IB Mathematics AHL 1.13 Modulus–argument (polar) form AI HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Markscheme
a) (i) Plot on the Argand diagram the point corresponding to \( \theta = \frac{\pi}{2} \):
The complex number is given by: \( z_{\theta} = \frac{1}{2} \times \theta \times e^{\theta i} \).
Using Euler’s formula, \( e^{\theta i} = \cos\theta + i \sin\theta \):
\[ z_{\theta} = \frac{1}{2} \times \theta \times (\cos\theta + i \sin\theta) \]
For \( \theta = \frac{\pi}{2} \):
\[ z_{\frac{\pi}{2}} = \frac{1}{2} \times \frac{\pi}{2} \times \left( \cos\frac{\pi}{2} + i \sin\frac{\pi}{2} \right) \]
\[ \cos\frac{\pi}{2} = 0, \quad \sin\frac{\pi}{2} = 1 \]
\[ z_{\frac{\pi}{2}} = \frac{1}{2} \times \frac{\pi}{2} \times (0 + i \times 1) = \frac{\pi}{4} \times i \]
In the Argand diagram, this corresponds to the point: \( \left(0, \frac{\pi}{4}\right) \approx \left(0, 0.7854\right) \). [1]
The point should be plotted on the positive imaginary axis at approximately \( 0.79 \) units from the origin.
Final Answer: Point at \( \left(0, \frac{\pi}{4}\right) \approx \left(0, 0.79\right) \). [1]
a) (ii) Plot on the Argand diagram the point corresponding to \( \theta = \pi \):
For \( \theta = \pi \):
\[ z_{\pi} = \frac{1}{2} \times \pi \times \left( \cos\pi + i \sin\pi \right) \]
\[ \cos\pi = -1, \quad \sin\pi = 0 \]
\[ z_{\pi} = \frac{1}{2} \times \pi \times (-1 + i \times 0) = -\frac{\pi}{2} \]
In the Argand diagram, this corresponds to the point: \( \left( -\frac{\pi}{2}, 0 \right) \approx \left(-1.5708, 0\right) \). [1]
The point should be plotted on the negative real axis at approximately \( -1.57 \) units from the origin.
Final Answer: Point at \( \left(-\frac{\pi}{2}, 0\right) \approx \left(-1.57, 0\right) \). [1]
a) (iii) Plot on the Argand diagram the point corresponding to \( \theta = \frac{3\pi}{2} \):
For \( \theta = \frac{3\pi}{2} \):
\[ z_{\frac{3\pi}{2}} = \frac{1}{2} \times \frac{3\pi}{2} \times \left( \cos\frac{3\pi}{2} + i \sin\frac{3\pi}{2} \right) \]
\[ \cos\frac{3\pi}{2} = 0, \quad \sin\frac{3\pi}{2} = -1 \]
\[ z_{\frac{3\pi}{2}} = \frac{1}{2} \times \frac{3\pi}{2} \times (0 + i \times (-1)) = -\frac{3\pi}{4} \times i \]
In the Argand diagram, this corresponds to the point: \( \left( 0, -\frac{3\pi}{4} \right) \approx \left(0, -2.3562\right) \). [1]
The point should be plotted on the negative imaginary axis at approximately \( -2.36 \) units from the origin.
Final Answer: Point at \( \left(0, -\frac{3\pi}{4}\right) \approx \left(0, -2.36\right) \). [1]
b) (i) Find the value of \( \theta \) when \( |z_{\theta}| = 4 \):
The modulus of the complex number is: \( |z_{\theta}| = \left| \frac{1}{2} \times \theta \times e^{\theta i} \right| \).
Since \( |e^{\theta i}| = 1 \):
\[ |z_{\theta}| = \frac{1}{2} \times \theta \times 1 = \frac{\theta}{2} \]
We need: \( \frac{\theta}{2} = 4 \). [1]
Solve for \( \theta \): \( \theta = 4 \times 2 = 8 \). [1]
Final Answer: \( \theta = 8 \). [2]
b) (ii) For this value of \( \theta \), plot the approximate position of \( z_{\theta} \) on the Argand diagram:
From part (b)(i), \( \theta = 8 \).
Calculate: \( z_{8} = \frac{1}{2} \times 8 \times e^{8i} = 4 \times e^{8i} = 4 \times (\cos 8 + i \sin 8) \). [1]
The modulus is: \( |z_{8}| = 4 \).
The argument is \( 8 \) radians. Convert to degrees for positioning: \( 8 \times \frac{180}{\pi} \approx 458.366^\circ \).
Reduce modulo \( 360^\circ \): \( 458.366 – 360 = 98.366^\circ \).
Thus, the point lies at a radius of \( 4 \) units and an angle of approximately \( 98.366^\circ \) (or \( 8 \) radians), in the second quadrant (since \( 90^\circ < 98.366^\circ < 180^\circ \)).
Approximate coordinates: \( x = 4 \times \cos 8 \approx 4 \times \cos(98.366^\circ) \approx 4 \times (-0.1392) \approx -0.5568 \).
\[ y = 4 \times \sin 8 \approx 4 \times \sin(98.366^\circ) \approx 4 \times 0.9903 \approx 3.9612 \]. [1]
So, plot the point at approximately \( (-0.56, 3.96) \) on the Argand diagram, on the circle of radius \( 4 \).
Final Answer: Point at approximately \( (-0.56, 3.96) \). [2]
Total: [7 marks]