2018 H2 Mathematics Paper 2 Question 2

Complex Numbers

Answers

(a)

s=69,t=13{s=69, t=13}
Other roots: 2+3i,{2 + 3 \mathrm{i}, } 12 (repeated){\frac{1}{2} \textrm{ (repeated)} }
(bi)
−32±332i{-\frac{3}{2}\pm\frac{3\sqrt{3}}{2}\mathrm{i}}
(bii)
3,  3e23πi,  3e−23πi{3, \; 3 \mathrm{e}^{\frac{2}{3} \pi \mathrm{i}}, \; 3 \mathrm{e}^{- \frac{2}{3} \pi \mathrm{i}}}
(biii)
Sum =0{=0}
Product =27{=27}

Full solutions

(a)

Since all coefficients are real, by the Conjugate Root Theorem, x=(2−3i)∗=2+3i{x= (2 - 3 \mathrm{i})^* =2 + 3 \mathrm{i}} is also a root
4x4−20x3+sx2−56x+t=(x−(2−3i))(x−(2+3i))(ax2+bx+c)=(x2−4x+13)(ax2+bx+c)\begin{align*} & 4x^4 - 20x^3 + sx^2 - 56x + t \\ & = \Big(x-(2 - 3 \mathrm{i})\Big)\Big(x-(2 + 3 \mathrm{i})\Big)(ax^2+bx+c) \\ & = (x^2 - 4 x + 13)(ax^2+bx+c) \\ \end{align*}
Comparing coefficients,
x4:a=4x3:b−4a=−20b=−4x1:−4c+13b=−56c=1x0:t=13ct=13  ■x2:s=c−4b+13as=69  ■\begin{align*} &x^4:&& a = 4 \\ &x^3:&& b-4a = -20 \\ && & b = -4 \\ &x^1:&& -4c+13b = -56 \\ && & c = 1 \\ &x^0:&& t = 13c \\ && & t = 13 \; \blacksquare \\ &x^2:&& s=c-4b+13a \\ && & s = 69 \; \blacksquare \\ \end{align*}
4x4−20x3+sx2−56x+t=(x2−4x+13)(4x2−4x+1)=(x2−4x+13)(2x−1)2\begin{align*} & 4x^4 - 20x^3 + sx^2 - 56x + t \\ &= (x^2 - 4 x + 13)(4 x^2 - 4 x + 1) \\ &= (x^2 - 4 x + 13)(2 x - 1)^2 \end{align*}
x=2±3i   or   x=12 (repeated)\begin{gather*} x = 2\pm3\mathrm{i} \; \textrm{ or } \; x = \frac{1}{2} \textrm{ (repeated)} \end{gather*}
Other roots:
2+3i,  12 (repeated)  ■2 + 3 \mathrm{i}, \; \frac{1}{2} \textrm{ (repeated)} \; \blacksquare
(bi)
Since w=3{w=3} is a root, (w−3){(w-3)} is a factor of w3−27{w^3-27}
w3−27=(w−3)(Aw2+Bw+C)\begin{align*} & w^3-27 \\ & = (w-3)(Aw^2+Bw+C) \\ \end{align*}
Comparing coefficients,
w3:A=1w0:−3C=−27C=9  ■w2:B−3A=0B=3  ■\begin{align*} &w^3:&& A = 1 \\ &w^0:&& -3C = -27 \\ && & C = 9 \; \blacksquare \\ &w^2:&& B-3A = 0 \\ && & B = 3 \; \blacksquare \\ \end{align*}
w3−27=(w−3)(w2+3w+9)\begin{align*} & w^3-27 \\ &= (w-3)(w^2 + 3 w + 9) \\ \end{align*}
w=−3±32−4(1)(9)2(1)=−3±−272=−32±332i  ■\begin{align*} w &= \frac{-3\pm\sqrt{3^2-4(1)(9)}}{2(1)}\\ &= \frac{- 3\pm\sqrt{- 27}}{2} \\ &= -\frac{3}{2}\pm\frac{3\sqrt{3}}{2}\mathrm{i} \; \blacksquare \end{align*}
(bii)
∣−32+332i∣=(32)2+(332)2=3\begin{align*} & \left| -\frac{3}{2}+\frac{3\sqrt{3}}{2}\mathrm{i} \right | \\ &= \sqrt{\left( \frac{3}{2} \right)^2 + \left(\frac{3\sqrt{3}}{2}\right)^2} \\ &= 3 \end{align*}
basic angle=tan⁡−133232=π3arg⁡(−32+332i)=π−π3=2π3\begin{align*} \textrm{basic angle} &= \tan^{-1} \frac{\frac{3\sqrt{3}}{2}}{\frac{3}{2}} \\ &= \frac{\pi}{3} \\ \arg \left(-\frac{3}{2}+\frac{3\sqrt{3}}{2}\mathrm{i} \right) &= \pi - \frac{\pi}{3} \\ &= \frac{2\pi}{3} \end{align*}
−32+332i=3e23πi  ■-\frac{3}{2}+\frac{3\sqrt{3}}{2}\mathrm{i} = 3 \mathrm{e}^{\frac{2}{3} \pi \mathrm{i}} \; \blacksquare
−32−332i=(3e23πi)∗=3e−23πi  ■\begin{align*}-\frac{3}{2}-\frac{3\sqrt{3}}{2}\mathrm{i} &= \left(3 \mathrm{e}^{\frac{2}{3} \pi \mathrm{i}}\right)^*\\ &= 3 \mathrm{e}^{- \frac{2}{3} \pi \mathrm{i}} \; \blacksquare\end{align*}
3=3  ■3 = 3 \; \blacksquare
(biii)
Sum=3+(−32+332i)+(−32−332i)=0  ■\begin{align*} & \textrm{Sum} \\ &= 3 + \left( -\frac{3}{2}+\frac{3\sqrt{3}}{2}\mathrm{i} \right) + \left( -\frac{3}{2}-\frac{3\sqrt{3}}{2}\mathrm{i} \right) \\ &= 0 \; \blacksquare \end{align*}
Product=(3)(3e23πi)(3e−23πi)=27  ■\begin{align*} & \textrm{Product} \\ &= (3)\left(3 \mathrm{e}^{\frac{2}{3} \pi \mathrm{i}}\right)\left(3 \mathrm{e}^{- \frac{2}{3} \pi \mathrm{i}}\right) \\ &= 27 \; \blacksquare \end{align*}