Traps in Complex Numbers & Quadratic Equations

10 mistake patterns students fall for. Each one shows the wrong approach vs the correct approach.

Forgetting conjugate when dividing

FORMULA

When dividing z₁/z₂, students forget to multiply numerator and denominator by the conjugate of the denominator.

✗ WRONG: Directly dividing real and imaginary parts separately: (a+bi)/(c+di) = a/c + (b/d)i

✓ RIGHT: Multiply by z̄₂/z̄₂: (a+bi)(c−di) / (c²+d²)

Wrong quadrant for argument

SIGN ERROR

Using tan⁻¹(b/a) directly gives principal value in (−π/2, π/2), which is wrong for z in Q2 or Q3.

✗ WRONG: arg(−1+i) = tan⁻¹(1/−1) = −π/4

✓ RIGHT: arg(−1+i) = π − π/4 = 3π/4 (Q2, so add π to the reference angle)

Writing z·z̄ = z² instead of |z|²

FORMULA

z·z̄ equals |z|² (a real number), not z². This is one of the most common errors.

✗ WRONG: z·z̄ = z² → (3+4i)(3−4i) = (3+4i)² = −7+24i

✓ RIGHT: z·z̄ = |z|² → (3+4i)(3−4i) = 9+16 = 25

Applying De Moivre without polar conversion

FORMULA

Students try to apply De Moivre's theorem directly to (a+bi)ⁿ without first converting to r(cosθ + isinθ) form.

✗ WRONG: (1+i)⁸ → trying binomial expansion (messy and error-prone)

✓ RIGHT: (1+i) = √2·e^(iπ/4) → (1+i)⁸ = (√2)⁸·e^(i·2π) = 16·1 = 16

Confusing |z−a|=|z−b| with a circle

CASE MISS

|z−a| = |z−b| is the perpendicular bisector (a line), not a circle. Students often set up the wrong locus.

✗ WRONG: Treating |z−(1+i)| = |z−(2−i)| as a circle equation

✓ RIGHT: This is a straight line: the perpendicular bisector of the segment joining (1,1) and (2,−1)

Forgetting conjugate pair rule has conditions

DOMAIN

Complex roots come in conjugate pairs only when the polynomial has REAL coefficients. Not true for complex coefficients.

✗ WRONG: If 2+3i is a root of z²+(1+i)z+c=0, assuming 2−3i is also a root

✓ RIGHT: Conjugate pair rule doesn't apply here because (1+i) is not a real coefficient. Use sum/product of roots instead.

Sign error in conjugate operations

SIGN ERROR

Students mess up signs when computing conjugates of sums, products, or quotients.

✗ WRONG: Conjugate of (3−2i)/(1+i) → computing (3+2i)/(1+i) instead of (3+2i)/(1−i)

✓ RIGHT: Conjugate distributes: conj(z₁/z₂) = z̄₁/z̄₂. So conj of (3−2i)/(1+i) = (3+2i)/(1−i)

Missing roots when solving zⁿ = w

CASE MISS

zⁿ = w has exactly n roots. Students often find only one root and forget the others.

✗ WRONG: z³ = 8 → z = 2 (only finding the real root)

✓ RIGHT: z³ = 8 → z = 2, 2ω, 2ω² where ω = e^(i2π/3). All three roots must be listed.

Squaring both sides of modulus equations incorrectly

FORMULA

When squaring |z₁ + z₂| = k, students expand incorrectly or forget the cross term.

✗ WRONG: |z₁+z₂|² = |z₁|² + |z₂|² (missing cross term)

✓ RIGHT: |z₁+z₂|² = |z₁|² + |z₂|² + 2Re(z₁z̄₂) = |z₁|² + |z₂|² + z₁z̄₂ + z̄₁z₂

Time-wasting: expanding instead of using properties

TIME WASTE

Many problems are solved faster using properties (|z|²=zz̄, arg rules, geometry) than brute-force x+iy substitution.

✗ WRONG: For |z−1|=|z+1|, substituting z=x+iy and expanding both sides

✓ RIGHT: Geometric approach: equidistant from 1 and −1 → lies on imaginary axis → Re(z)=0. Done in 5 seconds.