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Original question & working

Reduce a complex system to a quadratic · source reference

[2024/ACJC/Prelim/I/1] · Question pp. 1 · Working pp. 1, 2

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Question

The complex numbers zz and ww satisfy iw2+2wz=2iiw^2+2wz=2i and z+iz=2+iwz+iz=2+iw. Find zz and ww, giving your answers in the form a+iba+ib, where aa and bb are real numbers. [4]

[2024/ACJC/Prelim/I/1]

Source working

Printed working

iw2+2wz=2i(1)z+iz=2+iw(2)z(1+i)=2+iwz=2+iw1+i1−i1−i=2−2i+iw−i2w12−i2=12(2−2i+iw+w).\begin{aligned}iw^2+2wz&=2i\quad(1)\\z+iz&=2+iw\quad(2)\\z(1+i)&=2+iw\\z&=\frac{2+iw}{1+i}\frac{1-i}{1-i}\\&=\frac{2-2i+iw-i^2w}{1^2-i^2}=\frac12(2-2i+iw+w).\end{aligned}

Substitute into (1):

iw2+w(2−2i+iw+w)=2i,iw^2+w(2-2i+iw+w)=2i, (1+2i)w2+(2−2i)w−2i=0.(1+2i)w^2+(2-2i)w-2i=0. w=−(2−2i)±(2−2i)2−4(1+2i)(−2i)2(1+2i)=−2+2i±−162(1+2i)=−2+2i±4i2(1+2i)=1+ior−35+15i.\begin{aligned}w&=\frac{-(2-2i)\pm\sqrt{(2-2i)^2-4(1+2i)(-2i)}}{2(1+2i)}\\&=\frac{-2+2i\pm\sqrt{-16}}{2(1+2i)}=\frac{-2+2i\pm4i}{2(1+2i)}\\&=1+i\quad\text{or}\quad-\frac35+\frac15i.\end{aligned}

Thus (w,z)=(1+i,1)(w,z)=(1+i,1) or (−3/5+i/5,3/5−6i/5)(-3/5+i/5,3/5-6i/5).

Printed notation correction

The intermediate line printed as z(1+1)=2+iwz(1+1)=2+iw is a typographical error. The equation above it and its denominator require z(1+i)=2+iwz(1+i)=2+iw, used in the native mathematical working.

Printed answer key

z=1,w=1+iz=1,w=1+i or z=3/5−6i/5,w=−3/5+i/5z=3/5-6i/5,w=-3/5+i/5.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: COMPLEX NUMBERS. Question pages 1; solution pages 1, 2. These are original document page numbers, retained as text metadata.