IB Mathematics AHL 1.13 Modulus–argument (polar) form AI HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)(i)
Point at \( (1, -1) \) on Argand diagram.
(A1)
(a)(ii)
Calculate modulus \( a = |z| \):
\( a = \sqrt{1^2 + (-1)^2} = \sqrt{2} \) (A1)
Calculate argument \( b = \arg(z) \):
\( b = -\frac{\pi}{4} \) (or \( \frac{7\pi}{4} \)) (A1)
So \( z = \sqrt{2} e^{-i\frac{\pi}{4}} \)
Result:
\( z = \sqrt{2} e^{-i\frac{\pi}{4}} \)
(b)(i)
Express \( w_1 + w_2 \):
\( w_1 + w_2 = e^{i x} + e^{i(x – \frac{\pi}{2})} \) (M1)
Factor out \( e^{i x} \):
\( = e^{i x}(1 + e^{-i\frac{\pi}{2}}) \)
Simplify \( e^{-i\frac{\pi}{2}} = \cos\left(-\frac{\pi}{2}\right) + i \sin\left(-\frac{\pi}{2}\right) = 0 – i = -i \):
\( = e^{i x}(1 – i) \) (A1)
Result:
\( e^{i x}(1 – i) \)
(b)(ii)
Rewrite \( w_1 + w_2 \) in polar form:
\( w_1 + w_2 = e^{i x}(1 – i) \)
Modulus of \( 1 – i \): \( \sqrt{1^2 + (-1)^2} = \sqrt{2} \)
Argument of \( 1 – i \): \( \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4} \)
So \( 1 – i = \sqrt{2} e^{-i\frac{\pi}{4}} \)
\( w_1 + w_2 = e^{i x} \times \sqrt{2} e^{-i\frac{\pi}{4}} = \sqrt{2} e^{i(x – \frac{\pi}{4})} \) (M1)(A1)
Real part of \( \sqrt{2} e^{i(x – \frac{\pi}{4})} \):
\( \text{Re}(w_1 + w_2) = \sqrt{2} \cos(x – \frac{\pi}{4}) \) (M1)(A1)
Result:
\( \sqrt{2} \cos(x – \frac{\pi}{4}) \)
(c)(i)
Express total current \( I_T \):
\( I_A = 12 \cos(bt) \), \( I_B = 12 \cos\left(bt – \frac{\pi}{2}\right) \)
\( I_T = I_A + I_B = 12 \cos(bt) + 12 \cos\left(bt – \frac{\pi}{2}\right) \) (M1)
Using result from part (b), where amplitude is 12:
Maximum value = \( 12 \times \sqrt{2} \) (A1)
\( = 12\sqrt{2} \) (≈ 16.97)
Result:
\( 12\sqrt{2} \)
(c)(ii)
Phase shift from \( \cos\left(bt – \frac{\pi}{4}\right) \):
Phase shift = \( \frac{\pi}{4} \) (A1)
Result:
\( \frac{\pi}{4} \)