}\sin\theta - \tfrac{1}{4}\sin 3\theta$.

Show solution

Let $z = e^{\mathrm{i}\theta}$. Then $2\mathrm{i}\sin\theta = z - z^{-1}$, so $(2\mathrm{i})^3\sin^3\theta = (z - z^{-1})^3 = z^3 - 3z + 3z^{-1} - z^{-3}$.

$= (z^3 - z^{-3}) - 3(z - z^{-1}) = 2\mathrm{i}\sin 3\theta - 3 \cdot 2\mathrm{i}\sin\theta$.

So $-8\mathrm{i}\sin^3\theta = 2\mathrm{i}\sin 3\theta - 6\mathrm{i}\sin\theta$. Dividing by $-8\mathrm{i}$: $\sin^3\theta = \tfrac{3}{4}\sin\theta - \tfrac{1}{4}\sin 3\theta$.

Problem 7 — Conjugate Root Theorem

A quartic polynomial $P(z)$ with real coefficients has roots $3$ (multiplicity 2) and $2 - \mathrm{i}$. Write down all four roots and hence write $P(z)$ as a product of two real quadratic factors.

Show solution

By the conjugate root theorem, $2 + \mathrm{i}$ is also a root. The four roots are $3, 3, 2 - \mathrm{i}, 2 + \mathrm{i}$.

Real quadratic from root 3 (multiplicity 2): $(z-3)^2 = z^2 - 6z + 9$.

Real quadratic from complex pair: $(z-(2-\mathrm{i}))(z-(2+\mathrm{i})) = (z-2)^2 + 1 = z^2 - 4z + 5$.

$P(z) = (z^2 - 6z + 9)(z^2 - 4z + 5)$.

Problem 8 — nth Roots

Find all fifth roots of $32\,e^{\mathrm{i}\pi/5}$, expressing each in exponential form.

Show solution

$R = 32$, $\Phi = \pi/5$. $R^{1/5} = 2$. The roots are $z_k = 2\,e^{\mathrm{i}(\pi/5 + 2k\pi)/5} = 2\,e^{\mathrm{i}\pi(1+10k)/25}$ for $k = 0,1,2,3,4$.

  • $k=0$: $2e^{\mathrm{i}\pi/25}$
  • $k=1$: $2e^{\mathrm{i}11\pi/25}$
  • $k=2$: $2e^{\mathrm{i}21\pi/25}$
  • $k=3$: $2e^{\mathrm{i}31\pi/25}$ (principal argument: $2e^{-\mathrm{i}19\pi/25}$)
  • $k=4$: $2e^{\mathrm{i}41\pi/25}$ (principal argument: $2e^{-\mathrm{i}9\pi/25}$)

Problem 9 — Modulus-Argument Multiplication

Given $z_1 = 6(\cos(2\pi/3) + \mathrm{i}\sin(2\pi/3))$ and $z_2 = 3(\cos(\pi/6) + \mathrm{i}\sin(\pi/6))$, find $|z_1 z_2|$ and $\arg(z_1 z_2)$ without expanding into Cartesian form.

Show solution

$|z_1 z_2| = 6 \times 3 = 18$.

$\arg(z_1 z_2) = \tfrac{2\pi}{3} + \tfrac{\pi}{6} = \tfrac{4\pi}{6} + \tfrac{\pi}{6} = \tfrac{5\pi}{6}$. Since $\tfrac{5\pi}{6} \in (-\pi, \pi]$, this is already the principal argument.

Problem 10 — Proof Using de Moivre

Prove that $\cos 4\theta = 8\cos^4\!\theta - 8\cos^2\!\theta + 1$.

Show solution

Expand $(\cos\theta + \mathrm{i}\sin\theta)^4$ by the binomial theorem:

$= \cos^4\theta + 4\cos^3\theta(\mathrm{i}\sin\theta) + 6\cos^2\theta(\mathrm{i}\sin\theta)^2 + 4\cos\theta(\mathrm{i}\sin\theta)^3 + (\mathrm{i}\sin\theta)^4$

$= \cos^4\theta + 4\mathrm{i}\cos^3\theta\sin\theta - 6\cos^2\theta\sin^2\theta - 4\mathrm{i}\cos\theta\sin^3\theta + \sin^4\theta$.

By de Moivre, the real part equals $\cos 4\theta$:

$\cos 4\theta = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta$.

Substitute $\sin^2\theta = 1 - \cos^2\theta$:

$= \cos^4\theta - 6\cos^2\theta(1-\cos^2\theta) + (1-\cos^2\theta)^2$

$= \cos^4\theta - 6\cos^2\theta + 6\cos^4\theta + 1 - 2\cos^2\theta + \cos^4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$. $\square$