Argand Diagrams (AQA A-Level Further Maths): Revision Notes

Worked Example: Plotting a Complex Number and Its Conjugate

Given $z = 1 + 3i$, show $z$ and $z^*$ on an Argand diagram.

Solution:

First, identify the real and imaginary parts:

• $z = 1 + 3i$ has real part $a = 1$ and imaginary part $b = 3$
• Plot $z$ at the point $(1, 3)$

For the complex conjugate:

• $z^* = 1 - 3i$ has real part $a = 1$ and imaginary part $b = -3$
• Plot $z^*$ at the point $(1, -3)$

Notice that $z^*$ is the reflection of $z$ in the real axis.

Worked Example: Vector Addition

Given $z = 4 + i$ and $w = -2 + i$, show $z$, $w$ and $z + w$ on an Argand diagram.

Solution:

Plot each complex number as a position vector:

• $z = 4 + i$ is plotted at $(4, 1)$
• $w = -2 + i$ is plotted at $(-2, 1)$

To find $z + w$: $z + w = (4 + i) + (-2 + i) = 2 + 2i$

Plot $z + w$ at $(2, 2)$

The sum can be visualised using vector addition. Draw vectors from the origin to $z$ and to $w$, then complete the parallelogram. The diagonal from the origin represents $z + w$.

Worked Example: Scalar Multiples

Given $z = 1 + 2i$, show $z$, $3z$ and $-2z$ on an Argand diagram.

Solution:

First, plot $z = 1 + 2i$ at the point $(1, 2)$.

Calculate the scalar multiples:

• $3z = 3(1 + 2i) = 3 + 6i$, plot at $(3, 6)$
• $-2z = -2(1 + 2i) = -2 - 4i$, plot at $(-2, -4)$

All three points lie on a straight line through the origin. This is because scalar multiplication stretches or shrinks the position vector but keeps it along the same line.

Worked Example: Polynomial Roots and Geometry

Given: $f(x) = 2x^4 + x^3 - 15x^2 + 100x - 250$

a) Find all the roots of $f(x) = 0$ given that $x + 5$ is a factor.

Solution:

Using factor theorem or long division:

$2x^4 + x^3 - 15x^2 + 100x - 250 = (x + 5)(2x^3 - 9x^2 + 30x - 50) = 0$

Factorising further:

$2x^3 - 9x^2 + 30x - 50 = 0$

Using a calculator or further factorisation:

$x = \frac{5}{2}, 1 \pm 3i$

The four roots are: $x = -5$, $x = \frac{5}{2}$, $x = 1 + 3i$, $x = 1 - 3i$

b) Find the area of the quadrilateral formed by the points representing the four roots.

Solution:

Plot the roots on an Argand diagram:

• $A(-5, 0)$
• $B(1, 3)$
• $C(2.5, 0)$
• $D(1, -3)$

The quadrilateral is a kite, which can be split into two triangles.

Using the formula for area:

$\text{Area} = 2 \times \left(\frac{1}{2} \times 7.5 \times 3\right) = 22.5 \text{ square units}$

c) Prove that the quadrilateral contains two right angles.

Solution:

Calculate the gradient of line $AB$:

$\text{Gradient of } AB = \frac{3 - 0}{1 - (-5)} = \frac{3}{6} = \frac{1}{2}$

Calculate the gradient of line $BC$:

$\text{Gradient of } BC = \frac{0 - 3}{2.5 - 1} = \frac{-3}{1.5} = -2$

Check if lines are perpendicular:

$\frac{1}{2} \times (-2) = -1$

Since the product of gradients equals $-1$, angle $ABC$ is a right angle. Similarly, we can prove that angle $ADC$ is also a right angle.

Worked Example: Describing Transformations

Given $z = 1 - 3i$. The points $A$, $B$ and $C$ are represented by $z$, $z^2$ and $z^2 - 4z$ respectively.

a) Find $z^2 - 4z$ in the form $a + bi$.

Solution:

First calculate $z^2$:

$z^2 = (1 - 3i)^2 = 1 - 6i + 9i^2 = 1 - 6i - 9 = -8 - 6i$

Now find $z^2 - 4z$:

$z^2 - 4z = (-8 - 6i) - 4(1 - 3i) = -8 - 6i - 4 + 12i = -12 + 6i$

b) Describe the transformation that maps line segment $OA$ to $CB$.

Solution:

We have:

• $A$ at $(1, -3)$ representing $z = 1 - 3i$
• $B$ at $(-8, -6)$ representing $z^2 = -8 - 6i$
• $C$ at $(-12, 6)$ representing $z^2 - 4z = -12 + 6i$

Find the vector $\overrightarrow{CB}$:

$\overrightarrow{CB} = z^2 - (z^2 - 4z) = 4z = 4(1 - 3i) = 4 - 12i$

This shows that $\overrightarrow{CB} = 4\overrightarrow{OA}$.

Since $OA$ and $CB$ are parallel and $CB$ is four times the length of $OA$, the transformation is:

An enlargement with scale factor 4, centre the origin, followed by a translation by the vector $\begin{pmatrix} -12 \\ 6 \end{pmatrix}$

Alternatively: An enlargement with scale factor 4, centre $(4, -2)$