2017 H2 Mathematics Paper 1 Question 8

Complex Numbers

Answers

(a)

−1+2i,  2−i{- 1 + 2 \mathrm{i}, \; 2 - \mathrm{i}}
(bi)
ω2=−2i,{\omega^2 = - 2 \mathrm{i},}
ω3=−2−2i,{\omega^3 = - 2 - 2 \mathrm{i},}
ω4=−4.{\omega^4 = - 4.}
p=−6,q=−66{p=-6, q=-66}
(bii)
(ω2−2ω+2)(ω2−4ω+29){(\omega^2 - 2 \omega + 2)}\allowbreak{(\omega^2 - 4 \omega + 29)}

Full solutions

(a)

z=−b±b2−4ac2a=−(−2)±(−2)2−4(1−i)(5+5i)2(1−i)=2±4−4(10)2−2i=2±−362−2i=2±6i2−2i×2+2i2+2i=4+12i+4i−1222+22   or   4−12i+4i+1222+22=−1+2i   or   2−i  ■\begin{align*} z &= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ &= \frac{-(- 2)\pm\sqrt{(- 2)^2-4(1 - \mathrm{i})(5 + 5 \mathrm{i})}}{2(1 - \mathrm{i})}\\ &= \frac{2\pm\sqrt{4-4(10)}}{2 - 2 \mathrm{i}}\\ &= \frac{2\pm\sqrt{- 36}}{2 - 2 \mathrm{i}}\\ &= \frac{2\pm6\mathrm{i}}{2 - 2 \mathrm{i}} \times \frac{2 + 2 \mathrm{i}}{2 + 2 \mathrm{i}} \\ &= \frac{4+12\mathrm{i}+4\mathrm{i}-12}{2^2+2^2} \; \textrm{ or } \; \frac{4-12\mathrm{i}+4\mathrm{i}+12}{2^2+2^2} \\ &= - 1 + 2 \mathrm{i} \; \textrm{ or } \; 2 - \mathrm{i} \; \blacksquare \end{align*}
(bi)
ω2=(1−i)2=1−2i+i2=−2i  ■ω3=−2i(1−i)=−2i+2i2=−2−2i  ■ω4=(−2i)2=−4  ■\begin{align*} \omega^2 &= (1 - \mathrm{i})^2 \\ &= 1 - 2 \mathrm{i} + \mathrm{i}^2 \\ &= - 2 \mathrm{i} \; \blacksquare \\ \omega^3 &= - 2 \mathrm{i}(1 - \mathrm{i}) \\ &= -2 \mathrm{i} + 2 \mathrm{i}^2 \\ &= - 2 - 2 \mathrm{i} \; \blacksquare \\ \omega^4 &= (- 2 \mathrm{i})^2 \\ &= - 4 \; \blacksquare \end{align*}
Substituting them into the equation,
−4+p(−2−2i)+39(−2i)+q(1−i)+58=0(2p+q+54)+(−2p−78−q)i=0\begin{gather*} - 4 + p(- 2 - 2 \mathrm{i}) + 39(- 2 \mathrm{i}) + q(1 - \mathrm{i}) + 58 = 0 \\ (2p+q+54) + (-2p-78-q)\mathrm{i} = 0 \end{gather*}
Comparing real and imaginary parts,
−2p+q=−54−2p−q=78\begin{align} -2p + q &= -54 \\ -2p - q &= 78 \end{align}
Solving (1){(1)} and (2){(2)} simultaneously,
p=−6  ■q=−66  ■\begin{align*} p &= -6 \; \blacksquare \\ q &= -66 \; \blacksquare \end{align*}
(bii)
Since all coefficients are real, by the Conjugate Root Theorem, ω=(1−i)∗=1+i{\omega= (1 - \mathrm{i})^* =1 + \mathrm{i}} is also a root
ω4−6ω3+39ω2−66ω+58=(ω−(1−i))(ω−(1+i))(aω2+bω+c)=(ω2−2ω+2)(aω2+bω+c)\begin{align*} & \omega^4 - 6 \omega^3 + 39\omega^2 - 66 \omega+58 \\ & = \Big(\omega-(1 - \mathrm{i})\Big)\Big(\omega-(1 + \mathrm{i})\Big)(a\omega^2+b\omega+c) \\ & = (\omega^2 - 2 \omega + 2)(a\omega^2+b\omega+c) \\ \end{align*}
Comparing coefficients,
ω4:a=1ω0:2c=58c=29ω3:b−2a=−6b=−4\begin{align*} &\omega^4:&& a = 1 \\ &\omega^0:&& 2c = 58 \\ && & c = 29 \\ &\omega^3:&& b -2a = -6 \\ && & b = -4 \\ \end{align*}
ω4+pω3+39ω2+qω+58=(ω2−2ω+2)(ω2−4ω+29)  ■\begin{align*} & \omega^4 + p\omega^3 + 39\omega^2 + q\omega+58 \\ &= (\omega^2 - 2 \omega + 2)(\omega^2 - 4 \omega + 29) \; \blacksquare \end{align*}