2013 Preliminary Examinations QuestionCompilation

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 answersin the form x+ iy, where x and y are real. [4]

Solution. Note that the difference s − w is purely imaginary, whichimplies that Re (s) = Re (w). Let s = a+ bi and w = a+ ci, where a, band c are real and a > 0.

b− c = 6 (1)

(a+ bi) (a+ ci) = 10

a2 − bc+ a (b+ c) i = 10

∴ a2 − bc = 10 (2)

a (b+ c) = 0 (3)

∵ a > 0, ∴ b+ c = 0 (4)

Solving (1) and (4) simultaneously, b = 3 and c = −3. Substitutingthese 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 arereal. [2]

Solution. Let u = is and v = iw, which transforms the given equationsinto:

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 thereal parts of u and v here). Thus, u = −3+i and v = 3+i or u = −3−iand v = 3− i.

(b) Find, in the form z = reiθ, the three roots z1, z2 and z3 of the equationz3 = −3 +

√3i where arg (z1) < arg (z2) < arg (z3). Give your answers

in exact form. [3]

Solution.

z3 = −3 +√

3i

z3 = 1212 e

i(5π6

+2kπ), k ∈ Z

z = 1216 e

i

(5π18

+ 2kπ3

)

Letting k = −1, 0, 1, we obtain three distinct solutions:

z1 = 1216 e−

7π18i, z2 = 12

16 e

5π18i, z3 = 12

16 e

17π18

i

The points Z1, Z2 and Z3 represent z1, z2 and z3 respectively. Find thearea of the triangle formed by Z1Z2Z3. [2]

Solution. Note that triangle Z1Z2Z3 is an equilateral triangle made upof three congruent triangles Z1OZ2, Z2OZ3 and Z3OZ1.

Area of triangle Z1OZ2 =1

2·OZ1 ·OZ2 sin∠Z1OZ2

=1

2· 12

16 · 12

16 sin

2π

3

=

√3

4· 12

13 square units

∴ Area of triangle Z1Z2Z3 =3√

3

4· 12

13 square units

2

The constant c is a complex number such that the points representingcz1, cz2 and cz3 form another equilateral triangle which is congruent totriangle Z1Z2Z3, 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 con-gruent. Let c = eiθ, where θ ∈ (−π, π]. Therefore,

arg (cz1) = arg (c) + arg (z1)

= θ − 7π

18

Similarly, arg (cz2) = θ +5π

18

arg (cz3) = θ +17π

18

If one of the vertices of the new equilateral triangle lies on the positivereal axis, then one of the above arguments must equal zero. Hence,

θ =7π

18,−5π

18or − 17π

18and c = e

7π18i, e−

5π18i or e−

17π18

i.

3

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:

|z − 3− 4i| = |z + 4− 3i|

Locus of z

5

(3, 4)(−4, 3)

(10, 0)0

Re

Im

Find

(i) the range of values for arg(√

z − 10). [4]

Solution. The intersections of the perpendicular bisector and the circle

are (0, 0) and (−1, 7). Note that arg(√

z − 10)

=1

2arg (z − 10). With

reference to the Argand diagram drawn:

Maximum of arg (z − 10) = π

Minimum of arg (z − 10) = π − tan−1

[7− 0

10− (−1)

]

4

= π − tan−1 7

11

∴ Range:π

2− 1

2tan−1 7

11≤ arg

(√z − 10

)≤ π

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, indicatingclearly 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) = 2z4 + aiz3 + 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]

(c) One of the roots z1, is such that 0 < arg (z1) <π

2. The locus of points

representing z, where arg (z − z1) = k, passes through the origin. Findthe exact value of k, and the cartesian equation of this locus. [3]

5