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Original question & working

Eliminate with complex coefficients · source reference

[2017/MJC/Prelim/I/3] · Question pp. 1 · Working pp. 1

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Question

Do not use a calculator in answering this question. Showing your working, find the complex numbers zz and ww which satisfy the simultaneous equations

4iz−3w=1+5i,2z+(1+i)w=2+6i.4iz-3w=1+5i,\qquad2z+(1+i)w=2+6i.

[5]

[2017/MJC/Prelim/I/3]

Source working

Printed working — elimination

4iz−3w=1+5i(1)2z+(1+i)w=2+6i(2)(2)×2i:4iz+2i(1+i)w=2i(2+6i)4iz+2iw−2w=4i−12(3)(3)−(1):4iz+2iw−2w−(4iz−3w)=(4i−12)−(1+5i)w+2iw=−13−i(1+2i)w=−13−iw=−13−i1+2i1−2i1−2i=−13+26i−i−212−(2i)2=−15+25i5=−3+5i.\begin{aligned}4iz-3w&=1+5i\quad(1)\\2z+(1+i)w&=2+6i\quad(2)\\(2)\times2i:\quad4iz+2i(1+i)w&=2i(2+6i)\\4iz+2iw-2w&=4i-12\quad(3)\\(3)-(1):\quad4iz+2iw-2w-(4iz-3w)&=(4i-12)-(1+5i)\\w+2iw&=-13-i\\(1+2i)w&=-13-i\\w&=\frac{-13-i}{1+2i}\frac{1-2i}{1-2i}\\&=\frac{-13+26i-i-2}{1^2-(2i)^2}\\&=\frac{-15+25i}{5}=-3+5i.\end{aligned}

Substitute into (2):

2z=2+6i−(1+i)(−3+5i)=2+6i−(−3+5i−3i−5)=2+6i−(−8+2i)=10+4iz=5+2i.\begin{aligned}2z&=2+6i-(1+i)(-3+5i)\\&=2+6i-(-3+5i-3i-5)\\&=2+6i-(-8+2i)=10+4i\\z&=5+2i.\end{aligned}

Teacher note — alternative substitution

From (2), 2z=2+6i−(1+i)w2z=2+6i-(1+i)w. Substitute into (1):

2i(2+6i−(1+i)w)−3w=1+5i,2i(2+6i-(1+i)w)-3w=1+5i, 4i−12−2iw+2w−3w=1+5i,4i-12-2iw+2w-3w=1+5i, −(1+2i)w=13+i.-(1+2i)w=13+i.

Printed answer key

w=−3+5i,z=5+2iw=-3+5i,z=5+2i.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: COMPLEX NUMBERS. Question pages 1; solution pages 1. These are original document page numbers, retained as text metadata.