Complex Numbers – JEE Main MathematicsPractice Questions & PYQs

Set $A$ describes a circle with center $2+i$ and radius 3, given by $(x-2)^2 + (y-1)^2 = 9$. Set $B$ represents the condition $\operatorname{Re}((x+iy)(1-i)) = 2$, which expands to $\operatorname{Re}(x - ix + iy + y) = x+y = 2$. Substituting $y=2-x$ into the circle equation yields $(x-2)^2 + (2-x-1)^2 = 9$, simplifying to $2x^2 - 6x - 4 = 0$, or $x^2 - 3x - 2 = 0$. The roots $x_1, x_2$ satisfy $x_1+x_2 = 3$ and $x_1x_2 = -2$. The sum of the squares of the moduli is $\sum (x^{2+}y^2) = \sum (x^2 + (2-x)^2) = \sum (2x^2 - 4x + 4) = 2(x_1^{2+}x_2^2) - 4(x_1+x_2) + 8$. Using $x_1^{2+}x_2^2 = (x_1+x_2)^2 - 2x_1x_2 = 13$, the total sum is $2(13) - 4(3) + 8 = 22$.

The expression can be simplified by dividing the quadratic $2z^{2-}3z-2i=0$ by $z$ to obtain $z - \frac{i}{z} = \frac{3}{2}$. Squaring this relation gives $z^2 - 2i - \frac{1}{z^2} = \frac{9}{4}$, which simplifies to $z^2 - \frac{1}{z^2} = \frac{9}{4} + 2i$. Squaring again yields $z^4 + \frac{1}{z^4} - 2 = (\frac{9}{4} + 2i)^2$, leading to $z^4 + \frac{1}{z^4} = \frac{49}{16} + 9i$. The required expression reduces to $z^4 + \frac{1}{z^4}$, so the real part is $\frac{49}{16}$ and the imaginary part is $9$, resulting in $16 \times \frac{49}{16} \times 9 = 441$.

An equilateral triangle with centroid $z_0$ has its vertices at positions $z_k$ such that the set of shifted vectors $z_k - z_0$ forms a symmetric configuration around the origin, representing the roots of a complex polynomial. By rotating these vectors by $120^{\circ}$ or $\frac{2\pi}{3}$ radians, the sum of their squares $z_1^2 + z_2^2 + z_3^2$ evaluates to $R^2(e^{i4\pi/3} + e^{i8\pi/3} + e^{i12\pi/3})$. Since the sum of these phase factors is zero, the resulting sum of the squared differences is zero.

The nature of triangle $ABO$ is determined by calculating its side lengths and checking for properties like isosceles or obtuse angles. The lengths are $OA = |z_1| = \sqrt{11}$ and $OB = |z_1|/\sqrt{3} = \sqrt{11/3}$. Using the Law of Cosines with the given included angle of $\frac{\pi}{6}$, the side $AB$ is calculated as $AB = \sqrt{OA^2 + OB^2 - 2(OA)(OB)\cos(\frac{\pi}{6})} = \sqrt{11 + \frac{11}{3} - 11} = \sqrt{\frac{11}{3}}$. Since $OB = AB$, the triangle is isosceles, and calculating the interior angles reveals an obtuse angle of $120^\circ$ opposite to side $OA$.

Since $\alpha$ satisfies $x^{2+}x+1=0$, it is a primitive cube root of unity, meaning $\alpha^3=1$ and $\alpha^{2+}\alpha+1=0$. The general term of the summation, $(\alpha^k + \alpha^{-k})^2$, expands to $\alpha^{2k} + 2 + \alpha^{-2k}$, which simplifies to $\alpha^{2k} + 2 + \alpha^k$. If $k$ is a multiple of 3, the term equals $1 + 2 + 1 = 4$; otherwise, it equals $\alpha^2 + 2 + \alpha = 1$. The sum cycles in groups of three terms totaling $1 + 1 + 4 = 6$. For $n=11$, we have three complete cycles totaling 18, plus the terms for $k=10$ and $k=11$, which are both 1, resulting in a total sum of $18 + 1 + 1 = 20$.

The relationship between roots $\alpha, \beta$ and the polynomial coefficients $x^{2} - a x - b = 0$ is defined by Vieta's formulas, where $\alpha+\beta=a$ and $\alpha\beta=-b$. The sequence $P_{n} = \alpha^{n} - \beta^{n}$ satisfies the linear recurrence $P_{n} = a P_{n-1} + b P_{n-2}$. Substituting the given values $P_{3} = -5 \sqrt{7} i, P_{4} = -3 \sqrt{7} i, P_{5} = 11 \sqrt{7} i,$ and $P_{6} = 45 \sqrt{7} i$ into this recurrence produces the equations $45 = 11 a - 3 b$ and $11 = -3 a - 5 b$, which solve to $a=3$ and $b=-4$. The expression $\left| \alpha^{4} + \beta^{4} \right|$ is evaluated using the identity $\left| \alpha^{4} + \beta^{4} \right| = \sqrt{(\alpha^{4} - \beta^{4})^{2} + 4 \alpha^{4} \beta^{4}}$, which simplifies to $\sqrt{P_{4}^{2} + 4(-b)^{4}} = \sqrt{-63 + 1024} = \sqrt{961} = 31$.

Multiplying both sides of the given equation by the denominator $k+\overline{z}$ yields $2+k^2z = k^2z + kz\overline{z}$. Utilizing the property $|z|^2 = z\overline{z} = 1$, the equation simplifies to $2+k^2z = k^2z + k$, which directly implies $k=2$. The point $k+ik^2$ is therefore $2+4i$, and the distance to the circle $|z-(1+2i)|=1$ is the distance from $2+4i$ to the center $1+2i$ plus the radius, calculated as $\sqrt{(2-1)^2 + (4-2)^2} + 1 = \sqrt{1+4} + 1 = \sqrt{5}+1$.

Since $|z|=1$, the condition $\frac{z}{\bar{z}} + \frac{\bar{z}}{z} = 1$ is equivalent to $|z^2 + \frac{1}{z^2}| = 1$ because $\bar{z} = \frac{1}{z}$. Substituting $z = e^{i\theta}$ transforms this into $|e^{i2\theta} + e^{-i2\theta}| = |2\cos(2\theta)| = 1$, or $\cos(2\theta) = \pm \frac{1}{2}$. Solving for $\theta$ within the range $[0, 2\pi)$ yields 8 distinct solutions, corresponding to $2\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}, \text{ and } \frac{11\pi}{3}$.

The minimum distance between two disks in the complex plane, defined by $|z-c_1| \le r_1$ and $|z-c_2| \le r_2$, is the distance between their centers minus the sum of their radii, expressed as $|c_1 - c_2| - (r_1 + r_2)$. The centers are $8+2i$ and $2-6i$, giving a distance of $\sqrt{(8-2)^2 + (2-(-6))^2} = \sqrt{6^2 + 8^2} = 10$. Subtracting the sum of the radii $1+2=3$ from this distance results in $10 - 3 = 7$.

The curve $z(1+i) + \overline{z}(1-i) = 4$ simplifies to $2\operatorname{Re}(z(1+i)) = 4$, which equates to the line $x-y=2$. The region $|z-3| \leq 1$ describes a disk with center $(3,0)$ and radius $1$, and the perpendicular distance from the center to the line is $d = \frac{|3-0-2|}{\sqrt{1^{2+}(-1)^2}} = \frac{1}{\sqrt{2}}$. Since the line intersects the circle, it creates two segments with areas $\alpha$ and $\beta$, where the area of the smaller segment is $A = \cos^{-1}(\frac{d}{r}) - d\sqrt{r^{2-}d^2} = \frac{\pi}{4} - \frac{1}{2}$. The larger segment is then $\pi - (\frac{\pi}{4} - \frac{1}{2}) = \frac{3\pi}{4} + \frac{1}{2}$, and their absolute difference is $|\frac{3\pi}{4} + \frac{1}{2} - (\frac{\pi}{4} - \frac{1}{2})| = 1 + \frac{\pi}{2}$.

The expression simplifies using the property $\operatorname{Re}(w) + \operatorname{Re}(\overline{w}) = 2\operatorname{Re}(w)$, which reduces the given equation to $\operatorname{Re}\left(\frac{z-1}{2z+i}\right) = 1$. Substituting $z = x+iy$ and multiplying the numerator and denominator by the complex conjugate $2x - i(2y+1)$, we extract the real part as $\frac{2x^2 - 2x + 2y^2 + y}{4x^2 + (2y+1)^2} = 1$. This simplifies to the circle equation $2x^2 - 2x + 2y^2 + y = 4x^2 + 4y^2 + 4y + 1$, or $x^2 + y^2 + x + \frac{3}{2}y + \frac{1}{2} = 0$. Comparing this to the standard form $x^{2+}y^{2+}2gx+2fy+c=0$, the center $(a, b)$ is $(-1/2, -3/4)$ and the radius $r$ is $\sqrt{(-1/2)^2 + (-3/4)^2 - 1/2} = \sqrt{1/4 + 9/16 - 8/16} = \sqrt{5}/4$. Calculating the final expression, $\frac{15(-1/2)(-3/4)}{5/16}$ yields $15(3/8) \times (16/5) = 18$.

The locus of a complex number satisfying a modulus ratio equation corresponds to a circle in the complex plane. Expanding $\left|\frac{\overline{z}-i}{2\overline{z}+i}\right|=\frac{1}{3}$ by setting $z=x+iy$ leads to $9(x^{2+}(y+1)^2) = 4x^2 + (1-2y)^2$, which simplifies to $5x^2 + 5y^2 + 22y + 8 = 0$, indicating a center $C$ at $(0, -\frac{11}{5})$. The triangle with vertices $(0,0), C,$ and $(\alpha, 0)$ has a base of $|\alpha|$ along the x-axis and a height of $\frac{11}{5}$; thus, the area formula $\frac{1}{2} \times |\alpha| \times \frac{11}{5} = 11$ results in $|\alpha|=10$, making $\alpha^2 = 100$.

Expressing $z_1, z_2, z_3$ in polar form on the unit circle yields $z_1 = \frac{1-i}{\sqrt{2}}$, $z_2 = 1$, and $z_3 = \frac{1+i}{\sqrt{2}}$. Calculating the sum $S = z_1\overline{z}_2 + z_2\overline{z}_3 + z_3\overline{z}_1$ gives $\frac{1-i}{\sqrt{2}} + \frac{1-i}{\sqrt{2}} + \frac{(1+i)^2}{2}$, which simplifies to $\sqrt{2} - i\sqrt{2} + i = \sqrt{2} + i(1-\sqrt{2})$. The squared modulus $|S|^2 = (\sqrt{2})^2 + (1-\sqrt{2})^2 = 2 + 1 - 2\sqrt{2} + 2 = 5-2\sqrt{2}$, identifying $\alpha=5$ and $\beta=-2$, resulting in $\alpha^{2+}\beta^2 = 25+4 = 29$.

Simplifying the given expression by multiplying the numerator and denominator by the complex conjugate $\cos\theta+3i\sin\theta$ isolates the real part as $\frac{2\cos^2\theta-3\sin^2\theta}{\cos^2\theta+9\sin^2\theta}$. Inserting this into the equation $1+10\operatorname{Re}(\dots)=0$ leads to $21(\cos^2\theta-\sin^2\theta)=0$, which simplifies to $21\cos(2\theta)=0$. Solving for $\theta \in [0, 2\pi]$ yields four values: $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$, and calculating the sum of their squares provides $\frac{\pi^2}{16}(1+9+25+49) = \frac{21\pi^2}{4}$.

Define $u = 8 \sin \theta + 7 \cos \theta$ and $v = \sin \theta + 4 \cos \theta$, such that $\omega_1 = u + iv$ and $\omega_2 = v + iu$. Multiplying these expressions gives $\omega_1 \omega_2 = (u + iv)(v + iu) = uv + i(u^2) + iv^2 + i^2(vu) = i(u^2 + v^2)$, implying the real part $\alpha = 0$ and the imaginary part $\beta = u^2 + v^2$. Expanding $\beta = (8 \sin \theta + 7 \cos \theta)^2 + (\sin \theta + 4 \cos \theta)^2$ results in $65(\sin^2 \theta + \cos^2 \theta) + 60 \sin \theta \cos \theta = 65 + 30 \sin(2\theta)$. The maximum value $p$ is $65 + 30 = 95$ and the minimum value $q$ is $65 - 30 = 35$, resulting in $p+q = 130$.

The determinant equation simplifies to $z^3=1$, identifying the possible roots as $z=1, \omega, \omega^2$. The modulus condition $|\frac{z-a}{z+b}|=1$ is equivalent to $|z-a|=|z+b|$, meaning $z$ lies on the perpendicular bisector of the segment between $a$ and $-b$, which satisfies $Re(z) = \frac{a-b}{2}$. For $z=1$, $a-b=2$, yielding 4 valid integer pairs within $[-3,3]$ where $a+b \neq 0$, and for $z=\omega$ or $\omega^2$, $Re(z)=-1/2$, so $a-b=-1$, yielding 6 valid pairs. Summing these possibilities gives a total of 10 valid ordered pairs.

For the first statement, a complex number $w$ is real if and only if $w = \overline{w}$. Setting $\frac{z-i}{z+i} = \frac{\overline{z}+i}{\overline{z}-i}$ and utilizing the condition $|z|=1$, which implies $\overline{z} = 1/z$, we simplify the equation to $z+\overline{z}=0$, or $x=0$. On the unit circle, this allows $z=i$ and $z=-i$, but since the domain excludes $-i$, only $z=i$ satisfies the condition, making the first statement incorrect. For the second statement, $w$ is purely imaginary if $w + \overline{w} = 0$. Applying this to $\frac{z-1}{z+1} + \frac{\overline{z}-1}{\overline{z}+1} = 0$ results in $z\overline{z} - 1 = 0$, which is equivalent to $|z|=1$. Since any point on the unit circle (excluding $-1$) satisfies this, the second statement is correct.

Clearing the fraction by multiplying both sides by the denominator yields the equation $z^{2+}3i = (z-2+i)(2+3i)$. Expanding the right side results in $z^{2+}3i = 2z - 4 + 2i + 3iz - 6i - 3$, which rearranges into the quadratic equation $z^2 - (2+3i)z + (7+7i) = 0$. Given roots $z_1$ and $z_2$, the sum of their squares is $z_1^{2+}z_2^2 = (z_1+z_2)^2 - 2z_1z_2$, and substituting $z_1+z_2=2+3i$ and $z_1z_2=7+7i$ gives $(2+3i)^2 - 2(7+7i) = 4-9+12i - 14-14i = -19-2i$.

The modulus of the complex number $z+1$ where $z=\frac{1}{2}-2i$ is calculated as $|z+1|=|\frac{3}{2}-2i| = \sqrt{(\frac{3}{2})^{2+}(-2)^2} = 2.5$. By equating the real and imaginary components of $\alpha(\frac{1}{2}-2i)+\beta(1+i)$ to $2.5$, we derive the system of equations $0.5\alpha+\beta=2.5$ and $-2\alpha+\beta=0$. Substituting $\beta=2\alpha$ into the first equation yields $2.5\alpha=2.5$, resulting in $\alpha=1$ and $\beta=2$, thus their sum is 3.

The triangle inequality states that for any two complex numbers $u$ and $v$ with unit modulus, $|u+v| \le |u|+|v|=2$. Applying this property to the normalized vectors $\frac{z_1}{|z_1|}$ and $\frac{z_2}{|z_2|}$ confirms the inequality in the first statement is always true. For the second statement, the proportionality constant $\lambda$ allows us to write the terms as $a^2 = \lambda(y-z)(\overline{y}-\overline{z})$, and substituting this into the sum $\frac{a^2}{y-z} + \frac{b^2}{z-x} + \frac{c^2}{x-y}$ results in $\lambda(\overline{y}-\overline{z} + \overline{z}-\overline{x} + \overline{x}-\overline{y})$, which simplifies to $0$ rather than $1$, indicating the second statement is incorrect.