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Solve the complex equation.

Read the idea, work independently, then explain what changed.

TOPIC 01

2023 JM02

The root x=−5/3 introduced by squaring is extraneous.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

Solve the complex equation.

z−3zˉ+∣z∣=−1+16iz-3\bar z+|z|=-1+16i

Official paper · jm02-2023 · 4(a) · PDF 6

Official original and suggested answers ↗ · Suggested answer PDF page 10

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Set z=x+yi and compare imaginary parts first.
Hint 2
Before squaring the real equation, require its right side nonnegative.
Worked solution
  1. Separate the equations.

    4y=16  ⟹  y=4,x2+16=2x−1≥0  ⟹  x≥1/24y=16\implies y=4,\quad\sqrt{x^2+16}=2x-1\ge0\implies x\ge1/2
  2. Square and factor.

    x2+16=(2x−1)2  ⟺  (x−3)(3x+5)=0x^2+16=(2x-1)^2\iff(x-3)(3x+5)=0
  3. Only x=3 satisfies the unsquared equation.

    z=3+4i,∣z∣=5z=3+4i,\quad|z|=5

z=3+4i.

Checks and common pitfalls: The root x=−5/3 introduced by squaring is extraneous.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Separate the equations.
    4y=16  ⟹  y=4,x2+16=2x−1≥0  ⟹  x≥1/24y=16\implies y=4,\quad\sqrt{x^2+16}=2x-1\ge0\implies x\ge1/2
  • Square and factor.
    x2+16=(2x−1)2  ⟺  (x−3)(3x+5)=0x^2+16=(2x-1)^2\iff(x-3)(3x+5)=0
  • Only x=3 satisfies the unsquared equation.
    z=3+4i,∣z∣=5z=3+4i,\quad|z|=5

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Curriculum and source notes ↗