IBDP Maths AHL 1.13 Modulus-argument in polar form AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
(i) Expand \(\phi = (a + bi)^3\):
\((a + bi)^2 = a^2 + 2abi – b^2\), so \((a + bi)^3 = (a^2 – b^2 + 2abi)(a + bi) = a^3 + a^2bi – ab^2 + a^2bi + 2a b^2 i – b^3 i^2 = a^3 – 3ab^2 + (3a^2 b – b^3)i\).
Real part: \(a^3 – 3ab^2\).
(ii) Imaginary part: \(3a^2 b – b^3\).
(b) Using (a), for \(a = 1\), \(b = \sqrt{3}\):
Real: \(1^3 – 3 \cdot 1 \cdot (\sqrt{3})^2 = 1 – 9 = -8\).
Imaginary: \(3 \cdot 1^2 \cdot \sqrt{3} – (\sqrt{3})^3 = 3\sqrt{3} – 3\sqrt{3} = 0\).
Thus, \(\phi = (1 + \sqrt{3}\, i)^3 = -8 + 0i = -8\).
(c) Solve \(z^3 = -8 = 8 e^{i\pi}\). Cube roots:
\(z = 2 e^{i(\pi + 2k\pi)/3}\), for \(k = 0, 1, 2\):
– \(k = 0\): \(2 e^{i\pi/3} = 2 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = 1 + i\sqrt{3} = u\).
– \(k = 1\): \(2 e^{i\pi} = -2 = v\).
– \(k = 2\): \(2 e^{i5\pi/3} = 2 \left( \frac{1}{2} – i \frac{\sqrt{3}}{2} \right) = 1 – i\sqrt{3} = w\).
(d) Points: \(U(1, \sqrt{3})\), \(V(-2, 0)\), \(W(1, -\sqrt{3})\). Shoelace formula:
Area = \(\frac{1}{2} \left| 1 (0 – (-\sqrt{3})) + (-2) (-\sqrt{3} – \sqrt{3}) + 1 (\sqrt{3} – 0) \right| = \frac{1}{2} \cdot 6\sqrt{3} = 3\sqrt{3}\).
(e) Rotate by \(e^{i\pi/4}\):
– \(u = 2 e^{i\pi/3}\), \(u’ = 2 e^{i(\pi/3 + \pi/4)} = 2 e^{i7\pi/12}\).
– \(v = 2 e^{i\pi}\), \(v’ = 2 e^{i(\pi + \pi/4)} = 2 e^{i5\pi/4}\).
– \(w = 2 e^{i(-\pi/3)}\), \(w’ = 2 e^{i(-\pi/3 + \pi/4)} = 2 e^{i\pi/12}\).
(f) Compute \((u’)^3 = (2 e^{i7\pi/12})^3 = 8 e^{i7\pi/4} = 8 \left( \cos \frac{\pi}{4} – i \sin \frac{\pi}{4} \right) = 4\sqrt{2} – 4\sqrt{2}i\). Thus, \(c = 4\sqrt{2}\), \(d = -4\sqrt{2}\).
(g) Arguments: \(\frac{\pi}{3}, \pi, -\frac{\pi}{3}, \frac{7\pi}{12}, \frac{5\pi}{4}, \frac{\pi}{12}\). Common denominator (12): \(\frac{4\pi}{12}, \frac{12\pi}{12}, -\frac{4\pi}{12}, \frac{7\pi}{12}, \frac{15\pi}{12}, \frac{1\pi}{12}\). Smallest \(n\) for \(n \cdot \frac{\pi}{12} = 2k\pi\): \(n = 24\).
Markscheme
(a)
Expand \(\phi = (a + bi)^3\): \((a^3 – 3ab^2) + (3a^2 b – b^3)i \quad \mathbf{M1} \)
Real: \(a^3 – 3ab^2\), Imaginary: \(3a^2 b – b^3 \quad \mathbf{A1} \)
(b)
Substitute \(a = 1\), \(b = \sqrt{3}\): \(\phi = -8 \quad \mathbf{M1A1} \)
(c)
Roots: \(v = -2\), \(w = 1 – i\sqrt{3} \quad \mathbf{A1A1} \)
(d)
Shoelace: Area = \(3\sqrt{3} \quad \mathbf{M1A1} \)
(e)
Rotate: \(u’ = 2 e^{i7\pi/12}\), \(v’ = 2 e^{i5\pi/4}\), \(w’ = 2 e^{i\pi/12} \quad \mathbf{M1A1A1} \)
(f)
Compute \((u’)^3\): \(c = 4\sqrt{2}\), \(d = -4\sqrt{2} \quad \mathbf{M1A1} \)
(g)
Arguments differ by \(\frac{\pi}{12}\): \(n = 24 \quad \mathbf{M1A1} \)
[14 marks]
▶️ Answer/Explanation
(a) \( z_2^* = r_2 e^{-i\theta} \), so \( z_1 z_2^* = r_1 e^{i\alpha} \cdot r_2 e^{-i\theta} = r_1 r_2 e^{i(\alpha – \theta)} \).
(b) \(\text{Re} (z_1 z_2^*) = r_1 r_2 \cos(\alpha – \theta) = 0\), so \(\alpha – \theta = \frac{\pi}{2}\) (since \( 0 < \alpha – \theta < \pi \)), making \(\angle Z_1OZ_2 = 90^\circ\).
(c)
(i) For equilateral \( Z_1OZ_2 \), \( z_1 = z_2 e^{i\pi/3} = z_2 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) \).
(ii) \( z_1^2 + z_2^2 = z_2^2 \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) + z_2^2 = z_2^2 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = z_1 z_2 \).
(d) \( (z_1 + z_2)^2 – 2 z_1 z_2 = z_1 z_2 \), so \((-a)^2 = 3b \implies a^2 – 3b = 0\).
(e) For \( z^2 + az + 12 = 0 \), \( a^2 – 36 = 0 \implies a = \pm 6 \).
– \( a = -6 \): Roots \( z_1 = -3 + 3i \), \( z_2 = -3 – 3i \), angles \(\alpha – \theta = \frac{5\pi}{3} – \frac{\pi}{3} = \frac{4\pi}{3} > \pi\) (invalid).
– \( a = 6 \): Roots \( z_1 = 3 + 3i \), \( z_2 = 3 – 3i \), angles \(\frac{\pi}{4}, \frac{7\pi}{4}\), \(\alpha – \theta = \frac{7\pi}{12} – \frac{\pi}{4} = \frac{\pi}{3} < \pi\) (valid).
Only one pair satisfies \( 0 < \alpha – \theta < \pi \).
Markscheme
(a)
Conjugate and product \(\quad \mathbf{M1A1} \)
Show equality \(\quad \mathbf{A1} \)
(b)
Real part = 0 \(\quad \mathbf{M1} \)
Right angle \(\quad \mathbf{A1A1} \)
(c)
(i) \( z_1 = z_2 e^{i\pi/3} \quad \mathbf{M1A1} \)
(ii) Prove equality \(\quad \mathbf{M1A1} \)
(d)
Derive \( a^2 – 3b = 0 \quad \mathbf{M1A1} \)
(e)
Solve and check angles \(\quad \mathbf{M1} \)
One triangle \(\quad \mathbf{A1A1} \)
[14 marks]
▶️ Answer/Explanation
(a) \[ \begin{align*} z_1z_2 &= (1 + bi)\left((1 – b^2) – 2bi\right) \\ &= (1)(1 – b^2) + (1)(-2b)i + (bi)(1 – b^2) + (bi)(-2bi) \\ &= (1 – b^2) – 2bi + bi – b^3i – 2b^2i^2 \\ &= (1 – b^2 + 2b^2) + (-2b + b – b^3)i \\ &= (1 + b^2) + (-b – b^3)i \end{align*} \]
(b) Given arg(z1z2) = \(\frac{\pi}{4}\): \[ \tan\left(\frac{\pi}{4}\right) = \frac{\text{Im}(z_1z_2)}{\text{Re}(z_1z_2)} = \frac{-b – b^3}{1 + b^2} = 1 \] EITHER \[ \frac{-b(1 + b^2)}{1 + b^2} = 1 \implies -b = 1 \implies b = -1 \] OR \[ -b – b^3 = 1 + b^2 \implies b^3 + b^2 + b + 1 = 0 \implies (b + 1)(b^2 + 1) = 0 \] Since b ∈ ℝ, the only solution is b = -1. THEN The required value is \(\boxed{-1}\)
▶️ Answer/Explanation
a. i. Using De Moivre’s Theorem:
\[ w^2 = 2^2 \text{cis}\left(2 \times \frac{\pi}{3}\right) = 4\text{cis}\left(\frac{2\pi}{3}\right) \] \[ w^3 = 2^3 \text{cis}\left(3 \times \frac{\pi}{3}\right) = 8\text{cis}(\pi) \]Thus:
\[ \boxed{w^2 = 4\text{cis}\left(\frac{2\pi}{3}\right)} \quad \text{and} \quad \boxed{w^3 = 8\text{cis}(\pi)} \]a. ii. The points \(w^0 = 1\), \(w^1 = 2\text{cis}\left(\frac{\pi}{3}\right)\), \(w^2 = 4\text{cis}\left(\frac{2\pi}{3}\right)\), and \(w^3 = 8\text{cis}(\pi)\) form a quadrilateral on the Argand diagram:
The sketch shows these four points with their correct moduli and arguments.
b. The area of quadrilateral \(Q\) can be calculated using the formula for the area of triangles formed by consecutive points:
\[ \text{Area} = \frac{1}{2} \times 1 \times 2 \times \sin\frac{\pi}{3} + \frac{1}{2} \times 2 \times 4 \times \sin\frac{\pi}{3} + \frac{1}{2} \times 4 \times 8 \times \sin\frac{\pi}{3} \] \[ = \frac{\sqrt{3}}{2} + 2\sqrt{3} + 8\sqrt{3} = \frac{21\sqrt{3}}{2} \]Thus, the area is:
\[ \boxed{\frac{21\sqrt{3}}{2}} \]c. The area of polygon \(P_n\) is the sum of areas of triangles formed by consecutive points \(z^k\) and \(z^{k+1}\):
\[ \text{Area} = \sum_{k=0}^{n-1} \frac{1}{2} \times 2^k \times 2^{k+1} \times \sin\frac{\pi}{n} = \sin\frac{\pi}{n} \sum_{k=0}^{n-1} 2^{2k} \]The sum is a geometric series with first term 1 and common ratio 4:
\[ \sum_{k=0}^{n-1} 4^k = \frac{4^n – 1}{3} \]Thus, the area is:
\[ \boxed{\frac{1}{3}(4^n – 1)\sin\frac{\pi}{n}} \]