Complex Numbers – JEE Main MathematicsPractice Questions & PYQs

The set $A$ is a disk centered at $2$ with radius $4$, extending from $-2$ to $6$ along the real axis. The set $B$ defines an ellipse with foci at $\pm 2$ and a major axis length of $5$, placing its vertices at $-2.5$ and $2.5$. The maximum distance between the two sets is the distance between the endpoints $6$ and $-2.5$, which equals $6 - (-2.5) = 8.5 = \frac{17}{2}$.

The equation $z^2 + 4z - (1+12i) = 0$ yields roots $z = -2 \pm \sqrt{5+12i}$ when solved using the quadratic formula. Given that $(3+2i)^2 = 5+12i$, the roots are $z_1 = -2 + (3+2i) = 1+2i$ and $z_2 = -2 - (3+2i) = -5-2i$. Calculating the squared magnitudes, we find $|z_1|^2 = 1^2 + 2^2 = 5$ and $|z_2|^2 = (-5)^2 + (-2)^2 = 29$, resulting in a sum of $5+29 = 34$.

The first condition represents the perpendicular bisector of the segment connecting $6i$ and $2i$, which is the line $y=4$ in the complex plane. Writing $z$ as $x+4i$ and substituting into the second equation $|z-8+2i| = \frac{3}{5}|z+2i|$ leads to the equation $|(x-8)+6i| = \frac{3}{5}|x+6i|$, which squares to $(x-8)^2 + 36 = \frac{9}{25}(x^{2+}36)$. Solving this quadratic equation yields $x=8$ and $x=17$, so $|z|^2$ takes the values $8^{2+}4^2=80$ and $17^{2+}4^2=305$. The sum of these values is 385.

Applying the quadratic formula to $z^{2} + 4z + 16 = 0$ yields the roots $z = -2 \pm 2i\sqrt{3}$. Evaluating the expression $|z + \sqrt{3}i|^{2}$ for each root gives $|-2 + 3i\sqrt{3}|^{2} = (-2)^{2} + (3\sqrt{3})^{2} = 4 + 27 = 31$ and $|-2 - i\sqrt{3}|^{2} = (-2)^{2} + (-\sqrt{3})^{2} = 4 + 3 = 7$. Summing these values, $31 + 7 = 38$.

The modulus of a product is the product of the moduli, allowing the equation to be written as $\prod_{k=1}^n |1+ki|^2 = \prod_{k=1}^n (1+k^2) = 44200$. Evaluating this product for successive integers $n$, we find that for $n=1$, $2$; for $n=2$, $2 \times 5 = 10$; for $n=3$, $10 \times 10 = 100$; for $n=4$, $100 \times 17 = 1700$; and for $n=5$, $1700 \times 26 = 44200$. Thus, the condition is satisfied when $n=5$.

Expressing the complex number in polar form gives $z = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} = e^{i\pi/6}$, so $z^{201} = e^{i(201\pi/6)} = e^{i(67\pi/2)}$. Since $67\pi/2 = 33\pi + \pi/2$, which is coterminal with $3\pi/2$, $z^{201} = \cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2} = -i$. Therefore, the expression simplifies to $(-i-i)^8 = (-2i)^8 = 256 \times i^8 = 256 \times 1 = 256$.

The locus of points satisfying $|z - (4+8i)| = \sqrt{10}$ is a circle centered at $(4, 8)$ with radius $\sqrt{10}$, while the equation $|z-(3+5i)| + |z-(5+11i)| = 4\sqrt{5}$ represents an ellipse with foci at $(3, 5)$ and $(5, 11)$. The center of this ellipse is the midpoint of the foci, located at $(4, 8)$, which coincides with the center of the circle. Given the distance between the foci is $2\sqrt{10}$ and the sum of distances is $2a = 4\sqrt{5}$, the semi-major axis $a$ is $2\sqrt{5}$ and the semi-minor axis $b$ is $\sqrt{10}$, matching the radius of the circle. Because the circle's radius is equal to the semi-minor axis of the ellipse and both share the same center, they intersect exactly at the two endpoints of the ellipse's minor axis.

The set $C_{1}$ defines a circle centered at the origin with radius $r$, and $C_{2}$ defines a circle centered at $3+4i$ with radius $5$. Given the distance from the origin to the center of $C_{2}$ is $5$, the minimum distance between the two circles is $r - (5+5) = r-10$. Equating this to $2$ gives $r=12$, so the maximum distance between the circles is $r + (5+5) = 12+10 = 22$.

Given $x^{2+}x+1=0$, $x$ is a primitive cube root of unity, ensuring $x^3=1$ and $1/x = x^2$. The expression $x^n + 1/x^n$ repeats every three terms as $(x+x^2)^4 = (-1)^4 = 1$, $(x^{2+}x)^4 = (-1)^4 = 1$, and $(x^{3+}x^{-3})^4 = 2^4 = 16$. With 25 terms consisting of 8 full cycles of $1+1+16 = 18$ plus one additional term matching the first pattern, the total sum is $8 \times 18 + 1 = 145$.

The complex number satisfies two circular equations in the complex plane, defining the intersection points of two circles. By converting $|z-z_0|=r$ into $(x-x_0)^{2+}(y-y_0)^2=r^2$, one can solve the resulting system of linear equations to find $z=2+3i$. Substituting this value into the polynomial $z^{3+}3 z^{2-}15 z+141$ simplifies the expression to 50.

The equation $4z^2 + \overline{z} = 0$ implies that either $z=0$ or, for $z \neq 0$, the modulus satisfies $4|z|^2 = |\overline{z}| = |z|$. Since $|z|$ cannot be zero for non-zero solutions, we divide by $|z|$ to find $|z| = 1/4$, and thus $|z|^2 = 1/16$. Substituting $z = \frac{1}{4}e^{i\theta}$ into the original equation leads to the condition $e^{i3\theta} = -1$, which yields three distinct solutions for $\theta$ in the range $[0, 2\pi)$. The sum of $|z|^2$ over these solutions is $3 \times \frac{1}{16} = \frac{3}{16}$.

The condition $|z+2| = |z-2|$ signifies that $z$ lies on the perpendicular bisector of the segment connecting $-2$ and $2$, which is the imaginary axis where $z=iy$. Substituting $z=iy$ into the argument condition $\arg \left( \frac{z+3}{z-i} \right) = \frac{\pi}{4}$ identifies the specific complex number $z=-3i$. The value of $|z|^{2}$ is therefore $|-3i|^{2} = 9$.

The roots of the equation $z^2 + \sqrt{6}iz - 3 = 0$ are obtained using the quadratic formula as $z = \frac{-\sqrt{6}i \pm \sqrt{-6+12}}{2} = \frac{-\sqrt{6}i \pm \sqrt{6}}{2}$. This gives two roots, $\alpha = \frac{\sqrt{6}}{2}(-1-i)$ and $\beta = \frac{\sqrt{6}}{2}(1-i)$, which can be expressed in polar form as $\alpha = -\sqrt{3}e^{i\pi/4}$ and $\beta = \sqrt{3}e^{-i\pi/4}$. Raising these to the eighth power, we find $\alpha^8 = 81e^{i2\pi} = 81$ and $\beta^8 = 81e^{-i2\pi} = 81$, which sum to 162.

The values $\alpha$ and $\beta$ represent the complex cube roots of unity $\omega$ and $\omega^2$, which adhere to the properties $1+\alpha+\beta=0$ and $\alpha\beta=1$. By systematically substituting $\alpha+\beta=-1$ and simplifying the expressions within each of the four terms, the sum reduces to powers of $7$. Evaluating these terms confirms the total sum is equal to $7^{20}$, which is equivalent to $49^{10}$, identifying $m$ as $49$.

Given the roots $\alpha, \beta$ satisfy $\beta-\alpha = \sqrt{11}$ and $(\beta+\alpha)(\beta-\alpha) = 3i\sqrt{11}$, dividing the second equation by the first reveals $\beta+\alpha = 3i$. The product of the roots is $\alpha\beta = -5$, and using the identity $(\beta^{3-}\alpha^3)^2 = (\beta-\alpha)^2((\alpha+\beta)^{2-}\alpha\beta)^2$, we substitute the values to obtain $11 \cdot (-9 - (-5))^2$, which equals $11 \cdot 16 = 176$.

Substituting $z = x+iy$ into the equation $z(\overline{z}+2+i)+k(2+3i)=0$ and separating into real and imaginary parts results in the system $x^{2+}y^{2+}2x-y+2k=0$ and $x+2y+3k=0$. Eliminating $x$ by substituting $x = -2y-3k$ into the quadratic equation yields $5y^{2+}(12k-5)y+9k^{2-}4k=0$. For the existence of at least one solution, this quadratic in $y$ must have real roots, requiring its discriminant to be non-negative. Solving $(12k-5)^2 - 20(9k^{2-}4k) \geq 0$ leads to the inequality $36k^{2+}40k-25 \leq 0$, where the sum of the roots is $-40/36 = -10/9$, making $9(\alpha+\beta) = -10$.

For the region defined by $|z-5| \le 3$, the maximum positive principal argument occurs at the point of tangency from the origin to the circle, forming a right-angled triangle where the hypotenuse is the distance to the center, 5, and the opposite side is the radius, 3. This geometry gives the modulus $|z| = \sqrt{5^{2-}3^2} = 4$, and representing the complex number as $z = 4(\cos \alpha + i \sin \alpha) = \frac{16+12i}{5}$ allows direct substitution. Evaluating $34 \left| \frac{5z-12}{5iz+16} \right|^2$ with this $z$ yields $34 \left| \frac{1+3i}{1+4i} \right|^2 = 34 \left( \frac{1^{2+}3^2}{1^{2+}4^2} \right) = 34 \left( \frac{10}{17} \right) = 20$.

The condition $3 \le |2z-3(1+i)| \le 7$ defines an annulus centered at $z_0 = \frac{3}{2}(1+i)$ with inner radius $\frac{3}{2}$ and outer radius $\frac{7}{2}$. We want to minimize the distance $|z - z_1|$ where $z_1 = -\frac{1}{2}(5+3i)$. The distance between the center $z_0$ and the target point $z_1$ is $|z_0 - z_1| = |(\frac{3}{2}+\frac{3}{2}i) - (-\frac{5}{2}-\frac{3}{2}i)| = |4+3i| = 5$. Since the annulus extends to a maximum distance of $\frac{7}{2}$ from $z_0$, the minimum distance from $z_1$ to the set $S$ is $5 - \frac{7}{2} = \frac{3}{2}$.

Multiplying the complex fractions by their conjugates to simplify the expression leads to the equation $10x - 30xi - 10y - 20yi = 31 + 17i$. Comparing the real and imaginary parts results in the system $10x - 10y = 31$ and $-30x - 20y = 17$. Solving this linear system for $x$ and $y$ yields $x = 0.9$ and $y = -2.2$, making the final value $10(x - 3y) = 10(0.9 - 3(-2.2)) = 75$.

The properties of the roots of a quadratic equation $x^2 - Sx + P = 0$ are governed by the sum $S$ and product $P$ of its roots. Given the equation $x^2 - (3 - 2 i) x - (2 i - 2) = 0$, the sum of the roots is $3-2i$ and the product is $2-2i$, which factorize into the roots $1+0i$ and $2-2i$. Assigning these to $\alpha+i\beta$ and $\gamma+i\delta$, we identify $\alpha=1, \beta=0$ and $\gamma=2, \delta=-2$, so calculating $\alpha \gamma + \beta \delta$ yields $(1)(2) + (0)(-2) = 2$.