Complex Numbers and Quadratic Equation: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Complex Numbers and Quadratic Equation  
 
(January 2026) 
 
 
Complex Numbers 
 
Q1: Let 
 
Then the max  and {| z
1
 - z
2
| : z
1
 ? A and z
2
 ? B} is: 
(a) 17/2 
(b) 8 
(c) 9 
(d) 15/2 
Ans: (a) 
Sol:  
 
Analyze Set A: 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Complex Numbers and Quadratic Equation  
 
(January 2026) 
 
 
Complex Numbers 
 
Q1: Let 
 
Then the max  and {| z
1
 - z
2
| : z
1
 ? A and z
2
 ? B} is: 
(a) 17/2 
(b) 8 
(c) 9 
(d) 15/2 
Ans: (a) 
Sol:  
 
Analyze Set A: 
The set A is defined as {z ? C : |z - 2| = 4}. 
This represents a circle (and its interior) centered at (2, 0) with a radius r = 4. 
The circle extends from x = -2 to x = 6 along the real axis. 
Analyze Set B: 
The set B is defined as {z ? C : |z - 2| + |z + 2| = 5}. 
By the definition of an ellipse, the sum of distances from two fixed points (foci) is constant. 
Here, the foci are F 1(2, 0) and F 2(-2, 0), and the length of the major axis 2a = 5, so a = 2.5. 
By the definition of an ellipse, the sum of distances from two fixed points (foci) is constant. 
Here, the foci are F 1(2, 0) and F 2(-2, 0), and the length of the major axis 2a = 5, so a = 2.5. 
The distance between foci is 2ae = 4, so e = 4/5 = 0.8. 
The semi-minor axis b can be found using b² = a²(1 - e²) = (2.5)²(1 - 0.64) = 6.25(0.36) = 2.25, 
so b = 1.5. 
The ellipse is centered at the origin (0, 0) and extends from x = -2.5 to x = 2.5 along the real 
axis. 
To find max {|z 1 - z 2| : z1 ? A and z 2 ? B} This means maximize the distance between a point z1 
in the circle and a point z 2 on the ellipse. 
The furthest point in set A (the circle) is at the rightmost edge, z 1 = (2 + 4, 0) = (6, 0). 
The furthest point in set B (the ellipse) from z 1 is at its leftmost edge, z 2 = (-2.5, 0). 
Maximum Distance |z 1 - z 2| = |6 - (-2.5)| = |6 + 2.5| = 8.5 = 17/2 
 
Q2: Let z be a complex number such that |z - 6| = 5 and |z + 2 - 6i| = 5. Then the value of 
z
3
 + 3z
2
 - 15 z + 141 is equal to :  
(a) 61  
(b) 37  
(c) 42  
(d) 50  
Ans: (d) 
Sol: 
Let z = x + iy 
|z - (a + iß)| = r represents a circle in the complex plane with centre at (a, ß) and radius is r unit. 
Cartesian equation 
(x - a)² + (y - ß)² = r² 
The equations |z - 6| = 5 and |z - (-2 + 6i)| = 5 describe two circles with the same radius (r = 5). 
Circle 1 : Center C 1 = (6, 0) 
Circle 2 : Center C 2 = (-2, 6) 
Cartesian equation 
(x - 6)² + y² = 25 and (x + 2)² + (y - 6)² = 25 
Expanding both: 
x² - 12x + 36 + y² = 25 ? x² + y² - 12x + 11 = 0 . . . (i) 
x² + 4x + 4 + y² - 12y + 36 = 25 ? x² + y² + 4x - 12y + 15 = 0 . . . (ii) 
Subtracting the first equation from the second to find the common chord: 
(4x - (-12x)) - 12y + (15 - 11) = 0 
16x - 12y + 4 = 0 ? 4 - 3y + 1 = 0 ? y = (4x + 1) / 3 
Substitute y back into the first circle equation: 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Complex Numbers and Quadratic Equation  
 
(January 2026) 
 
 
Complex Numbers 
 
Q1: Let 
 
Then the max  and {| z
1
 - z
2
| : z
1
 ? A and z
2
 ? B} is: 
(a) 17/2 
(b) 8 
(c) 9 
(d) 15/2 
Ans: (a) 
Sol:  
 
Analyze Set A: 
The set A is defined as {z ? C : |z - 2| = 4}. 
This represents a circle (and its interior) centered at (2, 0) with a radius r = 4. 
The circle extends from x = -2 to x = 6 along the real axis. 
Analyze Set B: 
The set B is defined as {z ? C : |z - 2| + |z + 2| = 5}. 
By the definition of an ellipse, the sum of distances from two fixed points (foci) is constant. 
Here, the foci are F 1(2, 0) and F 2(-2, 0), and the length of the major axis 2a = 5, so a = 2.5. 
By the definition of an ellipse, the sum of distances from two fixed points (foci) is constant. 
Here, the foci are F 1(2, 0) and F 2(-2, 0), and the length of the major axis 2a = 5, so a = 2.5. 
The distance between foci is 2ae = 4, so e = 4/5 = 0.8. 
The semi-minor axis b can be found using b² = a²(1 - e²) = (2.5)²(1 - 0.64) = 6.25(0.36) = 2.25, 
so b = 1.5. 
The ellipse is centered at the origin (0, 0) and extends from x = -2.5 to x = 2.5 along the real 
axis. 
To find max {|z 1 - z 2| : z1 ? A and z 2 ? B} This means maximize the distance between a point z1 
in the circle and a point z 2 on the ellipse. 
The furthest point in set A (the circle) is at the rightmost edge, z 1 = (2 + 4, 0) = (6, 0). 
The furthest point in set B (the ellipse) from z 1 is at its leftmost edge, z 2 = (-2.5, 0). 
Maximum Distance |z 1 - z 2| = |6 - (-2.5)| = |6 + 2.5| = 8.5 = 17/2 
 
Q2: Let z be a complex number such that |z - 6| = 5 and |z + 2 - 6i| = 5. Then the value of 
z
3
 + 3z
2
 - 15 z + 141 is equal to :  
(a) 61  
(b) 37  
(c) 42  
(d) 50  
Ans: (d) 
Sol: 
Let z = x + iy 
|z - (a + iß)| = r represents a circle in the complex plane with centre at (a, ß) and radius is r unit. 
Cartesian equation 
(x - a)² + (y - ß)² = r² 
The equations |z - 6| = 5 and |z - (-2 + 6i)| = 5 describe two circles with the same radius (r = 5). 
Circle 1 : Center C 1 = (6, 0) 
Circle 2 : Center C 2 = (-2, 6) 
Cartesian equation 
(x - 6)² + y² = 25 and (x + 2)² + (y - 6)² = 25 
Expanding both: 
x² - 12x + 36 + y² = 25 ? x² + y² - 12x + 11 = 0 . . . (i) 
x² + 4x + 4 + y² - 12y + 36 = 25 ? x² + y² + 4x - 12y + 15 = 0 . . . (ii) 
Subtracting the first equation from the second to find the common chord: 
(4x - (-12x)) - 12y + (15 - 11) = 0 
16x - 12y + 4 = 0 ? 4 - 3y + 1 = 0 ? y = (4x + 1) / 3 
Substitute y back into the first circle equation: 
(x - 6)² + ((4x+1)/3)² = 25 
multiply both sides by 9 
9(x² + 36 - 12x) + 16x² + 8x + 1 = 225 
25x² - 100x + 100 = 0 
x² - 4x + 4 = 0 
(x - 2)² = 0 ? x = 2 
put x = 2 in y = (4x + 1)/3 
y = (4 × 2 + 1)/3 = 9/3 = 3 
so z = 2 + 3i 
calculate z³ + 3z² - 15z + 141 
Compute z³ 
z³ = z · z² = (2 + 3i)(-5 + 12i) 
= -10 + 24i - 15i + 36i² 
= -10 + 9i - 36 
= -46 + 9i 
Compute z² 
(2 + 3i)² = 4 + 12i + 9i² = 4 + 12i - 9 = -5 + 12i 
so 
z³ = -46 + 9i 
3z² = -15 + 36i 
-15z = -30 - 45i 
Add everything 
(-46 + 9i) + (-15 + 36i) + (-30 - 45i) + 141 
Real parts: 
-46 - 15 - 30 + 141 = 50 
Imaginary parts: 
9i + 36i - 45i = 0 
So z³ + 3z² - 15z + 141 = 50 
 
Q3: Let  
Then  is equal to : 
(a) 413 
(b) 398 
(c) 385 
(d) 423 
Ans: (c) 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Complex Numbers and Quadratic Equation  
 
(January 2026) 
 
 
Complex Numbers 
 
Q1: Let 
 
Then the max  and {| z
1
 - z
2
| : z
1
 ? A and z
2
 ? B} is: 
(a) 17/2 
(b) 8 
(c) 9 
(d) 15/2 
Ans: (a) 
Sol:  
 
Analyze Set A: 
The set A is defined as {z ? C : |z - 2| = 4}. 
This represents a circle (and its interior) centered at (2, 0) with a radius r = 4. 
The circle extends from x = -2 to x = 6 along the real axis. 
Analyze Set B: 
The set B is defined as {z ? C : |z - 2| + |z + 2| = 5}. 
By the definition of an ellipse, the sum of distances from two fixed points (foci) is constant. 
Here, the foci are F 1(2, 0) and F 2(-2, 0), and the length of the major axis 2a = 5, so a = 2.5. 
By the definition of an ellipse, the sum of distances from two fixed points (foci) is constant. 
Here, the foci are F 1(2, 0) and F 2(-2, 0), and the length of the major axis 2a = 5, so a = 2.5. 
The distance between foci is 2ae = 4, so e = 4/5 = 0.8. 
The semi-minor axis b can be found using b² = a²(1 - e²) = (2.5)²(1 - 0.64) = 6.25(0.36) = 2.25, 
so b = 1.5. 
The ellipse is centered at the origin (0, 0) and extends from x = -2.5 to x = 2.5 along the real 
axis. 
To find max {|z 1 - z 2| : z1 ? A and z 2 ? B} This means maximize the distance between a point z1 
in the circle and a point z 2 on the ellipse. 
The furthest point in set A (the circle) is at the rightmost edge, z 1 = (2 + 4, 0) = (6, 0). 
The furthest point in set B (the ellipse) from z 1 is at its leftmost edge, z 2 = (-2.5, 0). 
Maximum Distance |z 1 - z 2| = |6 - (-2.5)| = |6 + 2.5| = 8.5 = 17/2 
 
Q2: Let z be a complex number such that |z - 6| = 5 and |z + 2 - 6i| = 5. Then the value of 
z
3
 + 3z
2
 - 15 z + 141 is equal to :  
(a) 61  
(b) 37  
(c) 42  
(d) 50  
Ans: (d) 
Sol: 
Let z = x + iy 
|z - (a + iß)| = r represents a circle in the complex plane with centre at (a, ß) and radius is r unit. 
Cartesian equation 
(x - a)² + (y - ß)² = r² 
The equations |z - 6| = 5 and |z - (-2 + 6i)| = 5 describe two circles with the same radius (r = 5). 
Circle 1 : Center C 1 = (6, 0) 
Circle 2 : Center C 2 = (-2, 6) 
Cartesian equation 
(x - 6)² + y² = 25 and (x + 2)² + (y - 6)² = 25 
Expanding both: 
x² - 12x + 36 + y² = 25 ? x² + y² - 12x + 11 = 0 . . . (i) 
x² + 4x + 4 + y² - 12y + 36 = 25 ? x² + y² + 4x - 12y + 15 = 0 . . . (ii) 
Subtracting the first equation from the second to find the common chord: 
(4x - (-12x)) - 12y + (15 - 11) = 0 
16x - 12y + 4 = 0 ? 4 - 3y + 1 = 0 ? y = (4x + 1) / 3 
Substitute y back into the first circle equation: 
(x - 6)² + ((4x+1)/3)² = 25 
multiply both sides by 9 
9(x² + 36 - 12x) + 16x² + 8x + 1 = 225 
25x² - 100x + 100 = 0 
x² - 4x + 4 = 0 
(x - 2)² = 0 ? x = 2 
put x = 2 in y = (4x + 1)/3 
y = (4 × 2 + 1)/3 = 9/3 = 3 
so z = 2 + 3i 
calculate z³ + 3z² - 15z + 141 
Compute z³ 
z³ = z · z² = (2 + 3i)(-5 + 12i) 
= -10 + 24i - 15i + 36i² 
= -10 + 9i - 36 
= -46 + 9i 
Compute z² 
(2 + 3i)² = 4 + 12i + 9i² = 4 + 12i - 9 = -5 + 12i 
so 
z³ = -46 + 9i 
3z² = -15 + 36i 
-15z = -30 - 45i 
Add everything 
(-46 + 9i) + (-15 + 36i) + (-30 - 45i) + 141 
Real parts: 
-46 - 15 - 30 + 141 = 50 
Imaginary parts: 
9i + 36i - 45i = 0 
So z³ + 3z² - 15z + 141 = 50 
 
Q3: Let  
Then  is equal to : 
(a) 413 
(b) 398 
(c) 385 
(d) 423 
Ans: (c) 
Sol: 
 
equation (1) represent z is a point which is equidistant from point (0, 2) and (0, 6) so z lies on 
the line segment which is perpendicular bisector of the line segment joining point (0, 2) & (0, 6). 
If z = x + iy then solution of equation (1) is y = (6+2)/2 = 4 and x ? R. 
So, solution is z = x + 4i & x ? R. 
 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Complex Numbers and Quadratic Equation  
 
(January 2026) 
 
 
Complex Numbers 
 
Q1: Let 
 
Then the max  and {| z
1
 - z
2
| : z
1
 ? A and z
2
 ? B} is: 
(a) 17/2 
(b) 8 
(c) 9 
(d) 15/2 
Ans: (a) 
Sol:  
 
Analyze Set A: 
The set A is defined as {z ? C : |z - 2| = 4}. 
This represents a circle (and its interior) centered at (2, 0) with a radius r = 4. 
The circle extends from x = -2 to x = 6 along the real axis. 
Analyze Set B: 
The set B is defined as {z ? C : |z - 2| + |z + 2| = 5}. 
By the definition of an ellipse, the sum of distances from two fixed points (foci) is constant. 
Here, the foci are F 1(2, 0) and F 2(-2, 0), and the length of the major axis 2a = 5, so a = 2.5. 
By the definition of an ellipse, the sum of distances from two fixed points (foci) is constant. 
Here, the foci are F 1(2, 0) and F 2(-2, 0), and the length of the major axis 2a = 5, so a = 2.5. 
The distance between foci is 2ae = 4, so e = 4/5 = 0.8. 
The semi-minor axis b can be found using b² = a²(1 - e²) = (2.5)²(1 - 0.64) = 6.25(0.36) = 2.25, 
so b = 1.5. 
The ellipse is centered at the origin (0, 0) and extends from x = -2.5 to x = 2.5 along the real 
axis. 
To find max {|z 1 - z 2| : z1 ? A and z 2 ? B} This means maximize the distance between a point z1 
in the circle and a point z 2 on the ellipse. 
The furthest point in set A (the circle) is at the rightmost edge, z 1 = (2 + 4, 0) = (6, 0). 
The furthest point in set B (the ellipse) from z 1 is at its leftmost edge, z 2 = (-2.5, 0). 
Maximum Distance |z 1 - z 2| = |6 - (-2.5)| = |6 + 2.5| = 8.5 = 17/2 
 
Q2: Let z be a complex number such that |z - 6| = 5 and |z + 2 - 6i| = 5. Then the value of 
z
3
 + 3z
2
 - 15 z + 141 is equal to :  
(a) 61  
(b) 37  
(c) 42  
(d) 50  
Ans: (d) 
Sol: 
Let z = x + iy 
|z - (a + iß)| = r represents a circle in the complex plane with centre at (a, ß) and radius is r unit. 
Cartesian equation 
(x - a)² + (y - ß)² = r² 
The equations |z - 6| = 5 and |z - (-2 + 6i)| = 5 describe two circles with the same radius (r = 5). 
Circle 1 : Center C 1 = (6, 0) 
Circle 2 : Center C 2 = (-2, 6) 
Cartesian equation 
(x - 6)² + y² = 25 and (x + 2)² + (y - 6)² = 25 
Expanding both: 
x² - 12x + 36 + y² = 25 ? x² + y² - 12x + 11 = 0 . . . (i) 
x² + 4x + 4 + y² - 12y + 36 = 25 ? x² + y² + 4x - 12y + 15 = 0 . . . (ii) 
Subtracting the first equation from the second to find the common chord: 
(4x - (-12x)) - 12y + (15 - 11) = 0 
16x - 12y + 4 = 0 ? 4 - 3y + 1 = 0 ? y = (4x + 1) / 3 
Substitute y back into the first circle equation: 
(x - 6)² + ((4x+1)/3)² = 25 
multiply both sides by 9 
9(x² + 36 - 12x) + 16x² + 8x + 1 = 225 
25x² - 100x + 100 = 0 
x² - 4x + 4 = 0 
(x - 2)² = 0 ? x = 2 
put x = 2 in y = (4x + 1)/3 
y = (4 × 2 + 1)/3 = 9/3 = 3 
so z = 2 + 3i 
calculate z³ + 3z² - 15z + 141 
Compute z³ 
z³ = z · z² = (2 + 3i)(-5 + 12i) 
= -10 + 24i - 15i + 36i² 
= -10 + 9i - 36 
= -46 + 9i 
Compute z² 
(2 + 3i)² = 4 + 12i + 9i² = 4 + 12i - 9 = -5 + 12i 
so 
z³ = -46 + 9i 
3z² = -15 + 36i 
-15z = -30 - 45i 
Add everything 
(-46 + 9i) + (-15 + 36i) + (-30 - 45i) + 141 
Real parts: 
-46 - 15 - 30 + 141 = 50 
Imaginary parts: 
9i + 36i - 45i = 0 
So z³ + 3z² - 15z + 141 = 50 
 
Q3: Let  
Then  is equal to : 
(a) 413 
(b) 398 
(c) 385 
(d) 423 
Ans: (c) 
Sol: 
 
equation (1) represent z is a point which is equidistant from point (0, 2) and (0, 6) so z lies on 
the line segment which is perpendicular bisector of the line segment joining point (0, 2) & (0, 6). 
If z = x + iy then solution of equation (1) is y = (6+2)/2 = 4 and x ? R. 
So, solution is z = x + 4i & x ? R. 
 
 
? 25(x² + 64 - 16x + 36) = 9x² + 324 
? 25x² - 400x + 2500 = 9x² + 324 
? 16x² - 400x + 2176 = 0 
? x² - 25x + 136 = 0 
? x² - 17x - 8x + 136 = 0 
? x(x - 17) - 8(x - 17) = 0 
? (x - 17)(x - 8) = 0 
? x = 17, x = 8 
So set S contain two complex number z1 = 17 + 4i and z 2 = 8 + 4i 
 
= |17 + 4i|² + |8 + 4i|² 
= 17² + 4² + 8² + 4² 
= 289 + 16 + 64 + 16 = 385 
 
Q4:  is equal to 
(a) 1 
(b) 0