2016 H2 Mathematics Paper 1 Question 7

Complex Numbers

Answers

(a)

2+3i{2 + 3 \mathrm{i}}

(b)

a=3,k=−30{a=3, k=-30}

Full solutions

(a)

Substituting w=−1+5i{w=- 1 + 5 \mathrm{i}} into the equation
LHS=w2+(−1−8i)w+(−17+7i)=(−1+5i)2+(−1−8i)(−1+5i)+(−17+7i)=(1−10i−25)+(41+3i)+(−17+7i)=0=RHS\begin{align*} \textrm{LHS} &= w^2 + (- 1 - 8 \mathrm{i})w + (- 17 + 7 \mathrm{i}) \\ &= (- 1 + 5 \mathrm{i})^2 + (- 1 - 8 \mathrm{i})(- 1 + 5 \mathrm{i}) + (- 17 + 7 \mathrm{i}) \\ &= (1-10\mathrm{i}-25) + (41 + 3 \mathrm{i}) + (- 17 + 7 \mathrm{i}) \\ &= 0 \\ &= \textrm{RHS} \end{align*}
Hence −1+5i{- 1 + 5 \mathrm{i}} is a root of the equation ■{\blacksquare}
Let α=−1+5i{\alpha=- 1 + 5 \mathrm{i}} and β{\beta} be the second root of the equation
w2+(−1−8i)w+(−17+7i)=(w−α)(w−β)=w2−(α+β)w+(αβ)\begin{align*} & w^2 + (- 1 - 8 \mathrm{i})w + (- 17 + 7 \mathrm{i}) \\ &= (w-\alpha)(w-\beta) \\ &= w^2 -(\alpha+\beta)w + (\alpha\beta) \\ \end{align*}
Comparing coefficients and letting α=−1+5i,{\alpha = - 1 + 5 \mathrm{i},}
−(−1+5i+β)=−1−8iβ=2+3i  ■\begin{gather*} -(- 1 + 5 \mathrm{i}+\beta) = - 1 - 8 \mathrm{i} \\ \beta = 2 + 3 \mathrm{i} \; \blacksquare \end{gather*}

(b)

Since all coefficients are real, by the Conjugate Root Theorem, z=(1+ai)∗=1−ai{z= (1+a\mathrm{i})^* =1-a\mathrm{i}} is also a root
z3−5z2+16z+k=(z−(1+ai))(z−(1−ai))(bz+c)=(z2−2z+a2+1)(bz+c)\begin{align*} & z^3 - 5 z^2 + 16 z + k \\ & = \Big(z-(1+a\mathrm{i})\Big)\Big(z-(1-a\mathrm{i})\Big)(bz+c) \\ & = (z^2-2z+a^2+1)(bz+c) \\ \end{align*}
Comparing coefficients,
z3:b=1z2:c−2b=−5c=−3z1:(a2+1)b−2c=16a=3  ■z0:k=(a2+1)ck=−30  ■\begin{align*} &z^3:&& b = 1 \\ &z^2:&& c-2b = -5 \\ && & c = -3 \\ &z^1:&& (a^2+1)b -2c = 16 \\ && & a = 3 \; \blacksquare \\ &z^0:&& k = (a^2+1)c \\ && & k = -30 \; \blacksquare \\ \end{align*}