IB Mathematics AHL 1.12 Complex numbers AI HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Method 1: Exponential Form
\( z = \cos \theta + i \sin \theta = e^{i\theta} \)
Expression: \( \frac{1 + z}{1 – z} = \frac{1 + e^{i\theta}}{1 – e^{i\theta}} \)
Factorize: \( \frac{e^{i \frac{\theta}{2}} \cdot 2 \cos \frac{\theta}{2}}{e^{i \frac{\theta}{2}} \cdot (-2i \sin \frac{\theta}{2})} = -i \cot \frac{\theta}{2} \)
Real part: \( \text{Re} (-i \cot \frac{\theta}{2}) = 0 \)
Method 2: Trigonometric Form
\( 1 + z = 1 + \cos \theta + i \sin \theta \), \( 1 – z = 1 – \cos \theta – i \sin \theta \)
Rationalize: \( \frac{(1 + \cos \theta + i \sin \theta)(1 – \cos \theta + i \sin \theta)}{(1 – \cos \theta)^2 + \sin^2 \theta} \)
Denominator: \( 2 – 2 \cos \theta = 4 \sin^2 \frac{\theta}{2} \)
Numerator: \( 2i \sin \theta \cos \theta = i \cdot 4 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \)
Result: \( \frac{i \cdot 4 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{4 \sin^2 \frac{\theta}{2}} = i \cot \frac{\theta}{2} \)
Real part: \( \text{Re} (i \cot \frac{\theta}{2}) = 0 \)
Result: \( \text{Re} \left( \frac{1+z}{1-z} \right) = 0 \).
▶️ Answer/Explanation
a
Method 1: Polar Form
Convert: \( -2 + 2i = 2 \sqrt{2} e^{i \frac{3\pi}{4}} \)
Modulus: \( |z| = (2 \sqrt{2})^{1/3} = \sqrt{2} \)
Argument: \( \frac{\frac{3\pi}{4} + 2k\pi}{3} \), \( k = 0, 1, 2 \)
Solutions: \( z_1 = \sqrt{2} e^{i \frac{\pi}{4}} \)
\( z_2 = \sqrt{2} e^{i \frac{11\pi}{12}} \)
\( z_3 = \sqrt{2} e^{i \frac{-5\pi}{12}} \)
Method 2: Cartesian Form
Let \( z = a + bi \): \( (a + bi)^3 = -2 + 2i \)
Equate: \( a^3 – 3a b^2 = -2 \), \( 3a^2 b – b^3 = 2 \)
Test \( a = b \): \( a = 1 \), \( b = 1 \implies z = 1 + i \)
Convert: \( 1 + i = \sqrt{2} e^{i \frac{\pi}{4}} \)
Other roots via polar form: \( z_2, z_3 \)
Result: \( z_1 = \sqrt{2} e^{i \frac{\pi}{4}}, z_2 = \sqrt{2} e^{i \frac{11\pi}{12}}, z_3 = \sqrt{2} e^{i \frac{-5\pi}{12}} \) [6].
b
Convert \( z_1 = \sqrt{2} e^{i \frac{\pi}{4}} \):
\( \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = \sqrt{2} \left( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = 1 + i \)
Verify: \( (1 + i)^3 = -2 + 2i \)
Result: \( 1 + i \) [1].