Complex Numbers: JEE Main 2026 PYQs

Answer: Option (4)

Key idea: Complex number operations: rationalization and equating real and imaginary parts

1. Step 1: Rationalize $\frac{2x}{1+3i}$ by multiplying by $\frac{1-3i}{1-3i}$ to get $\frac{x(1-3i)}{5}$
2. Step 2: Rationalize $\frac{y}{1-2i}$ by multiplying by $\frac{1+2i}{1+2i}$ to get $\frac{y(1+2i)}{5}$
3. Step 3: Substitute and simplify: $50 \cdot \frac{1}{5}[x(1-3i) - y(1+2i)] = 31 + 17i$ gives $10[(x-y) + i(-3x-2y)] = 31 + 17i$
4. Step 4: Equate real parts: $10(x-y) = 31$, so $x - y = 3.1$
5. Step 5: Equate imaginary parts: $10(-3x-2y) = 17$, so $-3x - 2y = 1.7$
6. Step 6: Solve the system: From $x = y + 3.1$ and $-3(y+3.1) - 2y = 1.7$, get $y = -2.2$ and $x = 0.9$
7. Step 7: Calculate $10(x - 3y) = 10(0.9 - 3(-2.2)) = 10(7.5) = 75$

Answer: Option (4)

Key idea: Vieta's formulas and relationships between roots of quadratic equations

1. From $x^2 - 3x + r = 0$: $\alpha + \beta = 3$ and $\alpha\beta = r$
2. From $x^2 + 3x + r = 0$: $\frac{\alpha}{2} + 2\beta = -3$ and $\frac{\alpha}{2} \cdot 2\beta = r$
3. Solve the system: $\alpha + \beta = 3$ and $\alpha + 4\beta = -6$ to get $\beta = -3$ and $\alpha = 6$
4. Calculate $r = \alpha\beta = 6 \times (-3) = -18$
5. Find the roots of third equation: $2\alpha + \beta + 2r = 12 - 3 - 36 = -27$ and $\alpha - 2\beta - \frac{r}{2} = 6 + 6 + 9 = 21$
6. For $x^2 + 6x = m$, product of roots $= (-27)(21) = -567 = -m$
7. Therefore, $m = 567$

Answer: Option (3)

Key idea: Distance between points on two circles in complex plane

1. Step 1: Identify circle $C_1$ centered at origin with radius $r$, and $C_2$ centered at $3+4i$ with radius $5$. Distance between centers: $d = |3+4i| = 5$.
2. Step 2: Since $C_2$ lies within $C_1$, we need $r > d + 5 = 10$.
3. Step 3: Minimum distance between points on the circles: $\min|z_1-z_2| = r - (d+5) = r - 10$.
4. Step 4: Given $\min|z_1-z_2| = 2$, solve: $r - 10 = 2 \Rightarrow r = 12$.
5. Step 5: Maximum distance occurs when points are on opposite sides along the line joining centers: $\max|z_1-z_2| = r + d + 5 = 12 + 5 + 5 = 22$.

Answer: Option (1)

Key idea: Locus of complex numbers and argument properties

1. Step 1: From $|z + 2| = |z - 2|$, the complex number $z$ is equidistant from $-2$ and $2$, so $z$ lies on the perpendicular bisector of these points, which is the imaginary axis. Thus $z = iy$ for some real $y$.
2. Step 2: Substitute $z = iy$ into the second condition: $\frac{z + 3}{z - i} = \frac{3 + iy}{iy - i} = \frac{3 + iy}{i(y - 1)}$
3. Step 3: Simplify by multiplying numerator and denominator by $-i$: $\frac{(3 + iy)(-i)}{i(y-1)(-i)} = \frac{y - 3i}{y - 1} = \frac{y}{y-1} - \frac{3i}{y-1}$
4. Step 4: For $\arg\left(\frac{z+3}{z-i}\right) = \frac{\pi}{4}$, we need $\tan\left(\frac{\pi}{4}\right) = 1 = \frac{\text{Im}}{\text{Re}} = \frac{-3/(y-1)}{y/(y-1)} = \frac{-3}{y}$
5. Step 5: Solving $\frac{-3}{y} = 1$ gives $y = -3$, so $z = -3i$
6. Step 6: Calculate $|z|^2 = |-3i|^2 = 0^2 + (-3)^2 = 9$

Answer: Option (4)

Key idea: Complex numbers: Finding roots using quadratic formula and calculating modulus squared

1. Use quadratic formula: $z = \frac{-4 \pm \sqrt{16-64}}{2} = \frac{-4 \pm 4\sqrt{3}i}{2} = -2 \pm 2\sqrt{3}i$
2. So $S = \{-2 + 2\sqrt{3}i, -2 - 2\sqrt{3}i\}$
3. For $z_1 = -2 + 2\sqrt{3}i$: $z_1 + \sqrt{3}i = -2 + 3\sqrt{3}i$
4. Calculate $|z_1 + \sqrt{3}i|^2 = 4 + 27 = 31$
5. For $z_2 = -2 - 2\sqrt{3}i$: $z_2 + \sqrt{3}i = -2 - \sqrt{3}i$
6. Calculate $|z_2 + \sqrt{3}i|^2 = 4 + 3 = 7$
7. Sum = $31 + 7 = 38$

Answer: 2

Key idea: Conditions for positive roots of quadratic equation using Vieta's formulas and discriminant

1. For two positive roots, we need: (1) Product of roots $> 0$, (2) Sum of roots $> 0$, (3) Discriminant $\geq 0$
2. Product of roots: $\frac{4\lambda}{\lambda + 2} > 0$ gives $\lambda \in (-\infty, -2) \cup (0, \infty)$
3. Sum of roots: $\frac{3\lambda}{\lambda + 2} > 0$ gives $\lambda \in (-\infty, -2) \cup (0, \infty)$
4. Discriminant: $9\lambda^2 - 16\lambda^2 - 32\lambda \geq 0$ simplifies to $-7\lambda^2 - 32\lambda \geq 0$, giving $\lambda \in [-\frac{32}{7}, 0]$
5. Taking intersection of all conditions: $\lambda \in [-\frac{32}{7}, -2)$ where $-\frac{32}{7} \approx -4.57$
6. The integral values in $[-4.57, -2)$ are $\lambda = -4, -3$, giving **2 possible values**

Answer: Option (2)

Key idea: Complex Numbers, Vieta's Formulas, and Algebraic Identities

1. From $\beta^2 - \alpha^2 = (\beta - \alpha)(\beta + \alpha)$, substitute given values: $3i\sqrt{11} = \sqrt{11}(\beta + \alpha)$, so $\beta + \alpha = 3i$
2. Using $\beta - \alpha = \sqrt{11}$ and $\beta + \alpha = 3i$, solve to get $\beta = \frac{3i + \sqrt{11}}{2}$ and $\alpha = \frac{3i - \sqrt{11}}{2}$
3. Calculate $\alpha\beta = \frac{(3i)^2 - (\sqrt{11})^2}{4} = \frac{-9 - 11}{4} = -5$
4. Find $\beta^2 + \alpha^2 = (\beta + \alpha)^2 - 2\alpha\beta = (3i)^2 - 2(-5) = -9 + 10 = 1$
5. Use identity $\beta^3 - \alpha^3 = (\beta - \alpha)(\beta^2 + \alpha\beta + \alpha^2) = \sqrt{11}(1 - 5) = -4\sqrt{11}$
6. Calculate $(\beta^3 - \alpha^3)^2 = (-4\sqrt{11})^2 = 16 \times 11 = 176$

Answer: Option (1)

Key idea: Complex number equations with real parameter constraints using discriminant analysis

1. Let $z = x + iy$ where $x, y \in \mathbb{R}$. Substitute into the equation and expand $z(\bar{z} + 2 + i)$
2. Separate real and imaginary parts: Real: $x^2 + 2x + y^2 - y + 2k = 0$, Imaginary: $x + 2y + 3k = 0$
3. From imaginary part, get $x = -2y - 3k$ and substitute into real part to get $5y^2 + (12k - 5)y + (9k^2 - 4k) = 0$
4. For real solution to exist, discriminant $\Delta \geq 0$: $(12k - 5)^2 - 20(9k^2 - 4k) \geq 0$ gives $36k^2 + 40k - 25 \leq 0$
5. Solve quadratic: $k = \frac{-10 \pm 5\sqrt{13}}{18}$, so $\alpha = \frac{-10 - 5\sqrt{13}}{18}$ and $\beta = \frac{-10 + 5\sqrt{13}}{18}$
6. Calculate $\alpha + \beta = \frac{-20}{18} = \frac{-10}{9}$, therefore $9(\alpha + \beta) = -10$

Answer: Option (1)

Key idea: Vieta's formulas and recurrence relations for sum of powers of roots

1. Step 1: From the equation $z^2 + \sqrt{6}iz - 3 = 0$, use Vieta's formulas: $z_1 + z_2 = -\sqrt{6}i$ and $z_1z_2 = -3$
2. Step 2: Define $S_n = z_1^n + z_2^n$. From the equation, derive recurrence: $S_n = -\sqrt{6}i \cdot S_{n-1} + 3S_{n-2}$
3. Step 3: Calculate initial values: $S_0 = 2$, $S_1 = -\sqrt{6}i$, $S_2 = (z_1+z_2)^2 - 2z_1z_2 = -6 + 6 = 0$
4. Step 4: Use recurrence to find: $S_3 = -3\sqrt{6}i$, $S_4 = -18$, $S_5 = 9\sqrt{6}i$, $S_6 = 0$, $S_7 = 27\sqrt{6}i$
5. Step 5: Calculate $S_8 = -\sqrt{6}i \cdot 27\sqrt{6}i + 3(0) = -27 \cdot 6 \cdot i^2 = 162$

Answer: Option (2)

Key idea: Intersection of circle and ellipse in complex plane

1. Recognize that $|z - (4 + 8i)| = \sqrt{10}$ is a circle with center $(4, 8)$ and radius $\sqrt{10}$
2. The equation $|z - (3 + 5i)| + |z - (5 + 11i)| = 4\sqrt{5}$ represents an ellipse with foci at $A(3, 5)$ and $B(5, 11)$
3. Find the ellipse parameters: distance between foci $= 2\sqrt{10}$, so $c = \sqrt{10}$. Since $2a = 4\sqrt{5}$, we get $a = 2\sqrt{5}$ and $b = \sqrt{a^2 - c^2} = \sqrt{10}$
4. The ellipse center is the midpoint of $AB$: $\left(\frac{3+5}{2}, \frac{5+11}{2}\right) = (4, 8)$, which equals the circle center!
5. Since the circle radius $r = \sqrt{10}$ equals the ellipse semi-minor axis $b = \sqrt{10}$, the circle intersects the ellipse at exactly the two endpoints of the minor axis
6. Verify: The intersection points are $(1, 9)$ and $(7, 7)$, both satisfying both equations