2020 H2 Mathematics Paper 1 Question 6

Complex Numbers

Answers

t=−8,  k=2{t=-8, \; k=2}
Other root =−25+65i{= - \frac{2}{5} + \frac{6}{5} \mathrm{i}}

Full solutions

Substituting z=k+ki{z=k+k\mathrm{i}} into the equation
(k+ki)2(2+i)−8i(k+ki)+t=0(k2+2k2i−k2)(2+i)−8ki+8k+t=04k2i−2k2−8ki+8k+t=0(−2k2+8k+t)+i(4k2−8k)=0\begin{gather*} (k+k\mathrm{i})^2(2 + \mathrm{i}) - 8 \mathrm{i}(k+k\mathrm{i}) + t = 0 \\ (k^2+2k^2\mathrm{i}-k^2)(2 + \mathrm{i}) -8k\mathrm{i} +8k +t = 0\\ 4k^2\mathrm{i} - 2k^2 -8k\mathrm{i} +8k + t = 0 \\ (-2k^2+8k+t) + \mathrm{i}(4k^2-8k) = 0 \end{gather*}
Comparing imaginary parts,
4k2−8k=04k(k−2)=0\begin{align*} 4k^2 - 8k &= 0 \\ 4k(k-2) &= 0 \end{align*}
Since k≠0,{k \neq 0, }
k=2  ■k=2 \; \blacksquare
Comparing real parts,
−2k2+8k+t=0-2k^2 + 8k + t = 0
t=2(2)2−8(2)=−8  ■\begin{align*} t &= 2(2)^2-8(2) \\ &= -8 \; \blacksquare \end{align*}
Let α=k+ki{\alpha=k+k\mathrm{i}} and β{\beta} be the second root of the equation
z2(2+i)−8iz−8=(2+i)(w−α)(w−β)=(2+i)(w2−(α+β)w+(αβ))\begin{align*} & z^2(2 + \mathrm{i}) - 8 \mathrm{i}z -8 \\ &= (2 + \mathrm{i})(w-\alpha)(w-\beta) \\ &= (2 + \mathrm{i})\Big(w^2 -(\alpha+\beta)w + (\alpha\beta)\Big) \\ \end{align*}
Comparing the constants and letting α=k+ki,{\alpha = k+k\mathrm{i},}
(2+i)αβ=−8(2+i)(2+2i)β=−8(2+6i)β=−8\begin{align*} (2 + \mathrm{i})\alpha\beta &= -8 \\ (2 + \mathrm{i})(2 + 2 \mathrm{i})\beta &= -8 \\ (2 + 6 \mathrm{i})\beta &= -8 \\ \end{align*}
β=−82+6i×2−6i2−6i=−16+48i22+62=−25+65i  ■\begin{align*} \beta &= \frac{-8}{2 + 6 \mathrm{i}} \times \frac{2 - 6 \mathrm{i}}{2 - 6 \mathrm{i}} \\ &= \frac{- 16 + 48 \mathrm{i}}{2^2 + 6^2} \\ &= - \frac{2}{5} + \frac{6}{5} \mathrm{i} \; \blacksquare \end{align*}