A modulus equation imposes a sign

The complex numbers z,wz,w satisfy 2z+1=∣w∣2z+1=|w| and 2w−z=4+24i2w-z=4+24i. Find z,wz,w, giving answers in the form a+iba+ib, where a,ba,b are real numbers. marks

[2023/CJC/Prelim/I/1]

Solution

Find z and w

Square and reject

2z+1=∣w∣2z+1=|w| and 2w−z=4+24i2w-z=4+24i. Find zz and ww in the form a+iba+ib.

Explore this idea

Hint

  1. Use the first equation to show that zz is real.

  2. Split the second equation into real and imaginary parts.

  3. Square once, then check the sign condition.

Answer · Find z and w

z=6,w=5+12iz=6,w=5+12i.

The squared-equation candidate a=−19/15 violates 4 a−7≥0.

Explore the idea

Why one candidate must be rejected

Slide the real part aa of w=a+12iw=a+12i. The test compares 4a−74a-7 with ∣w∣=a2+144|w|=\sqrt{a^2+144} and needs them to be equal and non-negative.

Real part a of w5
Try an example
4a − 7 (equals 2z + 1)
13
|w|
13
verdict
equal, and non-negative: valid
  • Moves as the slider changes; a faint path shows where it goes
Key idea

Squaring both sides loses the sign. Always return to the original equation and check that the left-hand side is non-negative.

Ideas to try
  1. At a=5a=5 the left side is 1313 and ∣w∣=13|w|=13: a genuine solution.
  2. At a=−1915≈−1.27a=-\frac{19}{15}\approx-1.27 the left side is about −12.07-12.07, but ∣w∣≈12.07|w|\approx12.07. They match only after squaring.
Takeaway

Squaring can invent answers

When an equation involves a modulus or a square root, test every candidate in the original statement.