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Worked solution

Solve a conjugate equation without assuming conjugate roots

Do not use a calculator. One root of zz∗+2iz=a+6izz^*+2iz=a+6i, with a∈Ra\in\mathbb R, is z=3−7iz=3-7i. Find aa and the other root. marks

[2020/RI/Promo/7(a)]

CJC · Review problems · Question 16

Answer

a=72a=72; the other root is 3+9i3+9i.

Use the known root

(3−7i)(3+7i)+2i(3−7i)=58+14+6i=72+6i(3-7i)(3+7i)+2i(3-7i)=58+14+6i=72+6i. Hence a=72a=72.

How did this part go?

Find all possible coordinates

Let z=x+yiz=x+yi, with real x,yx,y. Then x2+y2+2i(x+yi)=72+6ix^2+y^2+2i(x+yi)=72+6i. Imaginary parts give 2x=62x=6, so x=3x=3. Real parts give 9+y2−2y=729+y^2-2y=72, hence y2−2y−63=(y+7)(y−9)=0y^2-2y-63=(y+7)(y-9)=0. The known root has y=−7y=-7; the other is z=3+9iz=3+9i. These exhaust the coordinate equations.

How did this part go?

Common mistakes

This equation includes both z and its conjugate and is not a real-coefficient polynomial in z. Do not infer 3+7i as a second root.

Solution notes

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

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