Complex Numbers & Quadratic Equations Previous Year Questions (JEE Main 2026)

$|z+2| = |z-2|$ means $z$ is equidistant from $-2$ and $2$, so $z$ lies on the perpendicular bisector — the imaginary axis. Write $z = iy$.

Then:

$$\frac{z+3}{z-i} = \frac{3 + iy}{iy - i} = \frac{3 + iy}{i(y-1)}$$

Multiply numerator and denominator by $-i$:

$$= \frac{-i(3+iy)}{y-1} = \frac{y - 3i}{y-1}$$

So the real part is $\dfrac{y}{y-1}$ and the imaginary part is $\dfrac{-3}{y-1}$.

For $\arg = \dfrac{\pi}{4}$ we need $\text{Im} = \text{Re}$ (both positive):

$$\frac{-3}{y-1} = \frac{y}{y-1} \;\Rightarrow\; y = -3$$

Check: real part $= \dfrac{-3}{-4} = 0.75 > 0$, imaginary part $= \dfrac{-3}{-4} = 0.75 > 0$, so $\arg = \dfrac{\pi}{4}$. ✓

$$|z|^2 = y^2 = (-3)^2 = 9$$

Answer: A

Let $z = x + iy$, so $z\bar z = x^2 + y^2$. Expanding:

$$z\bar z + z(2+i) + k(2+3i) = 0$$

$$x^2 + y^2 + (x+iy)(2+i) + k(2+3i) = 0$$

$$x^2 + y^2 + (2x - y) + i(x + 2y) + 2k + 3ki = 0$$

Separate real and imaginary parts:

$$\text{Real:}\quad x^2 + y^2 + 2x - y + 2k = 0$$

$$\text{Imag:}\quad x + 2y + 3k = 0 \;\Rightarrow\; x = -2y - 3k$$

Substitute into the real equation:

$$(-2y-3k)^2 + y^2 + 2(-2y-3k) - y + 2k = 0$$

$$5y^2 + (12k - 5)y + (9k^2 - 4k) = 0$$

For a real solution $y$ to exist, the discriminant must be non-negative:

$$(12k-5)^2 - 20(9k^2 - 4k) \geq 0$$

$$-36k^2 - 40k + 25 \geq 0 \;\Rightarrow\; 36k^2 + 40k - 25 \leq 0$$

The endpoints $\alpha, \beta$ are the roots, so by Vieta:

$$\alpha + \beta = -\frac{40}{36} = -\frac{10}{9}$$$$9(\alpha + \beta) = 9 \cdot \left(-\frac{10}{9}\right) = -10$$

Answer: A

Rationalize each term:

$$\frac{2x}{1+3i} = \frac{2x(1-3i)}{(1+3i)(1-3i)} = \frac{2x(1-3i)}{10} = \frac{x(1-3i)}{5}$$

$$\frac{y}{1-2i} = \frac{y(1+2i)}{(1-2i)(1+2i)} = \frac{y(1+2i)}{5}$$

So the bracket becomes $\dfrac{x(1-3i) - y(1+2i)}{5}$, and multiplying by $50$:

$$10\big[(x - y) + i(-3x - 2y)\big] = 31 + 17i$$

Equate real and imaginary parts:

$$10(x - y) = 31 \qquad 10(-3x - 2y) = 17$$

Solving: from the first $x - y = 3.1$; substituting gives $-50y = 110 \Rightarrow y = -2.2$, $x = 0.9$.

$$10(x - 3y) = 10\big(0.9 - 3(-2.2)\big) = 10(0.9 + 6.6) = 75$$

Answer: D

Solve the quadratic:

$$z = \frac{-4 \pm \sqrt{16 - 64}}{2} = \frac{-4 \pm \sqrt{-48}}{2} = -2 \pm 2\sqrt{3}\,i$$

So $S = \{-2 + 2\sqrt3\,i,\ -2 - 2\sqrt3\,i\}$.

For $z_1 = -2 + 2\sqrt3\,i$:

$$z_1 + \sqrt3\,i = -2 + 3\sqrt3\,i, \qquad |z_1 + \sqrt3\,i|^2 = 4 + 27 = 31$$

For $z_2 = -2 - 2\sqrt3\,i$:

$$z_2 + \sqrt3\,i = -2 - \sqrt3\,i, \qquad |z_2 + \sqrt3\,i|^2 = 4 + 3 = 7$$$$\sum_{z \in S} |z + \sqrt3\,i|^2 = 31 + 7 = 38$$

Answer: D

Since $a, b, c$ are real, complex roots occur in conjugate pairs, so the third root is $1 - i\sqrt2$. The three roots are $1,\ 1 + i\sqrt2,\ 1 - i\sqrt2$.

By Vieta’s formulas:

$$-a = 1 + (1 + i\sqrt2) + (1 - i\sqrt2) = 3 \;\Rightarrow\; a = -3$$

$$b = (1)(1+i\sqrt2) + (1)(1-i\sqrt2) + (1+i\sqrt2)(1-i\sqrt2) = 2 + 3 = 5$$

$$-c = (1)(1+i\sqrt2)(1-i\sqrt2) = 3 \;\Rightarrow\; c = -3$$

So the polynomial is $x^3 - 3x^2 + 5x - 3$. Over the symmetric interval $[-1,1]$ the odd terms integrate to $0$:

$$\int_{-1}^{1}\left(x^3 + 5x\right)dx = 0$$

$$\int_{-1}^{1}(-3x^2)\,dx = -3\cdot\frac{2}{3} = -2, \qquad \int_{-1}^{1}(-3)\,dx = -6$$$$\int_{-1}^{1}\left(x^3 - 3x^2 + 5x - 3\right)dx = -2 - 6 = -8$$

Answer: C

For two positive real roots of $ax^2 + bx + c = 0$ (here $a = \lambda+2$, $b = -3\lambda$, $c = 4\lambda$) we need three conditions.

1. Real roots — discriminant $\geq 0$:

$$9\lambda^2 - 16\lambda(\lambda+2) \geq 0 \;\Rightarrow\; -7\lambda^2 - 32\lambda \geq 0 \;\Rightarrow\; \lambda \in \left[-\tfrac{32}{7},\, 0\right]$$

2. Sum of roots $> 0$:

$$\frac{3\lambda}{\lambda+2} > 0 \;\Rightarrow\; \lambda > 0 \text{ or } \lambda < -2$$

3. Product of roots $> 0$:

$$\frac{4\lambda}{\lambda+2} > 0 \;\Rightarrow\; \lambda > 0 \text{ or } \lambda < -2$$

Intersecting all three (with $\lambda \neq -2$) gives:

$$\lambda \in \left[-\tfrac{32}{7},\, -2\right) \approx [-4.57,\, -2)$$

Integer values in this interval: $\lambda = -4$ and $\lambda = -3$.

Check $\lambda = -4$: roots are $2$ and $4$ (both positive). ✓ Check $\lambda = -3$: roots $\approx 1.63$ and $7.37$ (both positive). ✓

So there are 2 integral values.

Answer: B

From $x^2 - 3x + r = 0$: $\ \alpha + \beta = 3$ and $\alpha\beta = r$.

From $x^2 + 3x + r = 0$: $\ \dfrac{\alpha}{2} + 2\beta = -3$ (the product $\dfrac{\alpha}{2}\cdot 2\beta = \alpha\beta = r$ is automatically consistent).

Solve the two sum equations:

$$\alpha + \beta = 3, \qquad \frac{\alpha}{2} + 2\beta = -3$$

Substituting $\alpha = 3 - \beta$: $\ \dfrac{3-\beta}{2} + 2\beta = -3 \Rightarrow 3\beta = -9 \Rightarrow \beta = -3,\ \alpha = 6$.

Then $r = \alpha\beta = -18$.

Now the two given roots:

$$R_1 = 2\alpha + \beta + 2r = 12 - 3 - 36 = -27$$

$$R_2 = \alpha - 2\beta - \frac{r}{2} = 6 + 6 + 9 = 21$$

Check the sum against $x^2 + 6x - m = 0$: $R_1 + R_2 = -27 + 21 = -6 = -(\text{coefficient of }x)$. ✓

Product gives $-m$:

$$R_1 R_2 = (-27)(21) = -567 = -m \;\Rightarrow\; m = 567$$

Answer: D

$C_1$ has centre $O = 0$ and radius $r$; $C_2$ has centre $P = 3 + 4i$ and radius $5$. The distance between centres is:

$$d = |3 + 4i| = \sqrt{9 + 16} = 5$$

Since $C_2$ lies inside $C_1$, the closest a point of $C_1$ can get to a point of $C_2$ is:

$$\min|z_1 - z_2| = r - (d + 5) = r - 10$$

Given this equals $2$:

$$r - 10 = 2 \;\Rightarrow\; r = 12$$

The farthest distance is:

$$\max|z_1 - z_2| = r + (d + 5) = 12 + 10 = 22$$

Answer: C

Solve the quadratic:

$$z = \frac{-\sqrt6\,i \pm \sqrt{(\sqrt6\,i)^2 - 4(-3)}}{2} = \frac{-\sqrt6\,i \pm \sqrt{-6 + 12}}{2} = \frac{-\sqrt6\,i \pm \sqrt6}{2}$$

So the two roots are $z = \dfrac{\pm\sqrt6 - \sqrt6\,i}{2}$. Compute the modulus and square of each:

$$|z|^2 = \frac{6 + 6}{4} = 3$$

For $z_1 = \dfrac{\sqrt6 - \sqrt6\,i}{2}$:

$$z_1^2 = \frac{6(1 - i)^2}{4} = \frac{6(-2i)}{4} = -3i$$

For $z_2 = \dfrac{-\sqrt6 - \sqrt6\,i}{2}$:

$$z_2^2 = \frac{6(1 + i)^2}{4} = \frac{6(2i)}{4} = 3i$$

Now raise to the 8th power via $z^8 = (z^2)^4$:

$$z_1^8 = (-3i)^4 = 3^4 \cdot i^4 = 81, \qquad z_2^8 = (3i)^4 = 81$$$$\sum_{z \in S} z^8 = 81 + 81 = 162$$

Answer: A

Use $\beta^2 - \alpha^2 = (\beta - \alpha)(\beta + \alpha)$:

$$3i\sqrt{11} = \sqrt{11}\,(\alpha + \beta) \;\Rightarrow\; \alpha + \beta = 3i$$

Find $\alpha\beta$ from $(\beta - \alpha)^2 = (\alpha + \beta)^2 - 4\alpha\beta$:

$$11 = (3i)^2 - 4\alpha\beta = -9 - 4\alpha\beta \;\Rightarrow\; \alpha\beta = -5$$

Now $\beta^3 - \alpha^3 = (\beta - \alpha)(\alpha^2 + \alpha\beta + \beta^2)$, where

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = -9 + 10 = 1$$

$$\alpha^2 + \alpha\beta + \beta^2 = 1 + (-5) = -4$$

Therefore:

$$\beta^3 - \alpha^3 = \sqrt{11}\,(-4) = -4\sqrt{11}$$

$$(\beta^3 - \alpha^3)^2 = 16 \cdot 11 = 176$$

Answer: B

Solve for the roots:

$$z = \frac{-4 \pm \sqrt{16 + 4(1 + 12i)}}{2} = \frac{-4 \pm \sqrt{20 + 48i}}{2}$$

Find $\sqrt{20 + 48i} = p + qi$: then $p^2 - q^2 = 20$ and $2pq = 48$. Solving, $p = 6, q = 4$, so $\sqrt{20 + 48i} = 6 + 4i$.

$$z = \frac{-4 \pm (6 + 4i)}{2} \;\Rightarrow\; z_1 = 1 + 2i,\quad z_2 = -5 - 2i$$$$|z_1|^2 = 1 + 4 = 5, \qquad |z_2|^2 = 25 + 4 = 29$$$$|z_1|^2 + |z_2|^2 = 5 + 29 = 34$$

Answer: D

The first equation is a circle with centre $C = 4 + 8i$ and radius $R = \sqrt{10}$.

The second equation is an ellipse with foci $F_1 = 3 + 5i$ and $F_2 = 5 + 11i$, and constant sum $2a = 4\sqrt5 \Rightarrow a = 2\sqrt5$ (so $a^2 = 20$).

Distance between foci: $|F_1 - F_2| = |{-2} - 6i| = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$, so $c = \sqrt{10}$ and $c^2 = 10$. Since $2a = 4\sqrt5 > 2c = 2\sqrt{10}$, this is a valid ellipse.

Semi-minor axis: $b^2 = a^2 - c^2 = 20 - 10 = 10$, so $b = \sqrt{10}$.

The ellipse centre is the midpoint of the foci: $\dfrac{(3+5i)+(5+11i)}{2} = 4 + 8i = C$. The circle and ellipse share the same centre.

Key observation: the semi-minor axis $b = \sqrt{10}$ equals the circle radius $R = \sqrt{10}$, while the semi-major axis $a = 2\sqrt5 > R$. So the circle lies inside the ellipse everywhere except at the two ends of the minor axis, where they touch.

The two common points are $z = 1 + 9i$ and $z = 7 + 7i$ (both satisfy $|z - (4+8i)| = \sqrt{10}$ and the ellipse sum $= 4\sqrt5$).

$$\boxed{\text{Number of values} = 2}$$

Answer: B