2019 H2 Mathematics Paper 1 Question 1

Complex Numbers

Answers

b=−a{b=- a}
c=−7a{c=- 7 a}
d=15a{d=15 a}

Full solutions

Since a,b,c,d{a,b,c,d} are real numbers, by the conjugate root theorem, z=2−i{z=2 - \mathrm{i}} is the third root of the equation
az3+bz2+cz+d=a(z−(2+i))(z−(2−i))(z+3)=a((z−2)2−i2)(z+3)=a(z2−4z+5)(z+3)=a(z3−4z2+5z+3z2−12z+15)=a(z3−z2−7z+15)\begin{align*} & az^3 + bz^2 + cz + d \\ & = a \Big(z-(2 + \mathrm{i})\Big)\Big(z-(2 - \mathrm{i})\Big)(z + 3) \\ & = a\Big((z-2)^2 - \mathrm{i}^2\Big)(z + 3) \\ & = a(z^2 - 4 z + 5)(z + 3) \\ & = a(z^3 - 4 z^2 + 5 z + 3 z^2 - 12 z + 15) \\ & = a(z^3 - z^2 - 7 z + 15) \end{align*}
Comparing coefficients,
z2:b=−a  ■z:c=−7a  ■z0:d=15a  ■\begin{align*} &z^2:& b &=- a \; \blacksquare \\ &z:& c &=- 7 a \; \blacksquare \\ &z^0:& d &=15 a \; \blacksquare \\ \end{align*}