Common Mistakes

Traps in Complex Numbers & Quadratic Equations

10 mistake patterns students fall for. 4 high-frequency traps appear in almost every exam.

Forgetting conjugate when dividing

Very CommonFORMULA

When dividing z1/z2z_1/z_2, students forget to multiply numerator and denominator by the conjugate of the denominator.

Why: Division looks simple but requires rationalizing the denominator to separate real and imaginary parts.

WRONG: Directly dividing: a+bic+di=ac+bdi\frac{a+bi}{c+di} = \frac{a}{c} + \frac{b}{d}i
RIGHT: Multiply by zˉ2zˉ2\frac{\bar{z}_2}{\bar{z}_2}: (a+bi)(cdi)c2+d2\frac{(a+bi)(c-di)}{c^2+d^2}

Wrong quadrant for argument

Very CommonSIGN ERROR

Using tan1(b/a)\tan^{-1}(b/a) directly gives principal value in (π/2,π/2)(-\pi/2, \pi/2), which is wrong for zz in Q2 or Q3.

Why: The calculator gives the reference angle, but arg depends on which quadrant zz lies in.

WRONG: arg(1+i)=tan1(1/(1))=π/4\arg(-1+i) = \tan^{-1}(1/(-1)) = -\pi/4
RIGHT: arg(1+i)=ππ/4=3π/4\arg(-1+i) = \pi - \pi/4 = 3\pi/4 (Q2, so add π\pi to the reference angle)

Writing zzˉ=z2z \cdot \bar{z} = z^2 instead of z2|z|^2

Very CommonFORMULA

zzˉ=z2z \cdot \bar{z} = |z|^2 (a real number), not z2z^2. This is one of the most common errors.

Why: Students confuse multiplying by conjugate with squaring because both involve two copies of zz.

WRONG: zzˉ=z2(3+4i)(34i)=(3+4i)2=7+24iz \cdot \bar{z} = z^2 \Rightarrow (3+4i)(3-4i) = (3+4i)^2 = -7+24i
RIGHT: zzˉ=z2(3+4i)(34i)=9+16=25z \cdot \bar{z} = |z|^2 \Rightarrow (3+4i)(3-4i) = 9+16 = 25

Expanding instead of using properties

Very CommonTIME WASTE

Many problems are solved faster using properties (z2=zzˉ|z|^2 = z\bar{z}, arg rules, geometry) than brute-force x+iyx+iy substitution.

Why: Substituting z=x+iyz = x + iy feels safe but creates unnecessarily long algebra.

WRONG: For z1=z+1|z-1| = |z+1|, substituting z=x+iyz = x+iy and expanding both sides
RIGHT: Geometric approach: equidistant from 11 and 1-1 means Re(z)=0\text{Re}(z) = 0. Done in 5 seconds.

Applying De Moivre without polar conversion

CommonFORMULA

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

Why: Binomial expansion of (a+bi)n(a+bi)^n is tempting but extremely error-prone for large nn.

WRONG: (1+i)8(1+i)^8 via binomial expansion (messy and error-prone)
RIGHT: (1+i)=2eiπ/4(1+i)8=(2)8ei2π=16(1+i) = \sqrt{2}\,e^{i\pi/4} \Rightarrow (1+i)^8 = (\sqrt{2})^8 \cdot e^{i \cdot 2\pi} = 16

Confusing za=zb|z-a| = |z-b| with a circle

CommonCASE MISS

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

Why: za=r|z-a| = r is a circle, so students assume any modulus equation gives a circle.

WRONG: Treating z(1+i)=z(2i)|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)(1,1) and (2,1)(2,-1)

Conjugate pair rule has conditions

CommonDOMAIN

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

Why: The conjugate root theorem is taught without emphasizing the real-coefficient requirement.

WRONG: If 2+3i2+3i is a root of z2+(1+i)z+c=0z^2 + (1+i)z + c = 0, assuming 23i2-3i is also a root
RIGHT: Conjugate pair rule does not apply here because (1+i)(1+i) is not a real coefficient. Use sum/product of roots instead.

Sign error in conjugate operations

CommonSIGN ERROR

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

Why: Conjugate distributes over all operations, but the sign change applies to every ii in the expression.

WRONG: (32i1+i)=3+2i1+i\overline{\left(\frac{3-2i}{1+i}\right)} = \frac{3+2i}{1+i} (forgot to conjugate denominator)
RIGHT: z1/z2=zˉ1/zˉ2\overline{z_1/z_2} = \bar{z}_1/\bar{z}_2, so (32i1+i)=3+2i1i\overline{\left(\frac{3-2i}{1+i}\right)} = \frac{3+2i}{1-i}

Missing roots when solving zn=wz^n = w

CommonCASE MISS

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

Why: Real-number intuition (one nnth root) carries over incorrectly to complex numbers.

WRONG: z3=8z=2z^3 = 8 \Rightarrow z = 2 (only finding the real root)
RIGHT: z3=8z=2,2ω,2ω2z^3 = 8 \Rightarrow z = 2,\, 2\omega,\, 2\omega^2 where ω=ei2π/3\omega = e^{i2\pi/3}. All three roots must be listed.

Missing cross term in z1+z22|z_1 + z_2|^2

OccasionalFORMULA

When squaring z1+z2=k|z_1 + z_2| = k, students expand incorrectly or forget the cross term.

Why: The real identity (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 does not directly apply to modulus.

WRONG: z1+z22=z12+z22|z_1+z_2|^2 = |z_1|^2 + |z_2|^2 (missing cross term)
RIGHT: z1+z22=z12+z22+2Re(z1zˉ2)|z_1+z_2|^2 = |z_1|^2 + |z_2|^2 + 2\text{Re}(z_1\bar{z}_2)
Test yourself

Can you spot these traps under time pressure?

Take a timed quiz on Complex Numbers & Quadratic Equations and see if you avoid the mistakes above.