Complex Numbers – JEE Main MathematicsPractice Questions & PYQs

A complex number $w$ is purely real if its imaginary part equals zero. Expressing $z$ as $x+iy$ and simplifying rac{z^2}{z-1} by multiplying the numerator and denominator by the conjugate of $z-1$, the imaginary component is found to be proportional to $y(x^{2+}y^{2-}2x)$. Setting this to zero dictates that the locus is either the real axis where $y=0$ or the circle defined by $x^{2+}y^{2-}2x=0$, which passes through the origin.

The roots of a quadratic equation with real coefficients must occur in conjugate pairs. Since the roots lie on the line $ext{Re} z = 1$, they must be of the form $1+\pi$ and $1-\pi$ for some non-zero real number $p$. The product of these roots, given by the constant term \eta, is $(1+\pi)(1-\pi) = 1+p^2$, and since $p^2 > 0$, it follows that \eta > 1.

The identity $1+ ext{\omega }+ ext{\omega }^2=0$ for the cube root of unity allows the substitution $1+ ext{\omega } = - ext{\omega }^2$. Raising this to the seventh power results in $(- ext{\omega }^2)^7 = - ext{\omega }^{14}$, and because $ext{\omega }^3=1$, this further simplifies to $- ext{\omega }^2$. Applying the identity $- ext{\omega }^2 = 1+ ext{\omega }$ shows that the coefficients are $A=1$ and $B=1$.

The condition $|z-z_1| = |z-z_2|$ geometrically defines the perpendicular bisector of the line segment connecting $z_1$ and $z_2$ in the complex plane. The equation $|z-1|=|z+1|$ corresponds to the perpendicular bisector of the points $1$ and $-1$, which is the imaginary axis where $\operatorname{Re}(z) = 0$. Similarly, $|z-1|=|z-i|$ corresponds to the perpendicular bisector of $1$ and $i$, which is the line $y=x$. Since the only complex number satisfying both $x=0$ and $y=x$ is $z=0$, there is exactly $1$ such complex number.

The reverse triangle inequality $|a-b| \geq ||a|-|b||$ provides a lower bound for the difference of two complex numbers in terms of their magnitudes. Applying this to the given equation, we have $2 = |z - \frac{4}{z}| \geq |z| - |\frac{4}{z}|$, which simplifies to $|z| - \frac{4}{|z|} \leq 2$. Multiplying by $|z|$ creates the quadratic inequality $|z|^2 - 2|z| - 4 \leq 0$. Solving the equality $|z|^2 - 2|z| - 4 = 0$ using the quadratic formula yields $|z| = \frac{2 \pm \sqrt{4 - 4(1)(-4)}}{2} = 1 \pm \sqrt{5}$, and since $|z|$ must be positive, the maximum value is $1 + \sqrt{5}$.

The conjugate of a quotient is equivalent to the quotient of the conjugates, denoted as $\overline{(\frac{z_1}{z_2})} = \frac{\bar{z}_1}{\bar{z}_2}$. For the expression $\frac{1}{i-1}$, the numerator is $1$ (which is its own conjugate) and the conjugate of the denominator $i-1$ is $-i-1$. This results in $\frac{1}{-i-1}$, which can be rewritten as $\frac{1}{-(i+1)}$ or $\frac{-1}{i+1}$.

The inequality $|z - (-4)| \leq 3$ defines a closed disk of radius $3$ centered at $-4$ on the real axis in the complex plane. The expression $|z - (-1)|$ represents the distance from any point $z$ within this disk to the point $-1$. This distance is maximized at the boundary point of the disk furthest from $-1$, which is at $z = -4 - 3 = -7$. Calculating the distance at this point gives $|-7 - (-1)| = |-6| = 6$.

The sum of the $n$-th roots of unity $\sum_{k=0}^{n-1} e^{i \frac{2k\pi}{n}}$ is always zero. By rewriting the given expression as $i \sum_{k=1}^{10} (\cos \frac{2k\pi}{11} - i \sin \frac{2k\pi}{11})$, we recognize it as the geometric series $i \sum_{k=1}^{10} e^{-i \frac{2k\pi}{11}}$. Since the full sum from $k=0$ to $10$ equals zero, the sum from $k=1$ to $10$ must be $-1$ times the $k=0$ term, which is $1$. Multiplying this resulting sum of $-1$ by the factor $i$ yields the final value of $-i$.

The equation $z^{2+}z+1=0$ indicates that $z$ is a complex cube root of unity, either $\omega$ or $\omega^2$, where $z^3=1$ and $\frac{1}{z} = z^2$. For $z=\omega$, the terms evaluate to $(z+\frac{1}{z})^2 = (\omega+\omega^2)^2 = (-1)^2 = 1$, $(z^{2+}\frac{1}{z^2})^2 = (\omega^{2+}\omega)^2 = (-1)^2 = 1$, and $(z^{3+}\frac{1}{z^3})^2 = (1+1)^2 = 4$. Because these values repeat cyclically for higher powers, the sum becomes $1+1+4+1+1+4$, which equals $12$.

The condition $|\omega|=1$ implies that the modulus of the numerator equals the modulus of the denominator, specifically $|z| = |z - \frac{1}{\sqrt{3}}i|$. Geometrically, $|z|$ represents the distance of point $z$ from the origin, while $|z - \frac{1}{\sqrt{3}}i|$ represents the distance from the point $(0, \frac{1}{\sqrt{3}})$. Since $z$ is equidistant from these two distinct fixed points, it must lie on the perpendicular bisector of the line segment joining them, which forms a straight line.

Rearranging the equation $(x-1)^3 = -8$ allows us to solve for $x-1$ by taking the cube root, yielding $x-1 = -2\sqrt[3]{1}$. Since the cube roots of unity are $1, \omega$, and $\omega^2$, multiplying these by $-2$ gives the possible values for $x-1$ as $-2, -2\omega$, and $-2\omega^2$. Adding $1$ to each of these values reveals the solutions for $x$ are $-1, 1-2\omega$, and $1-2\omega^2$.

The equation $|z_1+z_2| = |z_1| + |z_2|$ represents the equality case of the triangle inequality, which occurs if and only if the vectors represented by $z_1$ and $z_2$ are collinear and point in the same direction. When two complex numbers have the same argument, they effectively lie on the same ray originating from the origin. Consequently, the difference between their arguments must be zero, as $\arg z_1 = \arg z_2$.

Starting from the given equation $\overline{z}+i \overline{w}=0$, taking the complex conjugate of both sides yields $z-i w=0$, which simplifies to $z=iw$. Substituting this relationship into the argument condition, we have $\arg(z w) = \arg(i w^2) = \arg(i) + 2\arg(w) = \pi$. Since $\arg(i) = \frac{\pi}{2}$, the equation becomes $2\arg(w) = \frac{\pi}{2}$, leading to $\arg(w) = \frac{\pi}{4}$. Because $z = iw = e^{i\pi/2} w$, its argument is $\arg(z) = \arg(i) + \arg(w) = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.

To express $z$ in terms of $p$ and $q$, we expand $z = (p+iq)^3$, which results in $p^3 + 3p^2(iq) + 3p(iq)^2 + (iq)^3 = p^3 + 3ip^2q - 3pq^2 - iq^3$. Grouping the real and imaginary parts, we obtain $z = p(p^2 - 3q^2) - iq(q^2 - 3p^2)$. Given that $z = x - iy$, equating the components gives $\frac{x}{p} = p^2 - 3q^2$ and $\frac{y}{q} = q^2 - 3p^2$. Substituting these into the requested expression $\frac{\frac{x}{p} + \frac{y}{q}}{p^{2+}q^2}$ simplifies to $\frac{p^2 - 3q^2 + q^2 - 3p^2}{p^{2+}q^2} = \frac{-2p^2 - 2q^2}{p^{2+}q^2} = -2$.

Squaring both sides of the equation $|z^{2-}1| = |z|^{2+}1$ allows the use of the property $|w|^2 = w\bar{w}$ to expand the modulus into $(z^{2-}1)(\bar{z}^{2-}1) = (|z|^{2+}1)^2$. Expanding the left side to $|z|^4 - z^2 - \bar{z}^2 + 1$ and the right side to $|z|^4 + 2|z|^2 + 1$ reveals that $-z^2 - \bar{z}^2 = 2|z|^2$, or $z^2 + 2z\bar{z} + \bar{z}^2 = 0$. This simplifies to $(z+\bar{z})^2 = 0$, implying $z = -\bar{z}$, which identifies $z$ as a purely imaginary number that lies on the imaginary axis.

Representing the complex numbers in polar form as $z = r_1 e^{i\theta_1}$ and $\omega = r_2 e^{i\theta_2}$, the condition $|z\omega|=1$ implies $r_1r_2=1$. The second condition $\arg(z) - \arg(\omega) = \theta_1 - \theta_2 = \frac{\pi}{2}$ implies that the difference in their angles is $\frac{\pi}{2}$. When calculating $\overline{z}\omega$, we get $(r_1 e^{-i\theta_1})(r_2 e^{i\theta_2}) = (r_1r_2) e^{i(\theta_2 - \theta_1)}$. Since $r_1r_2=1$ and $\theta_2 - \theta_1 = -\frac{\pi}{2}$, the expression evaluates to $e^{-i\pi/2}$, which corresponds to $-i$.

For any three complex numbers to form an equilateral triangle, they must satisfy the condition $z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1$. With the vertices at the origin, $z_1$, and $z_2$, the equation becomes $0 + z_1^2 + z_2^2 = 0 + z_1z_2 + 0$, or $z_1^2 + z_2^2 = z_1z_2$. Expressing the left side as $(z_1+z_2)^2 - 2z_1z_2 = z_1z_2$, we find that $(z_1+z_2)^2 = 3z_1z_2$. Applying Vieta's formulas for the quadratic equation $z^2 + az + b = 0$, where $z_1+z_2 = -a$ and $z_1z_2 = b$, we arrive at $(-a)^2 = 3b$, or $a^2 = 3b$.

Simplifying the term inside the parenthesis requires multiplying by the conjugate, leading to $\frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+2i+i^2}{1-i^2} = \frac{2i}{2} = i$. The equation then reduces to $i^x = 1$. The powers of the imaginary unit $i$ follow a cyclic pattern where $i^1=i$, $i^2=-1$, $i^3=-i$, and $i^4=1$. Therefore, for $i^x$ to equal 1, the exponent $x$ must be an integer multiple of 4, denoted as $x=4n$.

The modulus inequality $|z-4| < |z-2|$ signifies that the distance of $z$ from $4$ is less than its distance from $2$ on the complex plane. Substituting $z = x + iy$ yields the squared inequality $(x-4)^2 + y^2 < (x-2)^2 + y^2$, which simplifies to $x^2 - 8x + 16 < x^2 - 4x + 4$. Solving for $x$ results in $12 < 4x$, or simply $x > 3$, implying $ext{Re}(z) > 3$.

The locus of a point is defined as a hyperbola if the absolute difference of its distances from two fixed foci remains constant. By expressing the distance of the moving center $z$ from the centers of the two given circles $z_1$ and $z_2$, we find $|z-z_1| = r+a$ and $|z-z_2| = r+b$. Eliminating the variable radius $r$ through subtraction gives $|z-z_1| - |z-z_2| = a-b$, where $a-b$ is a constant. This geometric condition directly satisfies the definition of a hyperbola.