2013 H2 Mathematics Paper 1 Question 4

Complex Numbers

Answers

(i)

w3=−11−2i{w^3 = - 11 - 2 \mathrm{i}}

(ii)

a=27,  b=295{a=27, \; b=295}

(iii)

z=1+2i,{z=1 + 2 \mathrm{i}, } z=1−2i{z=1 - 2 \mathrm{i}} or z=−5927{z=- \frac{59}{27}}

Full solutions

(i)

w3=(1+2i)3=(1+2i)2(1+2i)=(1+4i−4)(1+2i)=(−3+4i)(1+2i)=−3+4i−6i−8=−11−2i  ■\begin{align*} w^3 &= (1 + 2 \mathrm{i})^3 \\ &= (1 + 2 \mathrm{i})^2(1 + 2 \mathrm{i}) \\ &= (1+4\mathrm{i}-4)(1 + 2 \mathrm{i}) \\ &= (- 3 + 4 \mathrm{i})(1 + 2 \mathrm{i}) \\ &= - 3 + 4 \mathrm{i} -6\mathrm{i} - 8 \\ &= - 11 - 2 \mathrm{i} \; \blacksquare \end{align*}

(ii)

Substituting z=w{z=w} into the equation,
aw3+5w2+17w+b=0a(−11−2i)+5(−3+4i)+17(1+2i)+b=0(−11a+2+b)+(−2a+54)i=0\begin{gather*} aw^3 + 5w^2 + 17w + b = 0 \\ a(- 11 - 2 \mathrm{i}) + 5(- 3 + 4 \mathrm{i}) + 17(1 + 2 \mathrm{i}) + b = 0 \\ (-11a+2+b) + (-2a+54)\mathrm{i} = 0 \\ \end{gather*}
Comparing real and imaginary parts,
−2a+54=0a=27  ■−11a+2+b=0b=295  ■\begin{gather*} -2a+54 = 0 \\ a = 27 \; \blacksquare \\ -11a + 2 + b = 0 \\ b = 295 \; \blacksquare \end{gather*}

(iii)

Since all coefficients are real, by the conjugate root theorem, w∗=1−2i{w^* = 1 - 2 \mathrm{i}} is also a root
27z3+5z2+17z+295=(z−(1+2i))(z−(1−2i))(cz+d)=((z−1)2−(2i)2)(cz+d)=(z2−2z+5)(cz+d)\begin{align*} & 27 z^3 + 5 z^2 + 17 z + 295 \\ & = \Big(z-(1 + 2 \mathrm{i})\Big)\Big(z-(1 - 2 \mathrm{i})\Big)(cz+d) \\ & = \Big((z-1)^2 - (2\mathrm{i})^2\Big)(cz+d) \\ &= (z^2 - 2 z + 5)(cz+d) \\ \end{align*}
Comparing coefficients,
c=27,  d=2955=59c=27, \; d=\frac{295}{5}=59
27z3+5z2+17z+295=0(z−(1+2i))(z−(1−2i))(27z+59)=0z=1+2i,  z=1−2i   or   z=−5927  ■\begin{gather*} 27 z^3 + 5 z^2 + 17 z + 295 = 0 \\ \Big(z-(1 + 2 \mathrm{i})\Big)\Big(z-(1 - 2 \mathrm{i})\Big)(27z+59) = 0 \\ z= 1 + 2 \mathrm{i}, \; z=1 - 2 \mathrm{i} \; \textrm{ or } \; z=- \frac{59}{27} \; \blacksquare \end{gather*}