Forgetting conjugate when dividing
Very CommonFORMULA
When dividing z1/z2, 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: c+dia+bi=ca+dbi RIGHT: Multiply by zˉ2zˉ2: c2+d2(a+bi)(c−di) Wrong quadrant for argument
Very CommonSIGN ERROR
Using tan−1(b/a) directly gives principal value in (−π/2,π/2), which is wrong for z in Q2 or Q3.
Why: The calculator gives the reference angle, but arg depends on which quadrant z lies in.
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⋅zˉ=z2 instead of ∣z∣2
Very CommonFORMULA
z⋅zˉ=∣z∣2 (a real number), not z2. This is one of the most common errors.
Why: Students confuse multiplying by conjugate with squaring because both involve two copies of z.
WRONG: z⋅zˉ=z2⇒(3+4i)(3−4i)=(3+4i)2=−7+24i RIGHT: z⋅zˉ=∣z∣2⇒(3+4i)(3−4i)=9+16=25 Expanding instead of using properties
Very CommonTIME WASTE
Many problems are solved faster using properties (∣z∣2=zzˉ, arg rules, geometry) than brute-force x+iy substitution.
Why: Substituting z=x+iy feels safe but creates unnecessarily long algebra.
WRONG: For ∣z−1∣=∣z+1∣, substituting z=x+iy and expanding both sides RIGHT: Geometric approach: equidistant from 1 and −1 means 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 without first converting to r(cosθ+isinθ) form.
Why: Binomial expansion of (a+bi)n is tempting but extremely error-prone for large n.
WRONG: (1+i)8 via binomial expansion (messy and error-prone) RIGHT: (1+i)=2eiπ/4⇒(1+i)8=(2)8⋅ei⋅2π=16 Confusing ∣z−a∣=∣z−b∣ with a circle
CommonCASE MISS
∣z−a∣=∣z−b∣ is the perpendicular bisector (a line), not a circle. Students often set up the wrong locus.
Why: ∣z−a∣=r is a circle, so students assume any modulus equation gives a circle.
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) 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+3i is a root of z2+(1+i)z+c=0, assuming 2−3i is also a root RIGHT: Conjugate pair rule does not apply here because (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 i in the expression.
WRONG: (1+i3−2i)=1+i3+2i (forgot to conjugate denominator) RIGHT: z1/z2=zˉ1/zˉ2, so (1+i3−2i)=1−i3+2i Missing roots when solving zn=w
CommonCASE MISS
zn=w has exactly n roots. Students often find only one root and forget the others.
Why: Real-number intuition (one nth root) carries over incorrectly to complex numbers.
WRONG: z3=8⇒z=2 (only finding the real root) RIGHT: z3=8⇒z=2,2ω,2ω2 where ω=ei2π/3. All three roots must be listed. Missing cross term in ∣z1+z2∣2
OccasionalFORMULA
When squaring ∣z1+z2∣=k, students expand incorrectly or forget the cross term.
Why: The real identity (a+b)2=a2+2ab+b2 does not directly apply to modulus.
WRONG: ∣z1+z2∣2=∣z1∣2+∣z2∣2 (missing cross term) RIGHT: ∣z1+z2∣2=∣z1∣2+∣z2∣2+2Re(z1zˉ2)