2013 Preliminary Examinations Question Compilation Complex Numbers
1. (2013/AJC/Prelim/P1/Q11) (a) The complex numbers s and w satisfy s − w = 6i and sw = 10. Given that Re (s) > 0, solve the equations for s and w, giving all answers in the form x + iy, where x and y are real. [4] Solution. Note that the difference s − w is purely imaginary, which implies that Re (s) = Re (w). Let s = a + bi and w = a + ci, where a, b and c are real and a > 0. b−c=6 (a + bi) (a + ci) = 10 a2 − bc + a (b + c) i = 10 ∴ a2 − bc = 10 a (b + c) = 0 ∵ a > 0, ∴ b + c = 0
(1)
(2) (3) (4)
Solving (1) and (4) simultaneously, b = 3 and c = −3. Substituting these values into (2), a2 + 9 = 10 a2 = 1 a = 1 ( ∵ a > 0) Hence, s = 1 + 3i and w = 1 − 3i. 1
Hence find the solution to the following equations u − v = −6 and uv = −10. Give your answers for u and v in the form x + iy, where x and y are real. [2] Solution. Let u = is and v = iw, which transforms the given equations into: s − w = 6i and sw = 10, which implies s = 1+ 3i and w = 1 − 3i, or s = −1 +3i and w = −1 − 3i (the second pair of values is required as there is no restriction to the real parts of u and v here). Thus, u = −3+i and v = 3+i or u = −3−i and v = 3 − i. (b) Find, in the√form z = reiθ , the three roots z1 , z2 and z3 of the equation z 3 = −3 + 3i where arg (z1 ) < arg (z2 ) < arg (z3 ). Give your answers in exact form. [3] Solution. z 3 = −3 +
√
3i
1 i 5π +2kπ ,k 12 2 e 6 1 i 5π + 2kπ 3 12 6 e 18
z3 = z=
∈Z
Letting k = −1, 0, 1, we obtain three distinct solutions: 1
7π
1
5π
1
z1 = 12 6 e− 18 i , z2 = 12 6 e 18 i , z3 = 12 6 e
17π i 18
The points Z1 , Z2 and Z3 represent z1 , z2 and z3 respectively. Find the area of the triangle formed by Z1 Z2 Z3 . [2] Solution. Note that triangle Z1 Z2 Z3 is an equilateral triangle made up of three congruent triangles Z1 OZ2 , Z2 OZ3 and Z3 OZ1 . 1 · OZ1 · OZ2 sin ∠Z1 OZ2 2 1 1 1 2π = · 12 6 · 12 6 sin 2√ 3 1 3 = · 12 3 square units 4√ 1 3 3 ∴ Area of triangle Z1 Z2 Z3 = · 12 3 square units 4 Area of triangle Z1 OZ2 =
2
The constant c is a complex number such that the points representing cz1 , cz2 and cz3 form another equilateral triangle which is congruent to triangle Z1 Z2 Z3 , and one of its vertices lies on the positive real axis. Find a suitable value for the complex constant c in exponential form. [2] Solution. Note that |c| = 1, since the two equilateral triangles are congruent. Let c = eiθ , where θ ∈ (−π, π]. Therefore, arg (cz1 ) = arg (c) + arg (z1 ) 7π =θ− 18 5π Similarly, arg (cz2 ) = θ + 18 17π arg (cz3 ) = θ + 18 If one of the vertices of the new equilateral triangle lies on the positive real axis, then one of the above arguments must equal zero. Hence, 7π 5π 17π 7π 5π 17π θ= ,− or − and c = e 18 i , e− 18 i or e− 18 i . 18 18 18
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2. (2013/AJC/Prelim/P2/Q3) The complex number z satisfies the following conditions: (a) |z − 3 − 4i| ≤ 5, (b) |z − 3 − 4i| ≥ |z + 4 − 3i|. On a single Argand diagram, sketch the locus of the points representing z. [4] Solution. The Argand diagram is show below: Im Locus of z 5 (3, 4) (−4, 3) Re 0
(10, 0) |z − 3 − 4i| = |z + 4 − 3i|
Find (i) the range of values for arg
√ z − 10 .
[4]
Solution. The intersections of the perpendicular bisector and the circle 1 √ are (0, 0) and (−1, 7). Note that arg z − 10 = arg (z − 10). With 2 reference to the Argand diagram drawn: Maximum of arg (z − 10) = π −1
Minimum of arg (z − 10) = π − tan
4
7−0 10 − (−1)
= π − tan−1 ∴ Range:
7 11
√ π 7 π 1 − tan−1 ≤ arg z − 10 ≤ 2 2 11 2
(ii) the least value of |iz + 7 + i|.
[2]
Solution. Taking out the factor i from the expression, |iz + 7 + i| = |i| |z + 1 − 7i| = (1) |z − (−1 + 7i)| = |z − (−1 + 7i)| Since the point (−1, 7) lies in the locus of z, the least value of |iz + 7 + i| would have to be 0. 3. (2013/ACJC/Prelim/P1/Q3) The complex number z satisfies the relations |z − 25| ≤ 15 and |z − 25| = |z − 35 − 20i|. (a) Illustrate both of these relations on a single Argand diagram, indicating clearly the intersection of the two loci. [3] (b) Find the greatest value of arg (z − 25).
[2]
4. (2013/ACJC/Prelim/P2/Q2) The polynomial P (z) = 2z 4 + aiz 3 + 2z + ai, a ∈ R, has factor 2z − i. (a) Find the exact value of a.
[2]
(b) Solve the equation P (z) = 0, leaving your answers in the form reiθ , where r > 0 and −π < θ ≤ π. [4] π (c) One of the roots z1 , is such that 0 < arg (z1 ) < . The locus of points 2 representing z, where arg (z − z1 ) = k, passes through the origin. Find the exact value of k, and the cartesian equation of this locus. [3]
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