2014 H2 Mathematics Paper 1 Question 5

Complex Numbers

Answers

(i)

z2=−3+4i{z^2 = - 3 + 4 \mathrm{i}}
1z3=−11125+2125i{{\displaystyle \frac{1}{z^3}} = - \frac{11}{125} + \frac{2}{125} \mathrm{i}}

(ii)

q=−250p{q=- 250p}
pz2+qz3=19p{\displaystyle pz^2+\frac{q}{z^3} = 19p}

Full solutions

(i)

z2=(1+2i)2=1+4i+4i2=−3+4i  ■\begin{align*} z^2 &= (1 + 2 \mathrm{i})^2 \\ &= 1 + 4\mathrm{i} + 4\mathrm{i}^2 \\ &= - 3 + 4 \mathrm{i} \; \blacksquare \end{align*}
1z3=1(1+2i)3=1(−3+4i)(1+2i)=1−11−2i×−11+2i−11+2i=−11+2i112+22=−11125+2125i  ■\begin{align*} \frac{1}{z^3} &= \frac{1}{(1 + 2 \mathrm{i})^3} \\ &= \frac{1}{(- 3 + 4 \mathrm{i})(1 + 2 \mathrm{i})} \\ &= \frac{1}{- 11 - 2 \mathrm{i}} \times \frac{- 11 + 2 \mathrm{i}}{- 11 + 2 \mathrm{i}} \\ &= \frac{- 11 + 2 \mathrm{i}}{11^2 + 2^2} \\ &= - \frac{11}{125} + \frac{2}{125} \mathrm{i} \; \blacksquare \end{align*}

(ii)

pz2+qz3=p(−3+4i)+q(−11125+2125i)=(−3p−11125q)+i(4p+2125q)\begin{align*} & pz^2 + \frac{q}{z^3} \\ & = p(- 3 + 4 \mathrm{i}) + q\left( - \frac{11}{125} + \frac{2}{125} \mathrm{i} \right) \\ &= \left( -3p - \frac{11}{125}q \right) + \mathrm{i}\left( 4p+ \frac{2}{125}q \right) \\ \end{align*}
Since pz2+qz3{pz^2 + \frac{q}{z^3}} is real,
4p+2125q=0q=−250p  ■\begin{gather*} 4p+ \frac{2}{125}q = 0 \\ q = - 250p \; \blacksquare \end{gather*}
pz2+qz3=−3p+11125(−250p)=19p  ■\begin{align*} & pz^2 + \frac{q}{z^3} \\ &= -3p + \frac{11}{125}\left( - 250p \right) \\ &= 19p \; \blacksquare \end{align*}