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Original question & working

A purely imaginary root unlocks a cubic · source reference

[2018/RI/Promo/11a] · Question pp. 3 · Working pp. 13, 14

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Question

(i) Solve z2−(1+2i)z+1+7i=0z^2-(1+2i)z+1+7i=0. [2]

(ii) One root of z3−(1+4i)z2−(3−9i)z+(14−2i)=0z^3-(1+4i)z^2-(3-9i)z+(14-2i)=0 is purely imaginary. Find all roots in Cartesian form a+iba+ib. [6]

[2018/RI/Promo/11a]

Source working

Printed working

(i) Quadratic formula:

z=1+2i±(1+2i)2−4(1+7i)2=1+2i±−7−24i2.z=\frac{1+2i\pm\sqrt{(1+2i)^2-4(1+7i)}}2=\frac{1+2i\pm\sqrt{-7-24i}}2.

Using GC, z=2−iz=2-i or −1+3i-1+3i.

(ii) Let the purely imaginary root be bibi, b∈Rb\in\mathbb R.

(ib)3−(1+4i)(ib)2−(3−9i)(ib)+(14−2i)=0,(ib)^3-(1+4i)(ib)^2-(3-9i)(ib)+(14-2i)=0, −ib3+b2+4ib2−3ib−9b+14−2i=0,-ib^3+b^2+4ib^2-3ib-9b+14-2i=0, (b2−9b+14)−i(b3−4b2+3b+2)=0.(b^2-9b+14)-i(b^3-4b^2+3b+2)=0.

Compare parts: b2−9b+14=0b^2-9b+14=0 (1), so b=2b=2 or 77; and b3−4b2+3b+2=0b^3-4b^2+3b+2=0 (2). Check: 23−4(22)+3(2)+2=02^3-4(2^2)+3(2)+2=0, whereas 73−4(72)+3(7)+2≠07^3-4(7^2)+3(7)+2\ne0. Purely imaginary root 2i2i.

z3−(1+4i)z2−(3−9i)z+(14−2i)=(z−2i)(z2+uz+v).z^3-(1+4i)z^2-(3-9i)z+(14-2i)=(z-2i)(z^2+uz+v).

For complex u,vu,v, constants give v=(14−2i)/(−2i)=1+7iv=(14-2i)/(-2i)=1+7i by GC; z2z^2 coefficients give −(1+4i)=−2i+u⇒u=−1−2i-(1+4i)=-2i+u\Rightarrow u=-1-2i.

z3−(1+4i)z2−(3−9i)z+(14−2i)=(z−2i)(z2−(1+2i)z+1+7i).z^3-(1+4i)z^2-(3-9i)z+(14-2i)=(z-2i)(z^2-(1+2i)z+1+7i).

From (i), all roots are 2i,2−i,−1+3i2i,2-i,-1+3i.

Printed answer key

(i) 2−i,−1+3i2-i,-1+3i. (ii) 2i,2−i,−1+3i2i,2-i,-1+3i.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: COMPLEX NUMBERS. Question pages 3; solution pages 13, 14. These are original document page numbers, retained as text metadata.