UK › Textbook › Further Pure 1

A-Level Further Mathematics: Further Pure 1

A-Level Further Mathematics · Edexcel / AQA / OCR / CIE · Updated March 2026 · 35 min read

Exam Board Note

Further Pure 1 content is examined across all major UK exam boards, though organisation varies. Edexcel covers this material in Further Pure 1 (FP1/8FM0:01). AQA includes most topics in its Further Mathematics Core modules. OCR A covers these topics in Further Pure A (Y541). CIE includes complex numbers and matrices in Further Mathematics Paper 1. The content on this page aligns with all four specifications — check your specific board's syllabus for exact topic listings.

Further Pure 1 forms the theoretical backbone of A-Level Further Mathematics. It extends the pure mathematics of A-Level with powerful new algebraic and analytical tools. Mastery of this material is essential before progressing to Further Pure 2 and the optional applied modules.

1. Complex Numbers

The need for complex numbers arises from attempting to solve equations like $x^2 + 1 = 0$. No real number satisfies this, because $x^2 \geq 0$ for all real $x$. We introduce the imaginary unit $i$ defined by $i^2 = -1$.

Definition: Complex Number

A complex number is an expression of the form $z = a + bi$, where $a, b \in \mathbb{R}$ and $i^2 = -1$. The number $a = \text{Re}(z)$ is the real part and $b = \text{Im}(z)$ is the imaginary part.

The complex conjugate of $z = a + bi$ is $\bar{z} = a - bi$. Note that $z\bar{z} = a^2 + b^2 = |z|^2$.

Arithmetic of Complex Numbers

Worked Example 1.1 — Complex Arithmetic

Given $z_1 = 3 + 2i$ and $z_2 = 1 - 4i$, find (a) $z_1 z_2$ and (b) $\dfrac{z_1}{z_2}$.

Part (a)

$$z_1 z_2 = (3 + 2i)(1 - 4i) = 3(1) + 3(-4i) + 2i(1) + 2i(-4i)$$ $$= 3 - 12i + 2i - 8i^2 = 3 - 10i - 8(-1) = 11 - 10i$$

Part (b) Multiply numerator and denominator by $\bar{z}_2 = 1 + 4i$:

$$\frac{z_1}{z_2} = \frac{3 + 2i}{1 - 4i} \cdot \frac{1 + 4i}{1 + 4i} = \frac{(3 + 2i)(1 + 4i)}{1^2 + 4^2}$$

Numerator: $(3+2i)(1+4i) = 3 + 12i + 2i + 8i^2 = 3 + 14i - 8 = -5 + 14i$

Denominator: $1 + 16 = 17$

$$\frac{z_1}{z_2} = \frac{-5 + 14i}{17} = -\frac{5}{17} + \frac{14}{17}i$$

Complex Roots of Polynomials

A key theorem in further mathematics: if the polynomial has real coefficients, then complex roots always occur in conjugate pairs. If $\alpha = a + bi$ is a root, then $\bar{\alpha} = a - bi$ is also a root.

Worked Example 1.2 — Finding Complex Roots

Given that $2 + i$ is a root of $p(x) = x^3 - 5x^2 + 9x - 5$, find all roots.

Step 1 Since the coefficients are real, $2 - i$ is also a root. So $(x - (2+i))(x - (2-i))$ is a factor.

$$(x - 2 - i)(x - 2 + i) = (x-2)^2 + 1 = x^2 - 4x + 5$$

Step 2 Divide $p(x)$ by $x^2 - 4x + 5$:

$$x^3 - 5x^2 + 9x - 5 = (x^2 - 4x + 5)(x - 1)$$

Check: $(x^2 - 4x + 5)(x-1) = x^3 - x^2 - 4x^2 + 4x + 5x - 5 = x^3 - 5x^2 + 9x - 5$ ✓

Step 3 The roots are $x = 2 + i$, $x = 2 - i$, and $x = 1$.

2. The Argand Diagram and Modulus-Argument Form

Complex numbers can be represented geometrically. The Argand diagram places the complex number $z = a + bi$ at the point $(a, b)$ in the complex plane, with the real axis horizontal and the imaginary axis vertical.

Definition: Modulus and Argument

For $z = a + bi$:

$$\tan\theta = \frac{b}{a} \quad \text{(adjust quadrant using signs of } a \text{ and } b\text{)}$$

The modulus-argument form (also called polar form) is:

$$z = r(\cos\theta + i\sin\theta)$$

where $r = |z|$ and $\theta = \arg(z)$. This is often abbreviated $r\,\text{cis}\,\theta$.

Key properties of modulus and argument:

Worked Example 2.1 — Modulus-Argument Form

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

Step 1 Find the modulus: $$|z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$$

Step 2 Find the argument. Since $a = -1 < 0$ and $b = \sqrt{3} > 0$, $z$ is in the second quadrant. $$\tan\alpha = \frac{\sqrt{3}}{1} = \sqrt{3} \implies \alpha = \frac{\pi}{3}$$ Therefore $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

Step 3 Write in polar form: $$z = 2\!\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$$

Loci in the Complex Plane

Equations and inequalities involving $|z - a|$ and $\arg(z - a)$ describe geometric loci:

FP1·1 — Argand diagram showing complex numbers in the complex plane. Blue point: $z_1 = 2 + i$ with $|z_1| = \sqrt{5}$. Red point: $z_2 = -1 + \sqrt{3}\,i$ (from Example 2.1) with $|z_2| = 2$ and $\arg(z_2) = \tfrac{2\pi}{3}$ (dashed). Green circle: locus $|z| = 2$.

3. De Moivre's Theorem

Theorem: De Moivre's Theorem

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

$$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$$

Equivalently, if $z = r(\cos\theta + i\sin\theta)$, then $z^n = r^n(\cos n\theta + i\sin n\theta)$.

De Moivre's theorem has two principal applications in A-Level Further Mathematics:

  1. Multiple angle identities: expressing $\sin(n\theta)$ and $\cos(n\theta)$ in terms of powers of $\sin\theta$ and $\cos\theta$.
  2. $n$th roots of complex numbers: finding all $n$ values of $z^{1/n}$.

Worked Example 3.1 — Triple Angle Formula

Use De Moivre's theorem to show that $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$.

Step 1 By De Moivre's theorem, $\cos 3\theta + i\sin 3\theta = (\cos\theta + i\sin\theta)^3$.

Step 2 Expand using the binomial theorem: $$(\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3\cos^2\theta(i\sin\theta) + 3\cos\theta(i\sin\theta)^2 + (i\sin\theta)^3$$ $$= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta$$

Step 3 Collect real and imaginary parts: $$\text{Real: } \cos^3\theta - 3\cos\theta\sin^2\theta = \cos\theta(\cos^2\theta - 3\sin^2\theta)$$ Use $\sin^2\theta = 1 - \cos^2\theta$: $$= \cos\theta(\cos^2\theta - 3(1-\cos^2\theta)) = \cos\theta(4\cos^2\theta - 3) = 4\cos^3\theta - 3\cos\theta$$

Therefore $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ as required.

Worked Example 3.2 — Cube Roots of Unity

Find all cube roots of $-8$.

Step 1 Write $-8 = 8(\cos\pi + i\sin\pi)$ in polar form.

Step 2 The cube roots have modulus $8^{1/3} = 2$ and arguments $\dfrac{\pi + 2k\pi}{3}$ for $k = 0, 1, 2$:

The three cube roots of $-8$ are $\mathbf{-2}$, $\mathbf{1 + \sqrt{3}\,i}$, and $\mathbf{1 - \sqrt{3}\,i}$.

4. Matrices: Operations, Determinants, and Inverses

A matrix is a rectangular array of numbers arranged in rows and columns. We write an $m \times n$ matrix $A$ for one with $m$ rows and $n$ columns, and denote its $(i,j)$ entry by $a_{ij}$.

Matrix Multiplication

The product $AB$ of an $m \times n$ matrix $A$ and an $n \times p$ matrix $B$ is the $m \times p$ matrix $C$ where: $$c_{ij} = \sum_{k=1}^n a_{ik} b_{kj}$$ Note that matrix multiplication is generally not commutative: $AB \neq BA$ in general.

Worked Example 4.1 — Matrix Multiplication

Compute $AB$ where $A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ -2 & 3 \end{pmatrix}$.

$$AB = \begin{pmatrix} 2(1)+(-1)(-2) & 2(0)+(-1)(3) \\ 3(1)+4(-2) & 3(0)+4(3) \end{pmatrix} = \begin{pmatrix} 4 & -3 \\ -5 & 12 \end{pmatrix}$$

The Determinant

Definition: 2×2 and 3×3 Determinants

For a $2 \times 2$ matrix: $\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc$

For a $3 \times 3$ matrix, expand along the first row: $$\det\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg)$$

A matrix is singular (non-invertible) if and only if its determinant is zero.

The Inverse Matrix

For a non-singular $2 \times 2$ matrix: $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$ For $3 \times 3$ matrices, the inverse is found using the matrix of cofactors transposed (the adjugate), divided by the determinant.

Worked Example 4.2 — Inverse of a 3×3 Matrix

Find the inverse of $M = \begin{pmatrix} 1 & 2 & 0 \\ -1 & 3 & 1 \\ 2 & 0 & 4 \end{pmatrix}$.

Step 1 Find $\det(M)$ expanding along row 1: $$\det(M) = 1\det\begin{pmatrix}3&1\\0&4\end{pmatrix} - 2\det\begin{pmatrix}-1&1\\2&4\end{pmatrix} + 0$$ $$= 1(12 - 0) - 2(-4 - 2) = 12 + 12 = 24$$

Step 2 Compute the matrix of cofactors $C$ where $C_{ij} = (-1)^{i+j}M_{ij}$: $$C = \begin{pmatrix} 12 & 6 & -6 \\ -8 & 4 & 4 \\ 2 & -1 & 5 \end{pmatrix}$$

Step 3 The inverse is $M^{-1} = \dfrac{1}{24}C^T$: $$M^{-1} = \frac{1}{24}\begin{pmatrix} 12 & -8 & 2 \\ 6 & 4 & -1 \\ -6 & 4 & 5 \end{pmatrix}$$

5. Matrix Transformations

In 2D, geometric transformations of the plane can be represented by $2 \times 2$ matrices. A transformation $T$ maps the point $\mathbf{x} = (x, y)^T$ to $T\mathbf{x}$.

Rotation by $\theta$ (anticlockwise) $\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}$
Reflection in $y = x$ $\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$
Reflection in $x$-axis $\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}$
Enlargement scale factor $k$ $\begin{pmatrix}k & 0 \\ 0 & k\end{pmatrix}$
Shear (parallel to $x$-axis) $\begin{pmatrix}1 & k \\ 0 & 1\end{pmatrix}$

When applying two transformations in succession, the combined matrix is the product. If $T_1$ is applied first then $T_2$, the combined transformation matrix is $M = T_2 T_1$ (note the order: the first transformation is on the right).

The area scale factor of a linear transformation represented by matrix $M$ equals $|\det(M)|$. If $\det(M) < 0$, the transformation involves a reflection.

Worked Example 5.1 — Combining Transformations

Find the matrix representing a rotation of $90°$ anticlockwise followed by a reflection in the $y$-axis. Apply this to the point $(3, 1)$.

Step 1 Rotation by $90°$: $R = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$

Step 2 Reflection in $y$-axis: $S = \begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}$

Step 3 Combined matrix (rotation first, so $R$ on right): $$M = SR = \begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$$ This is the reflection in $y = x$!

Step 4 Apply to $(3, 1)^T$: $$\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}3 \\ 1\end{pmatrix} = \begin{pmatrix}1 \\ 3\end{pmatrix}$$ The image is $(1, 3)$, as expected from the reflection $y = x$.

6. Eigenvalues and Eigenvectors

Definition: Eigenvalues and Eigenvectors

A non-zero vector $\mathbf{v}$ is an eigenvector of matrix $A$ with corresponding eigenvalue $\lambda$ if: $$A\mathbf{v} = \lambda\mathbf{v}$$ Geometrically, $A$ scales $\mathbf{v}$ by $\lambda$ without changing its direction (though it may reverse it if $\lambda < 0$).

Finding eigenvalues: Rearrange $A\mathbf{v} = \lambda\mathbf{v}$ as $(A - \lambda I)\mathbf{v} = \mathbf{0}$. For non-trivial solutions, we need $\det(A - \lambda I) = 0$. This is the characteristic equation.

Finding eigenvectors: Once $\lambda$ is known, solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$ to find the corresponding eigenvector(s).

Worked Example 6.1 — Eigenvalues and Eigenvectors

Find the eigenvalues and eigenvectors of $A = \begin{pmatrix} 5 & 2 \\ 2 & 2 \end{pmatrix}$.

Step 1 Form and solve the characteristic equation: $$\det(A - \lambda I) = \det\begin{pmatrix}5-\lambda & 2 \\ 2 & 2-\lambda\end{pmatrix} = (5-\lambda)(2-\lambda) - 4$$ $$= \lambda^2 - 7\lambda + 10 - 4 = \lambda^2 - 7\lambda + 6 = (\lambda - 1)(\lambda - 6) = 0$$ Eigenvalues: $\lambda_1 = 1$, $\lambda_2 = 6$.

Step 2 For $\lambda = 1$: solve $(A - I)\mathbf{v} = \mathbf{0}$: $$\begin{pmatrix}4 & 2 \\ 2 & 1\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix}$$ Row 1 gives $4x + 2y = 0$, so $y = -2x$. Eigenvector: $\mathbf{v}_1 = \begin{pmatrix}1 \\ -2\end{pmatrix}$.

Step 3 For $\lambda = 6$: solve $(A - 6I)\mathbf{v} = \mathbf{0}$: $$\begin{pmatrix}-1 & 2 \\ 2 & -4\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix}$$ Row 1 gives $-x + 2y = 0$, so $x = 2y$. Eigenvector: $\mathbf{v}_2 = \begin{pmatrix}2 \\ 1\end{pmatrix}$.

7. Proof by Mathematical Induction

The Method of Mathematical Induction

To prove a statement $P(n)$ is true for all integers $n \geq n_0$:

  1. Base case: Verify $P(n_0)$ is true.
  2. Inductive step: Assume $P(k)$ is true (the inductive hypothesis). Show that $P(k+1)$ must also be true.
  3. Conclusion: By the principle of induction, $P(n)$ is true for all $n \geq n_0$.

Worked Example 7.1 — Summation Formula

Prove by induction that $\displaystyle\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$ for all $n \geq 1$.

Base case $n = 1$: LHS $= 1^2 = 1$. RHS $= \dfrac{1 \cdot 2 \cdot 3}{6} = 1$. ✓

Inductive step Assume the result holds for $n = k$: $$\sum_{r=1}^k r^2 = \frac{k(k+1)(2k+1)}{6}$$ We must show it holds for $n = k+1$, i.e., $\displaystyle\sum_{r=1}^{k+1} r^2 = \dfrac{(k+1)(k+2)(2k+3)}{6}$.

$$\sum_{r=1}^{k+1} r^2 = \sum_{r=1}^k r^2 + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2$$ $$= \frac{k(k+1)(2k+1) + 6(k+1)^2}{6} = \frac{(k+1)[k(2k+1) + 6(k+1)]}{6}$$ $$= \frac{(k+1)(2k^2 + 7k + 6)}{6} = \frac{(k+1)(k+2)(2k+3)}{6}$$

Conclusion The result holds for $n = 1$, and if it holds for $n = k$ it holds for $n = k+1$. By the principle of mathematical induction, the result is true for all integers $n \geq 1$.

Worked Example 7.2 — Divisibility Proof

Prove by induction that $5^n - 1$ is divisible by 4 for all $n \geq 1$.

Base case $n = 1$: $5^1 - 1 = 4 = 4 \times 1$. Divisible by 4. ✓

Inductive step Assume $5^k - 1 = 4m$ for some integer $m$. Then $5^k = 4m + 1$. $$5^{k+1} - 1 = 5 \cdot 5^k - 1 = 5(4m + 1) - 1 = 20m + 5 - 1 = 20m + 4 = 4(5m + 1)$$ Since $5m + 1$ is an integer, $5^{k+1} - 1$ is divisible by 4.

Conclusion By mathematical induction, $5^n - 1$ is divisible by 4 for all $n \geq 1$.

8. Series: Method of Differences and Maclaurin Series

Method of Differences

The method of differences is used to evaluate finite sums by expressing the general term as a difference of a function: $f(r) = u(r) - u(r-1)$ (or $u(r+1) - u(r)$). The sum then telescopes:

$$\sum_{r=1}^n f(r) = \sum_{r=1}^n [u(r) - u(r-1)] = u(n) - u(0)$$

Worked Example 8.1 — Method of Differences

Show that $\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$, and hence find $\displaystyle\sum_{r=1}^n \frac{1}{r(r+1)}$.

Step 1 Partial fractions: $\dfrac{1}{r} - \dfrac{1}{r+1} = \dfrac{(r+1) - r}{r(r+1)} = \dfrac{1}{r(r+1)}$ ✓

Step 2 Telescope the sum: $$\sum_{r=1}^n \frac{1}{r(r+1)} = \sum_{r=1}^n\!\left(\frac{1}{r} - \frac{1}{r+1}\right)$$ $$= \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$$ $$= 1 - \frac{1}{n+1} = \frac{n}{n+1}$$

Maclaurin Series

The Maclaurin series expresses a function as an infinite power series centred at $x = 0$:

$$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$$

Standard Maclaurin series (valid for the stated values of $x$):

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \quad (\text{all } x)$$ $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \quad (\text{all } x)$$ $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \quad (\text{all } x)$$ $$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \quad (-1 < x \leq 1)$$ $$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots \quad (|x| < 1 \text{ if } n \notin \mathbb{Z}^+)$$

9. Roots of Polynomials (Vieta's Formulas)

Vieta's formulas relate the coefficients of a polynomial to symmetric functions of its roots, without requiring us to find the roots explicitly.

Vieta's Formulas

For the quadratic $ax^2 + bx + c = 0$ with roots $\alpha, \beta$:

$$\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}$$

For the cubic $ax^3 + bx^2 + cx + d = 0$ with roots $\alpha, \beta, \gamma$:

$$\alpha+\beta+\gamma = -\frac{b}{a}, \quad \alpha\beta+\beta\gamma+\gamma\alpha = \frac{c}{a}, \quad \alpha\beta\gamma = -\frac{d}{a}$$

For the quartic $ax^4 + bx^3 + cx^2 + dx + e = 0$ with roots $\alpha, \beta, \gamma, \delta$:

$$\sum\alpha = -\frac{b}{a},\quad \sum\alpha\beta = \frac{c}{a},\quad \sum\alpha\beta\gamma = -\frac{d}{a},\quad \alpha\beta\gamma\delta = \frac{e}{a}$$

Worked Example 9.1 — Symmetric Functions of Roots

The cubic $2x^3 - 5x^2 + x + 3 = 0$ has roots $\alpha, \beta, \gamma$. Find (a) $\alpha^2 + \beta^2 + \gamma^2$ and (b) $\alpha^3 + \beta^3 + \gamma^3$.

Vieta's formulas $\alpha+\beta+\gamma = \dfrac{5}{2}$, $\alpha\beta+\beta\gamma+\gamma\alpha = \dfrac{1}{2}$, $\alpha\beta\gamma = -\dfrac{3}{2}$.

Part (a) Use the identity $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)$: $$= \left(\frac{5}{2}\right)^2 - 2\cdot\frac{1}{2} = \frac{25}{4} - 1 = \frac{21}{4}$$

Part (b) Use Newton's identity: $p_3 = (\alpha+\beta+\gamma)p_2 - (\alpha\beta+\beta\gamma+\gamma\alpha)p_1 + 3\alpha\beta\gamma$ where $p_k = \alpha^k+\beta^k+\gamma^k$ and $p_1 = \frac{5}{2}$, $p_2 = \frac{21}{4}$: $$\alpha^3+\beta^3+\gamma^3 = \frac{5}{2}\cdot\frac{21}{4} - \frac{1}{2}\cdot\frac{5}{2} + 3\cdot\left(-\frac{3}{2}\right) = \frac{105}{8} - \frac{5}{4} - \frac{9}{2} = \frac{105-10-36}{8} = \frac{59}{8}$$

10. Practice Problems

Problem 1 — Complex Numbers

Find the complex number $z$ satisfying $z^2 = -5 + 12i$. Give your answer in the form $a + bi$.

Show Solution

Let $z = a + bi$. Then $z^2 = a^2 - b^2 + 2abi = -5 + 12i$.

Equating real parts: $a^2 - b^2 = -5$. Equating imaginary parts: $2ab = 12$, so $b = 6/a$.

Substituting: $a^2 - 36/a^2 = -5$. Multiply by $a^2$: $a^4 + 5a^2 - 36 = 0$, so $(a^2 - 4)(a^2 + 9) = 0$.

Since $a \in \mathbb{R}$, we have $a^2 = 4$, so $a = \pm 2$. If $a = 2$, $b = 3$; if $a = -2$, $b = -3$.

$$z = 2 + 3i \quad \text{or} \quad z = -2 - 3i$$

Problem 2 — Argand Diagram / Loci

Sketch the locus of $z$ satisfying $|z - 3| = |z + i|$ and find its Cartesian equation.

Show Solution

Let $z = x + iy$. Then $|z - 3| = |(x-3) + iy| = \sqrt{(x-3)^2 + y^2}$ and $|z + i| = |x + i(y+1)| = \sqrt{x^2 + (y+1)^2}$.

Setting equal and squaring: $(x-3)^2 + y^2 = x^2 + (y+1)^2$

$x^2 - 6x + 9 + y^2 = x^2 + y^2 + 2y + 1$

$-6x + 9 = 2y + 1$, giving $6x + 2y = 8$, i.e., $\mathbf{3x + y = 4}$.

This is the perpendicular bisector of the segment joining $(3, 0)$ to $(0, -1)$.

Problem 3 — De Moivre's Theorem

Use De Moivre's theorem to find all fifth roots of $32$ and plot them on an Argand diagram.

Show Solution

Write $32 = 32(\cos 0 + i\sin 0)$. The fifth roots have modulus $32^{1/5} = 2$ and arguments $\dfrac{0 + 2k\pi}{5} = \dfrac{2k\pi}{5}$ for $k = 0, 1, 2, 3, 4$.

$k=0$: $z = 2(\cos 0 + i\sin 0) = 2$

$k=1$: $z = 2\!\left(\cos\dfrac{2\pi}{5} + i\sin\dfrac{2\pi}{5}\right) \approx 0.618 + 1.902i$

$k=2$: $z = 2\!\left(\cos\dfrac{4\pi}{5} + i\sin\dfrac{4\pi}{5}\right) \approx -1.618 + 1.176i$

$k=3$: $z = 2\!\left(\cos\dfrac{6\pi}{5} + i\sin\dfrac{6\pi}{5}\right) \approx -1.618 - 1.176i$

$k=4$: $z = 2\!\left(\cos\dfrac{8\pi}{5} + i\sin\dfrac{8\pi}{5}\right) \approx 0.618 - 1.902i$

These 5 points lie equally spaced on a circle of radius 2 in the Argand diagram.

Problem 4 — Matrices

The matrix $A = \begin{pmatrix} 3 & 1 \\ k & 2 \end{pmatrix}$ is singular. Find $k$ and hence explain the geometric significance.

Show Solution

$A$ is singular when $\det(A) = 0$: $3(2) - (1)(k) = 6 - k = 0$, so $k = 6$.

When $k = 6$, $A = \begin{pmatrix}3 & 1 \\ 6 & 2\end{pmatrix}$. Since $\det(A) = 0$, the area scale factor is zero. Geometrically, the transformation maps the entire plane onto a line (or the origin), meaning all points are mapped to a 1-dimensional set — the transformation collapses the plane.

Problem 5 — Eigenvalues

Matrix $B = \begin{pmatrix} 4 & 1 \\ 3 & 2 \end{pmatrix}$ has eigenvalues $\lambda_1$ and $\lambda_2$. Find the eigenvalues, their corresponding eigenvectors, and verify that $\text{tr}(B) = \lambda_1 + \lambda_2$.

Show Solution

Characteristic equation: $(4-\lambda)(2-\lambda) - 3 = \lambda^2 - 6\lambda + 5 = (\lambda-1)(\lambda-5) = 0$.

Eigenvalues: $\lambda_1 = 1$, $\lambda_2 = 5$.

For $\lambda = 1$: $(B - I)\mathbf{v} = \begin{pmatrix}3 & 1 \\ 3 & 1\end{pmatrix}\mathbf{v} = \mathbf{0}$, giving $3x + y = 0$. Eigenvector: $\mathbf{v}_1 = \begin{pmatrix}1 \\ -3\end{pmatrix}$.

For $\lambda = 5$: $(B - 5I)\mathbf{v} = \begin{pmatrix}-1 & 1 \\ 3 & -3\end{pmatrix}\mathbf{v} = \mathbf{0}$, giving $x = y$. Eigenvector: $\mathbf{v}_2 = \begin{pmatrix}1 \\ 1\end{pmatrix}$.

$\text{tr}(B) = 4 + 2 = 6 = 1 + 5 = \lambda_1 + \lambda_2$ ✓

Problem 6 — Proof by Induction

Prove by induction that $\displaystyle\sum_{r=1}^n r \cdot r! = (n+1)! - 1$ for all positive integers $n$.

Show Solution

Base case ($n=1$): LHS $= 1 \cdot 1! = 1$. RHS $= 2! - 1 = 2 - 1 = 1$. ✓

Inductive step: Assume $\displaystyle\sum_{r=1}^k r \cdot r! = (k+1)! - 1$. Then:

$$\sum_{r=1}^{k+1} r \cdot r! = (k+1)! - 1 + (k+1)(k+1)! = (k+1)!\bigl(1 + (k+1)\bigr) - 1 = (k+2)! - 1$$

This is exactly the formula with $n = k+1$. By induction, the result holds for all $n \geq 1$.

Problem 7 — Series / Method of Differences

Show that $\dfrac{4}{r(r+2)} = \dfrac{2}{r} - \dfrac{2}{r+2}$, and hence find $\displaystyle\sum_{r=1}^n \frac{4}{r(r+2)}$ in simplified form.

Show Solution

$\dfrac{2}{r} - \dfrac{2}{r+2} = \dfrac{2(r+2) - 2r}{r(r+2)} = \dfrac{4}{r(r+2)}$ ✓

Writing out the telescoping sum: $$\sum_{r=1}^n \frac{4}{r(r+2)} = 2\sum_{r=1}^n\!\left(\frac{1}{r} - \frac{1}{r+2}\right)$$ $$= 2\left[\left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots\right]$$ Most terms cancel. The surviving terms are $1 + \dfrac{1}{2} - \dfrac{1}{n+1} - \dfrac{1}{n+2}$: $$= 2\left(\frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right) = 3 - \frac{2}{n+1} - \frac{2}{n+2}$$

Problem 8 — Roots of Polynomials

The quartic $x^4 - 6x^3 + 11x^2 - 2x - 4 = 0$ has roots $\alpha, \beta, \gamma, \delta$. Find $\alpha^2+\beta^2+\gamma^2+\delta^2$ and $\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}+\dfrac{1}{\delta}$.

Show Solution

From Vieta's formulas: $\sum\alpha = 6$, $\sum\alpha\beta = 11$, $\sum\alpha\beta\gamma = 2$, $\alpha\beta\gamma\delta = -4$.

$\alpha^2+\beta^2+\gamma^2+\delta^2 = (\sum\alpha)^2 - 2\sum\alpha\beta = 36 - 22 = \mathbf{14}$

$\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}+\dfrac{1}{\delta} = \dfrac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta} = \dfrac{2}{-4} = -\dfrac{1}{2}$

© 2026 MathHub Global. All rights reserved.