Complex Numbers — Locus + Rotation
Complex Numbers
42
JEE Qs
25%
Hard
80
min
Always visualize the geometric configuration on the Argand plane and translate complex number operations into geometric transformations to simplify problem-solving.
🧮 Key Formulas
✅ Key Points for JEE
- 1The modulus |z - z₀| represents the distance between the complex number z and z₀ on the Argand plane, which is fundamental for deriving loci of circles, perpendicular bisectors, ellipses, and hyperbolas.
- 2The argument Arg((z - z₁)/(z - z₂)) directly relates to the angle subtended by the line segment joining z₁ and z₂ at z, enabling derivation of circular arcs and straight lines.
- 3Geometric interpretation of multiplication by e^(iθ) as rotation and by a complex number r*e^(iθ) as rotation combined with scaling is crucial for solving rotation problems.
- 4Always identify the 'pivot point' for rotation: if it's not the origin, shift the system to the pivot point by subtracting it from the other complex numbers before applying the rotation operator.
- 5For complex number loci, frequently convert the given equation into standard coordinate geometry forms (x,y) by substituting z = x + iy to cross-verify or solve.
⚠️ Common Mistakes
- ✕Misinterpreting the direction of rotation (clockwise vs. anticlockwise) or not using the principal argument correctly, leading to incorrect signs in trigonometric terms.
- ✕Incorrectly identifying the center, radius, foci, or type of conic section when deriving loci from complex number equations, especially when dealing with variables or absolute values.
- ✕Algebraic errors when manipulating complex fractions involving arguments or moduli, particularly with squares and square roots.
- ✕Not considering degenerate cases for loci, e.g., when the sum/difference of distances equals the distance between foci (line segment instead of ellipse/hyperbola).
📝 Practice Questions
See allQ17.Let 2¯z+i ¯z−i = 13 , z ∈C , be the equation of a circle with center at C . If the area of the triangle, whose vertices are at the points (0, 0), C and (α, 0) is 11 square units, then α2 equals: (1) 50 (2) 100 (3) 81 (4) 121 25 25
Q25.Let α, β be the roots of the equation x2 −ax −b = 0 with Im(α) < Im(β). Let Pn = αn −βn . If P3 = −5√7i, P4 = −3√7i, P5 = 11√7i and P6 = 45√7i , then α4 + β4 is equal to . ∣∣ 2025 (23 Jan Shift 2) JEE Main Previous Year Paper
Q20.Let z1, z2 and z3 be three complex numbers on the circle |z| = 1 with arg (z1) = −π4 , arg (z2) = 0 and arg (z3) = π4 . If |z1¯z2 + z2¯z3 + z3¯z1|2 = α + β√2, α, β ∈Z, then the value of α2 + β2 is : (1) 24 (2) 29 (3) 41 (4) 31
Q19.Let the curve z(1 + i) + ¯z(1 −i) = 4, z ∈C, divide the region |z −3| ≤1 into two parts of areas α and β . Then |α −β| equals : (1) 1 + π2 (2) 1 + π3 (3) 1 + π6 (4) 1 + π4
Q14.The number of complex numbers z , satisfying |z| = 1 and z¯z + ¯zz = 1, is : (1) 4 (2) 8 (3) 10 (4) 6 Q15. ⎡ 0 ⎤ ⎡ 0 ⎤ ⎡4⎤ ⎡0⎤ ⎡2 ⎤ ⎡1 ⎤ Let A = [aij] be 3 × 3 matrix such that A 1 = 0 , A 1 = 1 and A 1 = 0 , then a23 equals : ⎣ 0 ⎦ ⎣ 1 ⎦ ⎣3⎦ ⎣0⎦ ⎣2 ⎦ ⎣0 ⎦ (1) -1 (2) 2 (3) 1 (4) 0 2 x sin 2 dx equals : 3 3
Q10.For a statistical data x1, x2, … , x10 of 10 values, a student obtained the mean as 5.5 and ∑10i=1 x2i = 371. He later found that he had noted two values in the data incorrectly as 4 and 5 , instead of the correct values 6 and 8 , respectively. The variance of the corrected data is (1) 9 (2) 5 (3) 7 (4) 4
NCERT Chapters
- Class 11 Maths Ch 5: Complex Numbers and Quadratic Equations
- Class 11 Maths Ch 10: Straight Lines
- Class 11 Maths Ch 11: Conic Sections