IBDP Maths AHL 1.13 Modulus-argument in polar form AA HL Paper 1- Exam Style Questions- New Syllabus
Question
(ii) \( \text{Re}\left( \frac{1}{z} \right). \)
Most-appropriate topic codes (IB Mathematics Analysis and Approaches 2021):
• AHL 1.12: Complex numbers; real and imaginary parts — parts (b)(i), (b)(ii)
• AHL 1.13: Modulus–argument and Euler form \( z = re^{i\theta} \); conversion between forms — part (b)
▶️ Answer/Explanation
(a) Express 3 as \( e^a \):
Using the definition of the natural logarithm:
\( 3 = e^{\ln 3} \)
Thus \( a = \ln 3 \).
(b) Express \( z \) in Euler’s form:
\( z = 3^{i-1} = 3^{-1} \cdot 3^i \)
From part (a), \( 3 = e^{\ln 3} \), so:
\( z = e^{-\ln 3} \cdot e^{i \ln 3} \)
\( z = e^{-\ln 3} \left( \cos(\ln 3) + i \sin(\ln 3) \right) \)
(b)(i) Real part of \( z \):
\( \text{Re}(z) = e^{-\ln 3} \cos(\ln 3) \)
Since \( e^{-\ln 3} = \frac{1}{3} \):
\( \text{Re}(z) = \frac{1}{3} \cos(\ln 3) \)
Thus \( p = \frac{1}{3} \), \( q = 3 \).
(b)(ii) Real part of \( \frac{1}{z} \):
\( \frac{1}{z} = 3 \cdot 3^{-i} = 3 \cdot e^{-i \ln 3} \)
\( \frac{1}{z} = 3 \left( \cos(-\ln 3) + i \sin(-\ln 3) \right) \)
\( \text{Re}\left( \frac{1}{z} \right) = 3 \cos(-\ln 3) = 3 \cos(\ln 3) \)
Thus \( p = 3 \), \( q = 3 \).
Final answers:
\( \text{(a) } \boxed{a = \ln 3} \)
\( \text{(b)(i) } \boxed{\frac{1}{3} \cos (\ln 3)} \)
\( \text{(b)(ii) } \boxed{3 \cos (\ln 3)} \)