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An equation such as |z| − z = 1 + 2i cannot be solved by moving terms around, because |z| and z are different kinds of quantity. The reliable way in is to write z = x + iy and turn one complex equation into two real ones. This post shows the method, four worked examples of rising difficulty, and the checks that catch lost or extra roots.
Where is this in the syllabus?
The NCERT Class 11 chapter Complex Numbers and Quadratic Equations (2026-27 reprint) covers the algebra of complex numbers, the modulus and conjugate in section 4.4, and the Argand plane in section 4.5; its examples compare real and imaginary parts, as in "If x + iy = (a + ib)/(a − ib), prove that x² + y² = 1". The rationalised chapter no longer contains a section on quadratic equations with complex roots, but the JEE (Main) 2026 syllabus still lists "Quadratic equations in real and complex number systems" alongside the algebra, modulus and argument of complex numbers. The JEE (Advanced) 2026 brochure lists conjugation, modulus and cube roots of unity. Check your year's syllabus.
What is the method?
Write z = x + iy, with x and y real. Then z̄ = x − iy and |z| = √(x² + y²).
Substitute and expand. Collect everything into (real part) + i(imaginary part).
Two complex numbers are equal only if their real parts are equal and their imaginary parts are equal. Write those two real equations.
Solve for x and y. Reject any solution where x or y is not real.
Substitute back to check.
Worked example 1: linear in z and z̄
Solve z + 2z̄ = 3 − i.
(x + iy) + 2(x − iy) = 3x − iy.
Real parts: 3x = 3, so x = 1. Imaginary parts: −y = −1, so y = 1.
z = 1 + i. Check: (1 + i) + 2(1 − i) = 3 − i.
Worked example 2: modulus and z together
Solve |z| − z = 1 + 2i.
|z| − z = √(x² + y²) − x − iy.
Imaginary parts: −y = 2, so y = −2.
Real parts: √(x² + 4) − x = 1, so √(x² + 4) = x + 1. Squaring: x² + 4 = x² + 2x + 1, so x = 3/2.
Squaring can create false roots, so check: x + 1 = 5/2 is positive, as a square root must be. z = 3/2 − 2i.
Check: |z| = √(9/4 + 4) = 5/2, and 5/2 − (3/2 − 2i) = 1 + 2i.
Worked example 3: z² = z̄
Solve z² = z̄ for all complex z.
z² = x² − y² + 2ixy and z̄ = x − iy.
Real parts: x² − y² = x. Imaginary parts: 2xy = −y, so y(2x + 1) = 0.
Case y = 0: x² = x, so x = 0 or 1. That gives z = 0 and z = 1.
Case x = −1/2: 1/4 − y² = −1/2, so y² = 3/4 and y = ±√3/2. That gives z = −1/2 ± (√3/2)i.
Four solutions: 0, 1 and −1/2 ± (√3/2)i. The last two are ω and ω², the complex cube roots of unity, which makes sense: if z ≠ 0, taking the modulus of both sides gives |z|² = |z|, so |z| = 1, and then z̄ = 1/z turns the equation into z³ = 1. Spotting that shortcut is good, but the x + iy method finds every root without it, including z = 0, which the shortcut divides away.
Worked example 4: z² + |z| = 0
z² + |z| = x² − y² + √(x² + y²) + 2ixy.
Imaginary parts: 2xy = 0, so x = 0 or y = 0.
If y = 0: x² + |x| = 0. Both terms are non-negative, so x = 0. That gives z = 0.
If x = 0: −y² + |y| = 0, so |y|(1 − |y|) = 0 and y = 0 or ±1. That gives z = 0, i and −i.
Three solutions: 0, i and −i. Check z = i: i² + 1 = 0.
Common mistakes and how to avoid them
Writing |z|² = z². In general |z|² = z z̄ = x² + y², while z² = x² − y² + 2ixy.
Dividing by y or x. Write y(2x + 1) = 0 and treat each factor as a case; dividing by y loses z = 0 and z = 1 in example 3.
Not checking after squaring. In example 2, a negative x + 1 would have produced a false root.
Allowing complex x or y. If solving gives x = i, it is not a valid solution, because x was defined as real.
For the special case of square roots of a complex number, see our post on finding square roots by comparing parts. For the geometry behind |z − a| = r and related loci, see complex numbers as an Argand-plane tool.
A practice routine
Day 1: Three linear equations in z and z̄, such as 2z − iz̄ = 3.
Day 2: Two equations mixing |z| and z, checking for false roots after squaring.
Day 3: Solve z² = −z̄ and z³ = z̄ by the x + iy method, and count roots.
Day 4: Redo one problem from Day 3 by taking modulus first, and compare.
If you can follow each step but cannot see the first move on a new problem, Fermi's AI tutor can work out what you already know and build a plan for the session, so you sit down to a clear next step.
Tonight, solve z² = −z̄ and check that you find four solutions, including z = 0.
