Question
Consider \(\phi =(a+bi)^{3}\), where \(a, b \in \mathbb{R}\).
(a) In terms of a and b, find
(i) the real part of \(\phi\);
(ii) the imaginary part of \(\phi\).
(b) Hence, or otherwise, show that \((1+\sqrt{3i})^{3}=-8\).
The roots of the equation \(z^{3}=-8\) are u, v, and w, where \(u=1+\sqrt{3i}\) and \(v\in \mathbb{R}\).
(c)Write down v and w, giving your answers in Cartesian form.
On an Argand diagram, u , v and w are represented by the points U , V and W respectively.
(d) Find the area of the triangle UVW .
Each of the points U , V and W is rotated counter-clockwise (anticlockwise) about 0 through an angle of \(\frac{\pi }{4}\) to form three new points U′ , V′ and W′ . These points represent the complex numbers u′ , v′ and w′ respectively.
(e) Find u′ , v′ and w′ , giving your answers in the form \(re^{i\theta }\) , where \(-\pi < \theta \leq \pi \).
(f)Given that u′ , v′ and w′ are the solutions of \(z^{3}=c+di\) , where \(c, d\in \mathbb{R}\) , find the value of c and the value of d .
It is given that u , v , w , u′ , v′ and w′ are all solutions of \(z^{n}=\alpha \) for some \(\alpha \in \mathbb{C}\) , where \(n\in \mathbb{N}\).
(g) Find the smallest positive value of n .
▶️Answer/Explanation
Ans:
(a) attempt to expand the brackets or attempt to find the modulus and argument of \(\phi\)
(b) attempt to substitute \(a=1\) and \(b=\sqrt{3}\) into their real or imaginary part found in (a) OR to expand the brackets OR to use polar form
(c) \(v=-2, w=1-\sqrt{3i}\)
(d) METHOD 1
triangle UVW has height \(h=3\) and base \(b=2\sqrt{3}\)
attempt to find area of triangle with their height and base
\(area=\frac{1}{2}\times 2\sqrt{3}\times 3\)
\(=3\sqrt{3}\) (square units)
METHOD 2
triangle UVW has sides of length \(\left ( \sqrt{3^{2}+\left ( \sqrt{3} \right )^{2}}= \right )\sqrt{12}\)
attempt to find area of equilateral triangle with their side length
METHOD 3
triangle UVO has sides of length \(\left ( \sqrt{1^{2}+\left ( \sqrt{3} \right )^{2}}= \right )2\)
attempt to find area of three isosceles triangles with their side length and angle \(\frac{2\pi }{3}\)
\(area=3\left ( \frac{1}{2}(2)^{2}sin\frac{2\pi }{3} \right )\)
\(=3\sqrt{3}\) (square units)
(e) attempt to express u, v or w in the form \(re^{i\theta }\) and multiply by \(e^{\frac{\pi }{4}i}\)
(f) EITHER
attempt to find one of (u’)3, (v’)3or (w’)3
OR
attempt to find product of their three roots u’, v’ and w’
THEN
(g) METHOD 1
attempt to write the arguments of u, v, w, u’, v’ and w’ over a common denominator OR to write the arguments in degrees
THEN
arguments of u, v, w, u’, v’, and w’ differ by \(\frac{3\pi }{12}\) and \(\frac{5\pi }{12}\) or \(45^{\circ }\) and \(75^{\circ }\)
so arguments of polygon vertices differ by \(\frac{\pi }{12}\) or \(15^{\circ }\)
\(n=24\)
METHOD 2
Let \(z=r cis\theta \Rightarrow z^{n}=r^{n}cis(n\theta )=r^{n}cis(n\theta )\), where \(\theta\) is the argument of u, v, w, u’, v’ and w’.
recognition to find \(n\theta\) where \(n=6,12,18,…\) and \(\theta =-\frac{3\pi }{4}, -\frac{\pi }{3}, -\frac{\pi }{12}, \frac{\pi }{3}, \frac{7\pi }{12}, \pi \)
when \(n=6\Rightarrow (n\theta =)-\frac{9\pi }{2}, -2\pi , \frac{7\pi }{2}, 6\pi \)
when \(n=12\Rightarrow (n\theta =)-9\pi , -4\pi , -\pi , 4\pi ,7\pi ,12\pi \) (which is not a multiple of \(2\pi\))
\(n=24\)
Detailed Solution
(a) To find the real and imaginary parts of
, we need to expand this expression
(b) For
,
, we get
Question:
In the following Argand diagram, the points Z1 , O and Z2 are the vertices of triangle Z1OZ2 described anticlockwise.
The point Z1 represents the complex number z1 = r1eiα , where r1 > 0. The point Z2 represents the complex number z2 = r2eiθ , where r2 > 0.
Angles α, θ are measured anticlockwise from the positive direction of the real axis such that 0 ≤ α, θ < 2π and 0 < α – θ < π.
(a) Show that z1z2 * = r1r2e i(α – θ) where z2* is the complex conjugate of z2 .
▶️Answer/Explanation
Ans: 
Note: Accept working in modulus-argument form
(b) Given that Re (z1 z2*) = 0, show that Z1OZ2 is a right-angled triangle.
In parts (c), (d) and (e), consider the case where Z1OZ2 is an equilateral triangle.
▶️Answer/Explanation
Ans: 
(c) (i) Express z1 in terms of z2 .
▶️Answer/Explanation
Ans: 
Note: Accept working in either modulus-argument form to obtain \(z_{1} = z_{2}\left ( cos\frac{\pi }{3} + isin\frac{\pi }{3} \right )\) or in Cartesian form to obtain \(z_{1} = z_{2}\left ( \frac{1}{2} + \frac{\sqrt{3}}{2} \right )i\)
(ii) Hence show that z12 + z22 = z1 z2 .
Let z1 and z2 be the distinct roots of the equation z2 + az + b = 0 where z ∈ C and a , b ∈ R.
▶️Answer/Explanation
Ans: 
(d) Use the result from part (c)(ii) to show that a2 – 3b = 0.
Consider the equation z2 + az + 12 = 0, where z ∈ C and a ∈ R.
▶️Answer/Explanation
Ans: 
(e) Given that 0 < α – θ < π, deduce that only one equilateral triangle Z1OZ2 can be formed from the point O and the roots of this equation.
▶️Answer/Explanation
Ans:
so (for 0 < α – θ < π), only one equilateral triangle can be formed from point O and the two roots of this equation
Question:
Consider the complex numbers z1 = 1 + bi and z2 = (1 – b2) – 2bi, where b ∈ R, b ≠ 0.
(a) Find an expression for z1z2 in terms of b.
▶️Answer/Explanation
Ans: z1z2 = (1 + bi) ((1-b2) – (2b)i)
= (1-b2 – 2i2b2) + i (-2b + b – b3)
= (1 + b2) + i (-b – b3)
Note: Award A1 for 1+ b2 and A1 for − bi – b3i .
(b) Hence, given that arg (z1 z2) = \(\frac{\pi }{4}\) , find the value of b.
▶️Answer/Explanation
Ans: arg (z1z2) = arctan \(\left ( \frac{-b-b^{3}}{1 + b^{2}} \right )= \frac{\pi }{4}\)
EITHER
arctan (-b) = \(\frac{\pi }{4}\) (since 1+b2 ≠, 0 for b ∈ R)
OR
-b – b3 = 1 + b2 (or equivalent)
THEN
b =−1
Question
Consider \(w = 2\left( {{\text{cos}}\frac{\pi }{3} + {\text{i}}\,{\text{sin}}\frac{\pi }{3}} \right)\)
a. These four points form the vertices of a quadrilateral, Q.
i. Express w2 and w3 in modulus-argument form.[3]
▶️Answer/Explanation
Ans: \({w^2} = 4\text{cis}\left( {\frac{{2\pi }}{3}} \right){\text{;}}\,\,{w^3} = 8{\text{cis}}\left( \pi \right)\) (M1)A1A1
Note: Accept Euler form.
Note: M1 can be awarded for either both correct moduli or both correct arguments.
Note: Allow multiplication of correct Cartesian form for M1, final answers must be in modulus-argument form.
[3 marks]
▶️Answer/Explanation
Ans:
A1A1
[2 marks]
▶️Answer/Explanation
Ans:
use of area = \(\frac{1}{2}ab\,\,{\text{sin}}\,C\) M1
\(\frac{1}{2} \times 1 \times 2 \times {\text{sin}}\frac{\pi }{3} + \frac{1}{2} \times 2 \times 4 \times {\text{sin}}\frac{\pi }{3} + \frac{1}{2} \times 4 \times 8 \times {\text{sin}}\frac{\pi }{3}\) A1A1
Note: Award A1 for \(C = \frac{\pi }{3}\), A1 for correct moduli.
\( = \frac{{21\sqrt 3 }}{2}\) AG
Note: Other methods of splitting the area may receive full marks.
[3 marks]
Let \(z = 2\left( {{\text{cos}}\frac{\pi }{n} + {\text{i}}\,{\text{sin}}\frac{\pi }{n}} \right),\,\,n \in {\mathbb{Z}^ + }\). The points represented on an Argand diagram by \({z^0},\,\,{z^1},\,\,{z^2},\, \ldots \,,\,\,{z^n}\) form the vertices of a polygon \({P_n}\).
▶️Answer/Explanation
Ans:
\(\frac{1}{2} \times {2^0} \times {2^1} \times {\text{sin}}\frac{\pi }{n} + \frac{1}{2} \times {2^1} \times {2^2} \times {\text{sin}}\frac{\pi }{n} + \frac{1}{2} \times {2^2} \times {2^3} \times {\text{sin}}\frac{\pi }{n} + \, \ldots \, + \frac{1}{2} \times {2^{n – 1}} \times {2^n} \times {\text{sin}}\frac{\pi }{n}\) M1A1
Note: Award M1 for powers of 2, A1 for any correct expression including both the first and last term.
\( = {\text{sin}}\frac{\pi }{n} \times \left( {{2^0} + {2^2} + {2^4} + \, \ldots \, + {2^{n – 2}}} \right)\)
identifying a geometric series with common ratio 22(= 4) (M1)A1
\( = \frac{{1 – {2^{2n}}}}{{1 – 4}} \times {\text{sin}}\frac{\pi }{n}\) M1
Note: Award M1 for use of formula for sum of geometric series.
\( = \frac{1}{3}\left( {{4^n} – 1} \right){\text{sin}}\frac{\pi }{n}\) A1
[6 marks]