Porisma

Worked solution

Use rotation to find area and real quadratic factors

In an Argand diagram, P,QP,Q represent z1=2+3iz_1=2+3i and z2=iz1z_2=iz_1.

(i) Find the area of triangle OPQOPQ, where OO is the origin. marks

(ii) These numbers are roots of (z2+az+b)(z2+cz+d)=0(z^2+az+b)(z^2+cz+d)=0, where a,b,c,d∈Ra,b,c,d\in\mathbb R. Find a,b,c,da,b,c,d. marks

[2017/TJC/I/8(a)]

CJC · Review problems · Question 13

Answer

(i) Area 13/213/2. (ii) (a,b,c,d)=(−4,13,6,13)(a,b,c,d)=(-4,13,6,13), or the swapped assignment (6,13,−4,13)(6,13,-4,13).

(i) Use a right angle and equal moduli

Multiplication by ii rotates 90° anticlockwise about OO without changing modulus. Thus z2=−3+2iz_2=-3+2i, OP=OQ=13OP=OQ=\sqrt{13} and ∠POQ=90∘\angle POQ=90^\circ. The area is 12∣z1∣∣z2∣=12∣2+3i∣2=13/2\tfrac12|z_1||z_2|=\tfrac12|2+3i|^2=13/2 square units. Equivalently the coordinate determinant is 2(2)−3(−3)=132(2)-3(-3)=13.

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(ii) Group the conjugate pairs

Real coefficients force the roots 2−3i2-3i and −3−2i-3-2i as well. Their monic quadratic factors are (z−2)2+9=z2−4z+13(z-2)^2+9=z^2-4z+13 and (z+3)2+4=z2+6z+13(z+3)^2+4=z^2+6z+13. The printed assignment is (a,b,c,d)=(−4,13,6,13)(a,b,c,d)=(-4,13,6,13). Since the prompt does not assign a particular root to the first factor, swapping the two quadratic factors also gives (6,13,−4,13)(6,13,-4,13). Both represent exactly the same polynomial.

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Common mistakes

Conjugate pairing follows from real coefficients. The product does not distinguish which real quadratic is written first.

Solution notes

Source filename says 2026; printed header says 2025/JC2. Both identities are retained; rendered pages determine radicals and modulus bars.

The source lists one coefficient assignment; the factor-order ambiguity is qualified in the reviewed working rather than replacing that printed answer.

Read the hint