2015 H2 Mathematics Paper 1 Question 9

Complex Numbers

Answers

(a)

a+13ai,  a−13ai{a+\frac{1}{\sqrt{3}}a\mathrm{i}, \; a-\frac{1}{\sqrt{3}}a\mathrm{i}}

(b)

Out of syllabus

Full solutions

(a)

w2w∗=(a+bi)2a−bi=a2−b2+2abia−bi×a+bia+bi=a3−ab2−2ab2+i(2a2b+a2b−b3)a2+b2=a3−3ab2+i(3a2b−b3)a2+b2\begin{align*} \frac{w^2}{w^*} &= \frac{(a+b\mathrm{i})^2}{a-b\mathrm{i}} \\ &= \frac{a^2-b^2+2ab\mathrm{i}}{a-b\mathrm{i}} \times \frac{a+b\mathrm{i}}{a+b\mathrm{i}} \\ &= \frac{a^3-ab^2-2ab^2 + \mathrm{i}\left( 2a^2b + a^2b - b^3 \right)}{a^2+b^2} \\ &= \frac{a^3-3ab^2 + \mathrm{i}\left( 3a^2b - b^3 \right)}{a^2+b^2} \\ \end{align*}
Given w2w∗{\displaystyle \frac{w^2}{w^*}} is purely imaginary,
a3−3ab2=0a(a2−3b2)=0a=0 (NA)   or   a2−3b2=0b=±13a\begin{gather*} a^3 - 3ab^2 = 0 \\ a(a^2 - 3b^2) = 0 \\ a = 0 \textrm{ (NA)} \; \textrm{ or } \; a^2 - 3b^2 = 0 \\ b = \pm \frac{1}{\sqrt{3}}a \end{gather*}
w=a+13ai   or   w=a−13ai  ■w=a+\frac{1}{\sqrt{3}}a\mathrm{i} \; \textrm{ or } \; w=a-\frac{1}{\sqrt{3}}a\mathrm{i} \; \blacksquare

(b)

Out of syllabus