IBDP Maths AHL 1.12 Complex numbers AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) (i) To show \( \arg z = \theta \), express \( z = 1 + \cos 2\theta + i \sin 2\theta \):
Real part: \( x = 1 + \cos 2\theta \), Imaginary part: \( y = \sin 2\theta \).
The argument is \( \arg z = \tan^{-1} \left( \frac{y}{x} \right) = \tan^{-1} \left( \frac{\sin 2\theta}{1 + \cos 2\theta} \right) \). M1
Use double-angle identities: \( \sin 2\theta = 2 \sin \theta \cos \theta \), \( \cos 2\theta = 2 \cos^2 \theta – 1 \).
Then: \( x = 1 + (2 \cos^2 \theta – 1) = 2 \cos^2 \theta \), \( y = 2 \sin \theta \cos \theta \).
\[ \frac{y}{x} = \frac{2 \sin \theta \cos \theta}{2 \cos^2 \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta \]
\[ \arg z = \tan^{-1} (\tan \theta) = \theta \] (since \( -\frac{\pi}{2} < \theta < \frac{\pi}{2} \)). A1
Verify: \( x = 2 \cos^2 \theta > 0 \), and \( y = 2 \sin \theta \cos \theta \) has the sign of \( \sin \theta \), consistent with \( \arg z = \theta \).
[2 marks]
(a) (ii) To show \( |z| = 2 \cos \theta \):
\[ |z| = \sqrt{(1 + \cos 2\theta)^2 + (\sin 2\theta)^2} \] M1
Expand: \( (1 + \cos 2\theta)^2 = 1 + 2 \cos 2\theta + \cos^2 2\theta \), \( (\sin 2\theta)^2 = \sin^2 2\theta \).
\[ |z|^2 = 1 + 2 \cos 2\theta + \cos^2 2\theta + \sin^2 2\theta = 1 + 2 \cos 2\theta + 1 = 2 (1 + \cos 2\theta) \]
Use \( 1 + \cos 2\theta = 2 \cos^2 \theta \):
\[ |z|^2 = 2 \cdot 2 \cos^2 \theta = 4 \cos^2 \theta \]
\[ |z| = \sqrt{4 \cos^2 \theta} = 2 \cos \theta \] (since \( \cos \theta > 0 \) for \( -\frac{\pi}{2} < \theta < \frac{\pi}{2} \)). A1
[2 marks]
(b) Find \( \theta \) such that \( \arg(z^2) = |z|^3 \):
\[ \arg(z^2) = 2 \arg z = 2 \theta \]
\[ |z|^3 = (2 \cos \theta)^3 = 8 \cos^3 \theta \]
Set up: \( 2 \theta = 8 \cos^3 \theta \Rightarrow \theta = 4 \cos^3 \theta \). M1
Define \( f(\theta) = \theta – 4 \cos^3 \theta \). Solve \( f(\theta) = 0 \):
Test values in \( (0, \frac{\pi}{2}) \): \( f(0.9) \approx -0.060 \), \( f(0.95) \approx 0.162 \). A root exists between 0.9 and 0.95.
Numerical approximation gives \( \theta \approx 0.913 \). A1 A1
[3 marks]
Total [7 marks]
Alternative Method for (a) (i) and (ii):
Rewrite \( z \):
\[ z = 1 + 2 \cos^2 \theta – 1 + 2 \sin \theta \cos \theta i = 2 \cos^2 \theta + 2 \sin \theta \cos \theta i \]
\[ z = 2 \cos \theta (\cos \theta + i \sin \theta) = 2 \cos \theta \text{cis} \theta \] M1
Thus: \( |z| = 2 \cos \theta \), \( \arg z = \theta \). A1
▶️ Answer/Explanation
Part A:
(a) For a positive real number \( z > 0 \), we have \( |z| = z \) and \( \arg(z) = 0 \) (since \( z \) lies on the positive real axis). A1
Thus: \( L(z) = \ln |z| + i \arg(z) = \ln z + i \cdot 0 = \ln z \). A1
[2 marks]
(b) (i) For \( z = -1 \):
\[ |z| = |-1| = 1, \quad \arg(-1) = \pi \quad (\text{since } -1 \text{ is on the negative real axis}) \]
\[ L(-1) = \ln 1 + i \pi = 0 + i \pi = i \pi \] A1 A1
[2 marks]
(b) (ii) For \( z = 1 – i \):
\[ |z| = \sqrt{1^2 + (-1)^2} = \sqrt{2}, \quad \arg(1 – i) = \tan^{-1} \left( \frac{-1}{1} \right) = -\frac{\pi}{4} \]
Since \( 0 \leq \arg(z) < 2\pi \), adjust: \( -\frac{\pi}{4} + 2\pi = \frac{7\pi}{4} \).
\[ L(1 – i) = \ln \sqrt{2} + i \frac{7\pi}{4} = \frac{1}{2} \ln 2 + i \frac{7\pi}{4} \] A1 A1
[2 marks]
(b) (iii) For \( z = -1 + i \):
\[ |z| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}, \quad \arg(-1 + i) = \tan^{-1} \left( \frac{1}{-1} \right) = -\frac{\pi}{4} + \pi = \frac{3\pi}{4} \quad (\text{second quadrant}) \]
\[ L(-1 + i) = \ln \sqrt{2} + i \frac{3\pi}{4} = \frac{1}{2} \ln 2 + i \frac{3\pi}{4} \] A1
[1 mark]
(c) To show \( L(z_1 z_2) \neq L(z_1) + L(z_2) \) for some \( z_1, z_2 \):
Use \( z_1 = -1 \), \( z_2 = 1 – i \), so \( z_1 z_2 = (-1)(1 – i) = -1 + i \). M1
From (b): \( L(-1) = i \pi \), \( L(1 – i) = \frac{1}{2} \ln 2 + i \frac{7\pi}{4} \), \( L(-1 + i) = \frac{1}{2} \ln 2 + i \frac{3\pi}{4} \).
Check: \( L(z_1) + L(z_2) = i \pi + \left( \frac{1}{2} \ln 2 + i \frac{7\pi}{4} \right) = \frac{1}{2} \ln 2 + i \left( \pi + \frac{7\pi}{4} \right) = \frac{1}{2} \ln 2 + i \frac{11\pi}{4} \).
Since \( \frac{11\pi}{4} > 2\pi \), adjust the argument: \( \frac{11\pi}{4} – 2\pi = \frac{3\pi}{4} \).
Thus: \( L(z_1) + L(z_2) = \frac{1}{2} \ln 2 + i \frac{3\pi}{4} = L(-1 + i) \).
This suggests the property holds here, so try another pair, e.g., \( z_1 = z_2 = -1 \):
\[ z_1 z_2 = (-1)(-1) = 1, \quad L(1) = \ln 1 + i \cdot 0 = 0 \]
\[ L(-1) + L(-1) = i \pi + i \pi = 2i \pi \neq 0 \] R1
Thus, \( L(z_1 z_2) \neq L(z_1) + L(z_2) \) in general. AG
[2 marks]
Total [9 marks]
Part B:
(a) Given \( f(x + y) = f(x)f(y) \), set \( x = y = 0 \):
\[ f(0 + 0) = f(0)f(0) \Rightarrow f(0) = f(0)^2 \] M1 A1
Since \( f(0) \neq 0 \), \( f(0)^2 = f(0) \Rightarrow f(0)(f(0) – 1) = 0 \Rightarrow f(0) = 1 \). R1
[3 marks]
(b) Method 1: Set \( y = -x \):
\[ f(x – x) = f(x)f(-x) \Rightarrow f(0) = f(x)f(-x) \] M1 A1
Since \( f(0) = 1 \), \( f(x)f(-x) = 1 \Rightarrow f(x) \neq 0 \). R1
Method 2: Suppose \( f(x) = 0 \) for some \( x \):
\[ f(x – x) = f(x)f(-x) \Rightarrow f(0) = 0 \cdot f(-x) = 0 \] M1 A1
This contradicts \( f(0) = 1 \), so \( f(x) \neq 0 \). R1
[3 marks]
(c) Using the derivative definition:
\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) – f(x)}{h} \] M1
\[ f(x + h) = f(x)f(h) \Rightarrow f'(x) = \lim_{h \to 0} \frac{f(x)f(h) – f(x)f(0)}{h} = f(x) \lim_{h \to 0} \frac{f(h) – f(0)}{h} \] A1 A1
\[ \lim_{h \to 0} \frac{f(h) – f(0)}{h} = f'(0) = k \Rightarrow f'(x) = k f(x) \] A1
[4 marks]
(d) Solve \( f'(x) = k f(x) \):
\[ \frac{f'(x)}{f(x)} = k \Rightarrow \int \frac{f'(x)}{f(x)} dx = \int k \, dx \Rightarrow \ln |f(x)| = kx + C \] M1 A1
Since \( f(x) \neq 0 \), \( \ln f(x) = kx + C \).
At \( x = 0 \): \( \ln f(0) = C \Rightarrow C = \ln 1 = 0 \). A1
\[ \ln f(x) = kx \Rightarrow f(x) = e^{kx} \] A1
[4 marks]
Total [14 marks]