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Practice Application Task

VCE Specialist Mathematics, Unit 3 - SAC 1
50 marks. 60 minutes working time. CAS permitted.
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Study Mode: This practice Application Task covers complex algebra, complex geometry (Argand diagram and loci), and rational functions. Work through each part with your CAS calculator and textbook nearby. Collapse the solutions and try each question before checking.

Specialist Mathematics Unit 3

SAC 1: Application Task
Unit:Specialist Mathematics Unit 3 Task format:Application Task Working time:60 minutes Reading time:5 minutes Total marks:50 marks CAS calculator:Permitted
Note: This is a practice paper. Your school's SAC may differ in length, marks, and time allocation. Some schools include a tech-free component. The format and difficulty are calibrated to be representative.
Scenario: Skylight Festival - Drone Light Show

A company designs a drone light show for a waterfront festival. The show uses a coordinate system where the main stage is at the origin. Engineers use complex numbers to plan drone positions and flight paths, and rational functions to model signal constraints between ground controllers and drones.

Part A: Drone Positioning (Complex Algebra)

10 marks
Each drone's position is represented as a complex number \(z = x + yi\), where distances are in metres. The show uses complex number operations to choreograph drone movements and formations.
Question 14 marks

Drone A is positioned at \(z_A = -3 + 3i\) metres.

a. Find \(|z_A|\) and \(\arg(z_A)\). [2 marks]
b. Express \(z_A\) in polar form. [1 mark]
c. Drone B is at \(z_B = 4\operatorname{cis}(\pi/6)\). Express \(z_B\) in Cartesian form. [1 mark]
Have you genuinely attempted this question on paper first?
a. [2 marks]
\[|z_A| = \sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\] \[\arg(z_A) = \pi - \arctan(3/3) = \pi - \pi/4 = 3\pi/4\]
\(|z_A| = 3\sqrt{2}\), \(\arg(z_A) = 3\pi/4\)
1 mark: correct modulus. 1 mark: correct argument.
b. [1 mark]
\(z_A = 3\sqrt{2}\operatorname{cis}(3\pi/4)\)
c. [1 mark]
\[z_B = 4\cos(\pi/6) + 4\sin(\pi/6)i = 4 \cdot \frac{\sqrt{3}}{2} + 4 \cdot \frac{1}{2}i = 2\sqrt{3} + 2i\]
\(z_B = 2\sqrt{3} + 2i\)
How did you go? (out of 4)
Question 23 marks

Using \(z_A = 3\sqrt{2}\operatorname{cis}(3\pi/4)\) and \(z_B = 4\operatorname{cis}(\pi/6)\):

a. Find \(z_A \times z_B\) in polar form. [1 mark]
b. Find \(z_A / z_B\) in polar form. [1 mark]
c. Interpret the product \(z_A \times z_B\) geometrically: what single transformation does multiplying by \(z_A\) apply to \(z_B\)? [1 mark]
a. [1 mark]
\[z_A \times z_B = 3\sqrt{2} \cdot 4 \cdot \operatorname{cis}\left(\frac{3\pi}{4} + \frac{\pi}{6}\right) = 12\sqrt{2}\operatorname{cis}\left(\frac{9\pi + 2\pi}{12}\right) = 12\sqrt{2}\operatorname{cis}(11\pi/12)\]
\(z_A \times z_B = 12\sqrt{2}\operatorname{cis}(11\pi/12)\)
b. [1 mark]
\[z_A / z_B = \frac{3\sqrt{2}}{4}\operatorname{cis}\left(\frac{3\pi}{4} - \frac{\pi}{6}\right) = \frac{3\sqrt{2}}{4}\operatorname{cis}\left(\frac{9\pi - 2\pi}{12}\right) = \frac{3\sqrt{2}}{4}\operatorname{cis}(7\pi/12)\]
\(z_A / z_B = \frac{3\sqrt{2}}{4}\operatorname{cis}(7\pi/12)\)
c. [1 mark]
Multiplying by \(z_A = 3\sqrt{2}\operatorname{cis}(3\pi/4)\) scales by a factor of \(3\sqrt{2}\) and rotates by \(3\pi/4\) (135 degrees) anticlockwise about the origin.
How did you go? (out of 3)
Question 33 marks

A formation sequence requires computing powers of complex positions.

a. Express \(z = \sqrt{3} + i\) in polar form. [1 mark]
b. Use De Moivre's theorem to find \(z^6\). [1 mark]
c. Hence show that \((\sqrt{3} + i)^6 = -64\). [1 mark]
a. [1 mark]
\[|z| = \sqrt{3 + 1} = 2, \quad \arg(z) = \arctan(1/\sqrt{3}) = \pi/6\]
\(z = 2\operatorname{cis}(\pi/6)\)
b. [1 mark]
\[z^6 = 2^6\operatorname{cis}(6 \cdot \pi/6) = 64\operatorname{cis}(\pi) = 64(-1 + 0i) = -64\]
\(z^6 = -64\)
c. [1 mark]
From parts (a) and (b), \((\sqrt{3} + i)^6 = z^6 = -64\). QED
How did you go? (out of 3)

Part B: Flight Paths (Complex Geometry)

12 marks
Safety regulations and artistic choreography define specific zones and paths in the complex plane using modulus and locus conditions.
Question 44 marks

Three drones form a triangle with vertices at \(z_1 = 2 + i\), \(z_2 = -1 + 4i\), \(z_3 = 5 + 4i\).

a. Plot the three points on an Argand diagram. [1 mark]
b. Find \(|z_2 - z_1|\), the distance from \(z_1\) to \(z_2\). [1 mark]
c. Find the midpoint of \(z_1\) and \(z_3\). [1 mark]
d. Find \(\arg(z_2 - z_1)\). [1 mark]
Graph space for Argand diagram
a. [1 mark]
Plot points at (2,1), (-1,4), and (5,4) on the Argand diagram.
b. [1 mark]
\[z_2 - z_1 = (-1 + 4i) - (2 + i) = -3 + 3i\] \[|z_2 - z_1| = \sqrt{9 + 9} = 3\sqrt{2}\]
\(|z_2 - z_1| = 3\sqrt{2}\)
c. [1 mark]
\[\frac{z_1 + z_3}{2} = \frac{(2+i) + (5+4i)}{2} = \frac{7 + 5i}{2} = 3.5 + 2.5i\]
Midpoint: \(3.5 + 2.5i\)
d. [1 mark]
\[\arg(-3 + 3i) = \pi - \arctan(3/3) = \pi - \pi/4 = 3\pi/4\]
\(\arg(z_2 - z_1) = 3\pi/4\)
How did you go? (out of 4)
Question 54 marks

A circular flight path is defined by \(|z - 3i| = 3\).

a. Write the Cartesian equation of this circle. [1 mark]
b. Determine whether \(z_0 = 5 + 3i\) lies on, inside, or outside this flight path. [1 mark]
c. Find the shortest distance from \(z_0\) to the flight path. [1 mark]
d. Sketch the locus \(|z - 4| = |z - 2i|\) and describe this locus geometrically. [1 mark]
Graph space for locus sketch
a. [1 mark]
\[|z - 3i| = 3 \Rightarrow |x + yi - 3i| = 3 \Rightarrow |x + (y-3)i| = 3\] \[\Rightarrow x^2 + (y-3)^2 = 9\]
\(x^2 + (y-3)^2 = 9\)
b. [1 mark]
\[|z_0 - 3i| = |5 + 3i - 3i| = |5| = 5 > 3\]
\(z_0\) lies outside the flight path.
c. [1 mark]
Distance from \(z_0\) to centre is 5. Radius is 3. Shortest distance to circle = 5 - 3 = 2 metres.
2 metres
d. [1 mark]
The locus \(|z - 4| = |z - 2i|\) is the perpendicular bisector of the line segment joining \((4, 0)\) and \((0, 2)\). Setting \(|x - 4 + yi| = |x + (y - 2)i|\) and squaring: \((x-4)^2 + y^2 = x^2 + (y-2)^2\), which simplifies to \(2x - y = 3\) or \(y = 2x - 3\).
The locus is the straight line \(y = 2x - 3\), the perpendicular bisector of the segment joining (4,0) and (0,2).
How did you go? (out of 4)
Question 64 marks

The show's main performance zone is defined by \(|z| \leq 6\) and \(\operatorname{Re}(z) \geq 0\).

a. Describe this region in words. [1 mark]
b. Sketch the region on an Argand diagram. [2 marks]
c. Find the area of the performance zone. [1 mark]
Graph space for region sketch
a. [1 mark]
A semicircle (right half) of radius 6 centred at the origin, including the boundary.
b. [2 marks]
The region is bounded by: the imaginary axis (vertical line \(x = 0\)) on the left, and a semicircular arc of radius 6 on the right, extending from (0,-6) to (0,6). Shade the entire right half-disc.
1 mark: correct semicircle shape and size. 1 mark: correct shading and boundary.
c. [1 mark]
Area = half of \(\pi r^2 = \frac{1}{2}\pi(6)^2 = 18\pi\)
\(18\pi\) square metres
How did you go? (out of 4)

Part C: Signal Modelling (Rational Functions)

14 marks
The signal strength between the ground controller and each drone depends on distance and interference patterns, modelled by rational functions.
Question 74 marks

The function \(f(x) = \dfrac{2x + 6}{x - 2}\) models a simplified signal response.

a. Find the x-intercept and y-intercept. [2 marks]
b. State the equations of the vertical and horizontal asymptotes. [2 marks]
a. [2 marks]
x-intercept: set \(2x + 6 = 0 \Rightarrow x = -3\).
y-intercept: set \(x = 0 \Rightarrow f(0) = 6/(-2) = -3\).
x-intercept: \(x = -3\), y-intercept: \(y = -3\)
b. [2 marks]
Vertical asymptote: denominator = 0 at \(x = 2\).
Horizontal asymptote: ratio of leading coefficients = \(2/1 = 2\).
VA: \(x = 2\), HA: \(y = 2\)
How did you go? (out of 4)
Question 85 marks

A signal interference model is given by \(g(x) = \dfrac{x^2 - 9}{x^2 - 4}\).

a. Factorise the numerator and denominator. [1 mark]
b. Find all asymptotes. [2 marks]
c. Find the intercepts and sketch the graph, showing all key features. [2 marks]
Graph space for rational function sketch
a. [1 mark]
\(g(x) = \dfrac{(x-3)(x+3)}{(x-2)(x+2)}\)
b. [2 marks]
Vertical asymptotes: \(x = 2\) and \(x = -2\).
Horizontal asymptote: ratio of leading coefficients (both degree 2) = 1/1 = 1, so \(y = 1\).
VA: \(x = 2, x = -2\). HA: \(y = 1\)
c. [2 marks]
x-intercepts: \(x = 3, x = -3\).
y-intercept: \(g(0) = (-9)/(-4) = 9/4 = 2.25\).
Sketch: symmetric about y-axis, vertical asymptotes at \(x = \pm 2\), horizontal asymptote at \(y = 1\), passes through \((\pm 3, 0)\) and (0, 2.25). Function approaches \(y = 1\) from above as \(x \to \pm\infty\).
1 mark: correct intercepts. 1 mark: correct shape with asymptotes.
How did you go? (out of 5)
Question 95 marks

The actual signal strength at distance \(d\) (in kilometres, \(d > 0\)) is given by \(S(d) = \dfrac{80d}{d^2 + 16}\).

a. Find \(S(4)\). [1 mark]
b. Show that \(d = 4\) is a stationary point of \(S(d)\). [2 marks]
c. Determine the nature of this stationary point and sketch \(S(d)\) for \(d > 0\). [2 marks]
Graph space for signal strength sketch
a. [1 mark]
\[S(4) = \frac{80 \cdot 4}{16 + 16} = \frac{320}{32} = 10\]
\(S(4) = 10\)
b. [2 marks]
Using the quotient rule:
\[S'(d) = \frac{80(d^2 + 16) - 80d \cdot 2d}{(d^2 + 16)^2} = \frac{80d^2 + 1280 - 160d^2}{(d^2 + 16)^2} = \frac{80(16 - d^2)}{(d^2 + 16)^2}\]
At \(d = 4\): \(S'(4) = \frac{80(16 - 16)}{(16 + 16)^2} = 0\). QED
1 mark: correct derivative. 1 mark: correct substitution and conclusion.
c. [2 marks]
For \(d < 4\): \(16 - d^2 > 0\), so \(S'(d) > 0\) (increasing).
For \(d > 4\): \(16 - d^2 < 0\), so \(S'(d) < 0\) (decreasing).
Therefore, \(d = 4\) is a local maximum.
As \(d \to 0^+\), \(S(d) \to 0\). As \(d \to \infty\), \(S(d) \to 0\). Maximum signal strength is \(S(4) = 10\).
1 mark: correct nature (maximum) with justification. 1 mark: correct sketch showing maximum at (4,10), endpoints at origin and approaching 0 as d increases.
How did you go? (out of 5)

Part D: Connecting Topics

14 marks
Advanced modelling requires combining techniques from complex number theory and function analysis.
Question 104 marks

The equation \(z^2 - 6z + 13 = 0\) arises in signal filter design.

a. Show that this equation has no real solutions. [1 mark]
b. Find the two complex solutions. [2 marks]
c. Plot both solutions on an Argand diagram and find the distance between them. [1 mark]
Graph space for Argand diagram
a. [1 mark]
Discriminant: \(\Delta = (-6)^2 - 4(1)(13) = 36 - 52 = -16 < 0\).
Since the discriminant is negative, there are no real solutions.
b. [2 marks]
\[z = \frac{6 \pm \sqrt{-16}}{2} = \frac{6 \pm 4i}{2} = 3 \pm 2i\]
\(z = 3 + 2i\) and \(z = 3 - 2i\)
c. [1 mark]
Points at (3,2) and (3,-2) on the Argand diagram. Distance = \(|(3+2i) - (3-2i)| = |4i| = 4\).
Distance between solutions: 4
How did you go? (out of 4)
Question 115 marks

Five drones are to be positioned at the 5th roots of 32.

a. Express 32 in polar form. [1 mark]
b. Find all five 5th roots of 32, expressing each in polar form. [3 marks]
c. Describe the geometric arrangement of the five drones. [1 mark]
Graph space for 5th roots arrangement
a. [1 mark]
\(32 = 32\operatorname{cis}(0)\)
b. [3 marks]
The 5th roots are \(z = 32^{1/5}\operatorname{cis}(2k\pi/5) = 2\operatorname{cis}(2k\pi/5)\) for \(k = 0, 1, 2, 3, 4\).

\(z_0 = 2\operatorname{cis}(0) = 2\)
\(z_1 = 2\operatorname{cis}(2\pi/5)\)
\(z_2 = 2\operatorname{cis}(4\pi/5)\)
\(z_3 = 2\operatorname{cis}(6\pi/5)\)
\(z_4 = 2\operatorname{cis}(8\pi/5)\)
1 mark: correct modulus (2). 1 mark: correct general form. 1 mark: all 5 roots listed.
c. [1 mark]
The five drones form a regular pentagon of radius 2 metres, centred at the origin, with one vertex on the positive real axis.
How did you go? (out of 5)
Question 125 marks

A drone path height is modelled by \(h(x) = \dfrac{x^2 + kx + 4}{x - 1}\) where \(k\) is a real parameter.

a. State the vertical asymptote. [1 mark]
b. By polynomial long division, show that \(h(x) = x + (k+1) + \dfrac{k+5}{x-1}\). [2 marks]
c. Find the value of \(k\) such that \(h(2) = 10\). [1 mark]
d. For \(k = -5\), determine the exact simplified form of \(h(x)\) and describe what happens to the graph. [1 mark]
a. [1 mark]
\(x = 1\)
b. [2 marks]
Dividing \(x^2 + kx + 4\) by \(x - 1\):
\[x^2 + kx + 4 = (x - 1) \cdot q(x) + r\]
First: \(x^2 / x = x\). Then \(x(x-1) = x^2 - x\).
Remainder: \(x^2 + kx + 4 - (x^2 - x) = (k+1)x + 4\).
Next: \((k+1)x / x = k+1\). Then \((k+1)(x-1) = (k+1)x - (k+1)\).
Final remainder: \((k+1)x + 4 - ((k+1)x - (k+1)) = k + 5\).
Therefore: \(h(x) = x + (k+1) + \dfrac{k+5}{x-1}\)
1 mark: correct division process. 1 mark: complete verification.
c. [1 mark]
\[h(2) = \frac{4 + 2k + 4}{2 - 1} = \frac{8 + 2k}{1} = 8 + 2k = 10\]
\[2k = 2 \Rightarrow k = 1\]
\(k = 1\)
d. [1 mark]
For \(k = -5\): \(h(x) = \dfrac{x^2 - 5x + 4}{x - 1} = \dfrac{(x-1)(x-4)}{x-1} = x - 4\) for \(x \neq 1\).
The graph is the straight line \(y = x - 4\) with a hole (point discontinuity) at \(x = 1\), where \(y = -3\).
How did you go? (out of 5)
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VCE Specialist Mathematics Unit 3. Skylight Festival drone light show context. Axiom Mathematics