Modulus-Argument Form and Loci (AQA A-Level Further Maths): Revision Notes

Worked Example: Converting to modulus-argument form

Let's convert $z = 7 - i$ to modulus-argument form.

Step 1: Find the modulus:

$|z| = \sqrt{7^2 + (-1)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}$

Step 2: Find the argument. We calculate:

$\tan^{-1}\left(\frac{1}{7}\right) = 0.142 \text{ radians}$

Step 3: Since the complex number is in the fourth quadrant (positive real part, negative imaginary part), the argument measured clockwise from the positive real axis is negative:

$\arg(z) = -0.142 \text{ radians}$

Step 4: Therefore, the modulus-argument form is:

$z = 5\sqrt{2}(\cos(-0.142) + i\sin(-0.142))$

Exam tip: Always ensure your argument is in the correct range $-\pi < \theta \leq \pi$. Draw a quick sketch to verify which quadrant your complex number is in.

Worked Example: Finding argument and modulus of a product

Given that $z_1 = \sqrt{3} + i$ and $z_1z_2 = -4 - 4i$, find the argument and modulus of $z_2$.

Step 1: Find $|z_1|$:

$|z_1| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2$

Step 2: Find $|z_1z_2|$:

$|z_1z_2| = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = 4\sqrt{2}$

Step 3: Using the property $|z_1z_2| = |z_1||z_2|$:

$4\sqrt{2} = 2|z_2|$

$|z_2| = 2\sqrt{2}$

Step 4: For the argument, we can use a calculator to find:

$\arg(z_1) = \frac{\pi}{6} \text{ and } \arg(z_1z_2) = -\frac{3\pi}{4}$

Step 5: Using the property $\arg(z_1z_2) = \arg(z_1) + \arg(z_2)$:

$-\frac{3\pi}{4} = \frac{\pi}{6} + \arg(z_2)$

$\arg(z_2) = -\frac{3\pi}{4} - \frac{\pi}{6} = -\frac{11\pi}{12}$

Worked Example: Sketching a circle locus

Sketch the locus of points that satisfy $|z - 2 + 3i| = 2$.

Step 1: Rewrite in the standard form $|z - z_1| = r$:

$|z - (2 - 3i)| = 2$

Step 2: This tells us the locus is a circle with:

• Center at $z_1 = 2 - 3i$, which is the point $(2, -3)$ on the Argand diagram
• Radius $r = 2$

Exam tip: When you see $|z - (a + bi)| = r$, immediately identify it as a circle. The point inside the modulus (with opposite signs) is the center.

Worked Example: Sketching a half-line locus

Sketch the locus of $z$ where $\arg(z + 2 + 3i) = \frac{\pi}{6}$.

Step 1: Rewrite in the standard form $\arg(z - z_1) = \theta$:

$\arg(z - (-2 - 3i)) = \frac{\pi}{6}$

Step 2: This tells us:

• The half-line starts at the point $z_1 = -2 - 3i$, which is $(-2, -3)$
• The angle with the positive real axis is $\frac{\pi}{6}$

Step 3: Draw a horizontal line in the positive real direction from $(-2, -3)$, then rotate it anticlockwise by $\frac{\pi}{6}$ radians.

Exam tip: The notation $\arg(z - z_1) = \theta$ means "the angle from $z_1$ to $z$". Start at $z_1$ and draw the ray at angle $\theta$.

Worked Example: Sketching a perpendicular bisector

Sketch the locus of points satisfying $|z - 2| = |z + 3i|$.

Step 1: Rewrite in standard form:

$|z - 2| = |z - (-3i)|$

Step 2: This means the locus includes all points equidistant from $z_1 = 2$ (the point $(2, 0)$) and $z_2 = -3i$ (the point $(0, -3)$).

Step 3: Plot both points and connect them with a dotted line. The locus is the perpendicular bisector of this line segment.

Worked Example: Shading regions

Part 1: Shade the region where $|z - 5| \geq 3$.

Step 1: The boundary is a circle with center $(5, 0)$ and radius 3.

Step 2: Test the origin: $|0 - 5| = 5$, which is greater than 3, so the origin satisfies the inequality.

Step 3: Since we want $|z - 5| \geq 3$, shade the exterior of the circle (including the boundary with a solid line).

Part 2: For the region where $0 < \arg(z - 4i) < \frac{3\pi}{4}$:

Step 1: The boundaries are half-lines from the point $(0, 4)$. One boundary is at angle 0 (the positive real direction) and the other at angle $\frac{3\pi}{4}$.

Step 2: Test the origin: $\arg(0 - 4i) = \arg(-4i) = -\frac{\pi}{2}$, which is outside the required range.

Step 3: Shade the wedge-shaped region between the two half-lines, excluding the boundaries (use dotted lines) and excluding the point $(0, 4)$.

Worked Example: Circle locus to Cartesian form

Find the Cartesian equation of $|z - 3 + 4i| = \sqrt{5}$.

Step 1: Let $z = x + iy$:

$|x + iy - 3 + 4i| = \sqrt{5}$

Step 2: Simplify:

$|(x - 3) + i(y + 4)| = \sqrt{5}$

Step 3: Find the modulus:

$\sqrt{(x - 3)^2 + (y + 4)^2} = \sqrt{5}$

Step 4: Square both sides:

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

This is the equation of a circle with center $(3, -4)$ and radius $\sqrt{5}$.

Worked Example: Half-line locus to Cartesian form

Find the Cartesian equation of $\arg(z + i) = -\frac{\pi}{6}$.

Step 1: Let $z = x + iy$:

$\arg(x + iy + i) = -\frac{\pi}{6}$

$\arg(x + i(y + 1)) = -\frac{\pi}{6}$

Step 2: Since $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$:

$\frac{y + 1}{x} = \tan\left(-\frac{\pi}{6}\right)$

$\frac{y + 1}{x} = -\frac{\sqrt{3}}{3}$

Step 3: Rearranging:

$\sqrt{3}x + 3y + 3 = 0$

Since the locus is a half-line, we only need the part where $x > 0$ and $y < -1$ (based on the angle and starting point).

Exam tip: When finding Cartesian equations of half-lines, remember to state any restrictions on $x$ and $y$ that define which part of the line forms the half-line.