Complex Numbers – JEE Main MathematicsPractice Questions & PYQs

Representing z=x+iy and Re(conj(z))=x transforms the given equation into x-iy=i(x^{2-}y^{2+}2ixy+x), which leads to the system of equations x=-2xy and -y=x^{2-}y^{2+}x. The first equation x(1+2y)=0 yields two cases: if x=0, then y=0 or y=1; if y=-1/2, then x^{2+}x-3/4=0, resulting in x=1/2 or x=-3/2. The squared moduli of these points |z|^2 are 0, 1, (1/4+1/4)=1/2, and (9/4+1/4)=10/4, which sum to 0+1+0.5+2.5=4.

Using the property that $\operatorname{Re}(w)=\frac{w+\overline{w}}{2}$, the given conditions imply $az^{2+}bz+a\overline{z}^{2+}b\overline{z}=2a$ and $bz^{2+}az+b\overline{z}^{2+}a\overline{z}=2b$. Subtracting these equations gives $(a-b)(z^{2-}\overline{z}^{2-}(z-\overline{z}))=0$. Since $a \neq b$, we have $z^{2-}\overline{z}^{2-}(z-\overline{z})=0$, which simplifies to $(z-\overline{z})(z+\overline{z}-1)=0$. This implies either $z=\overline{z}$ (real $z$) or $z+\overline{z}=1$ (real part is $\frac{1}{2}$). Testing these in the original real part equations leads to a contradiction, showing there are no such complex numbers satisfying the conditions.

The expression $\frac{z^{2+}8iz-15}{z^{2-}3iz-2}$ belongs to $\mathbb{R}$ when its imaginary part is zero, which simplifies to the condition $1 + \frac{11iz-13}{z^{2-}3iz-2} \in \mathbb{R}$. Requiring the imaginary part of the denominator term to vanish leads to the relation $x^2 - y^2 + 3y - 2 = 0$ for $z = x+iy$. Substituting the given values $x = \alpha$ and $y = -\frac{13}{11}$ into this equation gives $\alpha^2 = (-\frac{13}{11}-1)(-\frac{13}{11}-2)$, which simplifies to $\alpha^2 = \frac{840}{121}$. Multiplying by 242 results in $242\alpha^2 = 1680$.

The intersection of two circles $S_n$ and $T_n$ is empty if the distance $d$ between their centers $(3, -2)$ and $(2, -3)$ satisfies $d > r_1+r_2$ or $d < |r_1-r_2|$. With $d=\sqrt{2}$, the inequality $\sqrt{2} > n/4 + 1/n$ simplifies to $n^2 - 4\sqrt{2}n + 4 < 0$, which holds for $n \in \{1, 2, 3, 4\}$. These four integer values for $n$ result in an empty intersection, matching the specified count.

Substituting $z=x+iy$ into the equation $\overline{z} = iz^2 + z^2 - z$ results in the system $x = x^{2-}y^{2-}2xy$ and $-y = x^{2-}y^{2+}2xy$ when comparing real and imaginary components. Subtracting these equations leads to $2x(1+2y)=0$, which implies $x=0$ or $y=-1/2$. Solving for these cases produces the three complex roots $0$, $\frac{1+\sqrt{2}}{2} - \frac{1}{2}i$, and $\frac{1-\sqrt{2}}{2} - \frac{1}{2}i$. The sum of the squares of the moduli for these points is $0^2 + |\frac{1+\sqrt{2}}{2} - \frac{1}{2}i|^2 + |\frac{1-\sqrt{2}}{2} - \frac{1}{2}i|^2$, which simplifies to 2.

The equation $|z|=3$ represents a circle centered at the origin with radius 3, and the given argument relation describes a circular arc belonging to a circle with center $(0,1)$ and radius $\sqrt{2}$. The distance between these two centers is 1, which is less than the difference of their radii, $3-\sqrt{2}$. Consequently, the smaller circle lies entirely within the larger one, meaning they share no common points.

A complex number $w$ is purely imaginary if $w+\overline{w}=0$ and purely real if $w-\overline{w}=0$. Solving the first condition forces $\sin^2\alpha=1/2$, yielding $\alpha=5\pi/4, 7\pi/4$, while the second condition forces $\cos\beta=0$, yielding $\beta=3\pi/2$. Using these values to determine the set $S$ and evaluating the sum of the complex expressions $(iZ_{\alpha \beta} + 1/(i\overline{Z}_{\alpha \beta}))$ for each pair leads to the total value of $1$.

The set $S_1$ represents a circle with center 3 and radius $1/2$. For $S_2$, expanding the equation $|z_2 - |z_2+1|| = |z_2 + |z_2-1||$ reveals that $z_2$ must either lie on the imaginary axis where $x=0$, or on the real line segment between $-1$ and $1$ where $|z_2+1| + |z_2-1| = 2$. Calculating the distance between these sets, the closest distance between the circle $S_1$ and the segment $[-1, 1]$ occurs between $z_2=1$ and the closest point on the circle, $z_1=2.5$. The minimum distance is $2.5 - 1 = 1.5$, or $3/2$.

Given the relation $\frac{z_1}{z_2} = -i$, the arguments satisfy $\arg(z_1) - \arg(z_2) = -\frac{\pi}{2}$. The condition $\arg\left(\frac{z_1}{\bar{z}_2}\right)=\pi$ simplifies to $\arg(z_1) + \arg(z_2) = \pi$ because $\arg(\bar{z}_2) = -\arg(z_2)$. Solving this system of two linear equations leads to $\arg(z_1) = \frac{\pi}{4}$.

The roots of the equation $z^{2+}z+1=0$ are the complex cube roots of unity, $\omega$ and $\omega^2$. Expanding the summation yields $\sum_{n=1}^{15} (z^{2n} + z^{-2n} + 2(-1)^n)$, where the power sums of $\omega$ over fifteen terms evaluate to zero. Computing the remaining alternating series $2\sum_{n=1}^{15}(-1)^n$ results in $-2$, and taking the absolute value gives $2$.

The substitution $z=x+iy$ transforms the equation $z^{2+}\overline{z}=0$ into $(x^{2-}y^{2+}x)+i(2xy-y)=0$. Equating the real and imaginary parts to zero produces the system $x^{2-}y^{2+}x=0$ and $y(2x-1)=0$. Solving these equations leads to the points $z=0$, $z=-1$, and $z=1/2 \pm i\sqrt{3}/2$. Summing the real and imaginary components of all these roots yields a final total of $0$.

The inequality $1 < |z - (3-2i)| < 4$ describes an annular region centered at $3-2i$ with an inner radius of 1 and an outer radius of 4. Finding the number of complex numbers $z=a+ib$ with integer coordinates $a$ and $b$ involves determining the integer pairs $(X, Y)$ satisfying $1 < X^{2+}Y^2 < 16$, where $X=a-3$ and $Y=b+2$. Counting all valid integer points that satisfy this distance inequality results in a total of 40 elements.

The minimum value of the sum of distances from $z$ to two points $A$ and $B$ is the length of the segment $AB$, which is achieved when $z$ lies on this segment. Given $A(3 ightarrow 2, 0)$ and $B(0, p ightarrow 2)$, the distance squared is $AB^2 = (3 ightarrow 2)^2 + (p ightarrow 2)^2 = 18+2p^2$. Equating this to $(5 ightarrow 2)^2 = 50$ leads to $2p^2=32$, so $p^2=16$, giving $p=4$ as a solution.

For any complex number $z$, the expression $z^n + \bar{z}^n$ simplifies to $2\text{Re}(z^n)$. To find $z^5$ for $z=2+3i$, we first compute the squares: $z^2 = (2+3i)^2 = 4-9+12i = -5+12i$ and $z^4 = (-5+12i)^2 = 25-144-120i = -119-120i$. Multiplying $z^4$ by $z$ yields $z^5 = (-119-120i)(2+3i) = -238 - 357i - 240i + 360 = 122 - 597i$. Taking twice the real part, we get $2(122) = 244$.

The modulus condition $|z|^2=|z| imes 2^{1-|z|}$ requires $|z|=1$ since the complex number $z$ is non-zero. Substituting this into the original equation $z^2=\overline{z} \cdot 2^{1-|z|}$ yields $z^2=\overline{z}$, and multiplying both sides by $z$ results in $z^3=|z|^2=1$. Given the condition $b \neq 0$, the valid solutions for $z$ are $e^{i2\pi/3}$ and $e^{i4\pi/3}$. Substituting these into $(1+1/z)^n=1$ leads to $(1+e^{-i2\pi/3})^n=1$, which simplifies to $(e^{-i\pi/3})^n=1$, requiring the smallest positive integer $n=6$.

The points $z_1, z_2, z_3$ lie on the circle $x^{2+}y^2=25$, and the segment joining $z_2$ and $z_3$ has a slope of $-\frac{1}{2}$. Since the line through $z$ and $z_1$ is perpendicular to this segment, its slope must be $2$, which defines the line $y=2x-2$. Solving for the intersection with the circle identifies the point $z$ as $\left(-\frac{7}{5}, -\frac{24}{5}\right)$, yielding an argument of $\tan^{-1}\left(\frac{24}{7}\right)-\pi$ in the third quadrant.

The set $A$ defines an annulus centered at $1+i$ with inner radius $1$ and outer radius $2$, while the set $B$ represents a circle centered at $1-i$ with radius $1$. Given the distance between the two centers $(1,1)$ and $(1,-1)$ is exactly $2$, the circle $B$ is internally tangent to the boundary of the outer circle of the annulus and intersects the region, meaning the intersection is an arc containing infinitely many points.

The roots of the quadratic equation $x^{2+}(2i-1)=0$ satisfy the relation $x^2 = 1-2i$. The problem asks for $\left| \alpha^8 + \beta^8 \right|$, which is equivalent to $\left| (\alpha^2)^4 + (\beta^2)^4 \right|$. Substituting $\alpha^2 = \beta^2 = 1-2i$ into this expression yields $\left| 2(1-2i)^4 \right|$. Applying the property of the modulus $\left| z^n \right| = \left| z \right|^n$, the value becomes $2 \left| 1-2i \right|^4 = 2(\sqrt{1^{2+}(-2)^2})^4 = 2(\sqrt{5})^4 = 2(25) = 50$.

The set $|z-3| \le 1$ represents a disk centered at $(3,0)$ with radius $1$, while the inequality simplifies to $4x-3y \le 12$. The point within this disk closest to $4i$ is found by moving from the center $(3,0)$ toward $(0,4)$ by a distance of one, which results in the coordinate $\left(\frac{12}{5}, \frac{4}{5}\right)$. Computing $25(\alpha + \beta)$ for $\alpha = \frac{12}{5}$ and $\beta = \frac{4}{5}$ equals $80$.

The expression $v=|z|^{2+}|z-3|^{2+}|z-6i|^2$ is a sum of squared distances that reaches its minimum when $z$ is at the centroid $z_0=1+2i$. Substituting $z=x+iy$ allows $v$ to be written in the form $3(x-1)^{2+}3(y-2)^{2+}30$, which gives $v_0=30$. Evaluating the complex term $|2z_0^{2-}\overline{z}_0^{3+}3|^2$ at $z_0=1+2i$ results in $|8+6i|^2=100$. Adding $v_0^2=900$ to this result gives a total value of $1000$.