IB Mathematics SL 1.3 Geometric sequences and series AI HL Paper 2- Exam Style Questions- New Syllabus

Question

A geometric fractal structure is represented on an Argand diagram.

The construction of the fractal follows a specific rule: An initial vertical line segment from the origin \(0\) to \(5i\) undergoes two rotations of \(\frac{\pi}{12}\) radians about the origin—one clockwise and one anti-clockwise—to create the first pair of branches. The endpoint of one of these initial branches is labeled as \(A\), as illustrated in Diagram 1.

(a) Determine the complex number corresponding to point \(A\), expressing the result in exponential form.

In subsequent stages, two smaller branches are appended to the tip of every existing branch. Each new segment is scaled to 70% of the length of its parent branch. For every pair, one branch is rotated \(\frac{\pi}{12}\) clockwise and the other \(\frac{\pi}{12}\) anti-clockwise relative to the orientation of the previous segment. This growth pattern is shown in Diagram 2.

(b) (i) Demonstrate that the cumulative length of all line segments in the fractal after the second stage (Diagram 2) totals 24 units.
(ii) Calculate the total length of the fractal structure after 10 full stages of growth.
(iii) Determine whether the total length of the branches approaches a finite limit if the process continues indefinitely. Justify your conclusion.

Consider the complex number \(b\), which represents point \(B\) in Diagram 3, located at the extremity of a specific third-stage branch originating from \(A\).

(c) (i) Express \(b\) as the sum of three complex numbers, each written in exponential form (\(re^{i\theta}\)).
(ii) Hence, find the coordinates of \(b\) in Cartesian form, \(a + bi\), where \(a, b \in \mathbb{R}\).

Most-appropriate topic codes:

• TOPIC AHL 1.13: Complex numbers in polar and exponential form — part (a)
• TOPIC SL 1.3: Geometric sequences and series — part (b)
• TOPIC AHL 1.13: Geometric interpretation of complex numbers — part (c)

▶️ Answer/Explanation

Detailed solution

(a)
The initial segment on the imaginary axis has an argument of \(\frac{\pi}{2}\).
Point A is formed by a clockwise rotation of \(\frac{\pi}{12}\).
Argument \(= \frac{\pi}{2} – \frac{\pi}{12} = \frac{5\pi}{12}\).
Modulus is 5.
\(A = \mathbf{5e^{\frac{5\pi}{12}i}}\).

(b)
(i) Stage 1: 2 branches of length 5. Total = \(10\).
Stage 2: 4 branches of length \(5 \times 0.7 = 3.5\). Total = \(14\).
Total Length \(= 10 + 14 = \mathbf{24}\) (AG).
(ii) This is a geometric series where \(u_1 = 10\) and the common ratio \(r = 2 \times 0.7 = 1.4\).
\(S_{10} = \frac{10(1.4^{10} – 1)}{1.4 – 1} \approx \mathbf{698}\).
(iii) Since \(|r| = 1.4 > 1\), the geometric series diverges. There is no finite limit.

(c)
(i) To find \(b\), we sum the vectors for each stage of the branch leading to \(B\):
Correct moduli: \(5\), \(5 \times 0.7 = 3.5\), \(5 \times 0.7^2 = 2.45\).
Arguments are calculated by adding/subtracting \(\frac{\pi}{12}\) at each junction: \(\frac{5\pi}{12}\), \(\frac{4\pi}{12}\) (or \(\frac{\pi}{3}\)), and \(\frac{5\pi}{12}\).
\(b = 5e^{\frac{5\pi}{12}i} + 3.5e^{\frac{\pi}{3}i} + 2.45e^{\frac{5\pi}{12}i}\).
(Using 3 s.f. decimals: \(b = 5e^{1.31i} + 3.5e^{1.05i} + 2.45e^{1.31i}\)).

(ii) Converting to Cartesian form:
\(b = (3.67820… + 10.2272…i)\).
\(b = \mathbf{3.68 + 10.2i}\).
Note: \(3.66 + 10.2i\) is also acceptable if 3 s.f. values were used during calculation.

IB Mathematics SL 1.3 Geometric sequences and series AI HL Paper 2- Exam Style Questions- New Syllabus

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