Complex Numbers – JEE Main MathematicsPractice Questions & PYQs

Given the relationship $3|z_1| = 2|z_2|$, the ratio $a = \frac{3z_1}{2z_2}$ has a magnitude of $|a| = 1$. By expressing $a$ in polar form as $\cos \theta + i \sin \theta$, the complex number $z = a + \frac{1}{a}$ becomes $(\cos \theta + i \sin \theta) + (\cos \theta - i \sin \theta)$, which simplifies to $2 \cos \theta$. As this resulting expression is purely real, its imaginary part is $0$.

Start by simplifying the expression for $z$ to $z = \frac{(1+i)^2}{a-i} = \frac{2i}{a-i}$, and use the modulus property to set $\frac{|2i|}{|a-i|} = \frac{2}{\sqrt{a^{2+}1}} = \sqrt{2/5}$. Squaring both sides yields $20 = 2(a^{2+}1)$, which implies $a^2=9$, and since $a>0$, we find $a=3$. Substituting $a=3$ back into the simplified expression gives $z = \frac{2i}{3-i} = \frac{-2+6i}{10} = -0.2+0.6i$, so the conjugate $\bar{z}$ is $-\frac{1}{5}-\frac{3i}{5}$.

A complex number is defined as purely imaginary when its real component equals zero. By multiplying the numerator and denominator of $\frac{3+2i\sin\theta}{1-2i\sin\theta}$ by the conjugate of the denominator, $1+2i\sin\theta$, the expression is transformed into a form where the real and imaginary parts can be easily distinguished.

The denominator becomes $1^2 + (2\sin\theta)^2$, simplifying to $1+4\sin^2\theta$, while the numerator expands to $3 + 6i\sin\theta + 2i\sin\theta - 4\sin^2\theta$, which resolves to $(3-4\sin^2\theta) + i(8\sin\theta)$. Equating the real part $3-4\sin^2\theta$ to zero results in $\sin^2\theta = \frac{3}{4}$, meaning $\sin\theta = \pm\frac{\sqrt{3}}{2}$. Within the interval $(-\frac{\pi}{2}, \pi)$, these conditions are satisfied at $\theta = -\frac{\pi}{3}$, $\theta = \frac{\pi}{3}$, and $\theta = \frac{2\pi}{3}$. Summing these values yields $-\frac{\pi}{3} + \frac{\pi}{3} + \frac{2\pi}{3} = \frac{2\pi}{3}$.

A complex number w = rac{z- ext{\alpha}}{z+ ext{\alpha}} is purely imaginary if w + ar{w} = 0, where ar{w} is the complex conjugate. Setting rac{z- ext{\alpha}}{z+ ext{\alpha}} + rac{ar{z}- ext{\alpha}}{ar{z}+ ext{\alpha}} = 0 and clearing denominators results in $2|z|^2 - 2 ext{\alpha}^2 = 0$. Given $|z|=2$, we solve $|z|^2 = ext{\alpha}^2$ to find $ext{\alpha}^2 = 4$, yielding the positive value 2.

Given $z = x + 10i$, substituting this into $\frac{2z-n}{2z+n} = 2i-1$ leads to $(2x-n) + 20i = (2i-1)((2x+n) + 20i)$. Expanding the right side yields $4ix + 2in - 40 - 2x - n - 20i$. Equating real and imaginary parts respectively gives $2x - n = -2x - n - 40$ and $20 = 4x + 2n - 20$. Solving these equations reveals $x = -10$ and $n = 40$, confirming $\operatorname{Re}(z) = -10$ and $n = 40$.

Completing the square for $x^2 + 2x + 2 = 0$ results in $(x+1)^2 = -1$, yielding roots $\alpha = -1+i$ and $\beta = -1-i$. In polar form, these are $\sqrt{2}e^{i3\pi/4}$ and $\sqrt{2}e^{-i3\pi/4}$. Raising these to the fifteenth power, $\alpha^{15} = (\sqrt{2})^{15} e^{i45\pi/4} = 128\sqrt{2} e^{i(11\pi + \pi/4)} = -128 - 128i$, and similarly $\beta^{15} = 128\sqrt{2} e^{-i45\pi/4} = -128 + 128i$. Summing these results gives $\alpha^{15} + \beta^{15} = -256$.

The roots of the quadratic equation $x^2 + x + 1 = 0$ are the non-real cube roots of unity, $\omega$ and $\omega^2$, which satisfy $\omega^3 = 1$. Substituting $\omega$ into the expression $z = 3 + 6i\omega^{81} - 3i\omega^{93}$ simplifies to $z = 3 + 6i(1) - 3i(1) = 3 + 3i$. The argument is found using $\arg(z) = \arctan\left(\frac{3}{3}\right) = \arctan(1) = \frac{\pi}{4}$.

The set of points satisfying $|Z_1|=9$ describes a circle centered at the origin with a radius of 9, while $|Z_2-3-4i|=4$ represents a circle centered at $(3,4)$ with a radius of 4. The distance between the centers is $ext{\sqrt{3^{2+}4^2}}=5$, which is equal to the difference between the two radii, $9-4=5$. Since the distance between centers equals the difference of radii, the circles touch internally, making the minimum distance between any point on the two circles 0.

The geometric interpretation of the equation $|z-z_1|=|z-z_2|$ is the perpendicular bisector of the line segment connecting the two points $z_1$ and $z_2$ in the complex plane. With $z_1=i$ and $z_2=1$, the segment joins $(0,1)$ and $(1,0)$, which has a slope of $-1$. Consequently, the perpendicular bisector must have a slope of $1$ and, as the origin satisfies the original equation, the resulting line passes through the origin.

A complex number $w$ is purely imaginary when $w + \bar{w} = 0$. Substituting $|z|=1$ into the expression $w + \bar{w} = \frac{1+(1-8\alpha)z}{1-z} + \frac{1+(1-8\alpha)/z}{1-1/z}$ and simplifying algebraically leads to the equation $8\alpha = 0$. This condition is satisfied exclusively when $\alpha = 0$.

The condition $|z - (3 - 2i)| \leq 4$ defines a disk in the complex plane centered at $3 - 2i$ with a radius of $4$. The distance from the origin to the center is $|3 - 2i| = \sqrt{3^2 + (-2)^2} = \sqrt{13}$, which is less than the radius of $4$, meaning the origin lies inside the disk. The maximum distance from the origin to any point in the disk is $\sqrt{13} + 4$, and the minimum distance is $0$ because the origin is included, making the difference $\sqrt{13} + 4$.

Simplifying the complex number $\frac{1+i\sqrt{3}}{1-i\sqrt{3}}$ by multiplying the numerator and denominator by the conjugate $1+i\sqrt{3}$ results in $-\frac{1}{2} + i\frac{\sqrt{3}}{2}$, which corresponds to $e^{i2\pi/3}$. Requiring $(e^{i2\pi/3})^n = 1$ implies $n\frac{2\pi}{3}$ must be a multiple of $2\pi$. The smallest positive integer $n$ satisfying this condition is $3$.

The roots of the quadratic equation $x^{2-}x+1=0$ are the complex numbers $e^{i\pi/3}$ and $e^{-i\pi/3}$, which satisfy the property $\alpha^6 = 1$ and $\beta^6 = 1$. The expression $\alpha^{101} + \beta^{107}$ reduces to $\alpha^5 + \beta^5$. Since $\alpha^5 = \beta$ and $\beta^5 = \alpha$, the sum becomes $\alpha + \beta$, which is the sum of the roots of the quadratic equation, equaling $1$.

The determinant involving powers of $\omega$ is simplified using the property $1+\omega+\omega^2=0$ and $\omega^3=1$. Performing the row operation $R_1 \to R_1+R_2+R_3$ transforms the first row into $3, 0, 0$, which allows the determinant to be calculated as $3(\omega^2 - \omega^4) = 3(\omega^2 - \omega)$. Substituting $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $\omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$ results in $3(-i\sqrt{3}) = -3z$, and equating this to $3k$ yields $k = -z$.

Expressing the fraction $\frac{iz-2}{z-i}$ as $i - \frac{3}{z-i}$ allows the given imaginary part equation to be written as $\operatorname{Im}(i - \frac{3}{z-i}) = -1$, which simplifies to $\operatorname{Im}(\frac{3}{z-i}) = 2$. Substituting $z=x+iy$ leads to the equation $x^2 + y^2 - y - 1 = 0$, representing a circle $x^2 + (y - 1/2)^2 = 9/16$. This identifies the radius of the circle as $\sqrt{9/16} = 3/4$.

The locus of a complex number is derived by substituting $z = x+iy$ into the equation $2|x+i(y+3)| = |x+i(y-1)|$ and squaring both sides, which gives $4(x^2 + (y+3)^2) = x^2 + (y-1)^2$. Expanding the terms leads to $4x^2 + 4y^2 + 24y + 36 = x^2 + y^2 - 2y + 1$, which simplifies to $3x^2 + 3y^2 + 26y + 35 = 0$ or $x^2 + y^2 + \frac{26}{3}y + \frac{35}{3} = 0$. The radius of this circle is calculated using $\sqrt{g^2 + f^2 - c}$ as $\sqrt{0^2 + (\frac{13}{3})^2 - \frac{35}{3}} = \sqrt{\frac{169}{9} - \frac{105}{9}} = \sqrt{\frac{64}{9}} = \frac{8}{3}$.

A complex number $w$ is purely imaginary if its real part is zero, which is determined by multiplying the numerator and denominator of the given expression by the conjugate of the denominator, $1+2i\sin\theta$. Expanding the product gives a numerator where the real part is $2 - 6\sin^2\theta$, and setting this real part to zero leads to $6\sin^2\theta = 2$. Consequently, $\sin^2\theta = \frac{1}{3}$, which results in $\theta = \sin^{-1}(\frac{1}{\sqrt{3}})$.

For a complex number to be unimodular, its magnitude must be 1, so squaring the given condition $\left|\frac{z_1-2z_2}{2-\overline{z_1}z_2}\right|=1$ allows us to equate the square of the magnitude to 1, or $|z_1-2z_2|^2 = |2-\overline{z_1}z_2|^2$. Expanding both sides leads to $|z_1|^2 + 4|z_2|^2 - 2z_1\overline{z_2} - 2\overline{z_1}z_2 = 4 + |z_1|^2|z_2|^2 - 2z_1\overline{z_2} - 2\overline{z_1}z_2$. Simplifying this equation results in $(|z_1|^{2-}4)(1-|z_2|^2)=0$, and because $z_2$ is not unimodular, we conclude $|z_1|=2$, representing a circle of radius 2.

The triangle inequality $|z_1 + z_2| extgreater= ||z_1| - |z_2||$ provides the lower bound for the magnitude of a complex sum. Given $|z| extgreater= 2$, substituting into this inequality gives |z + rac{1}{2}| extgreater= ||z| - |-rac{1}{2}|| extgreater= 2 - rac{1}{2} = rac{3}{2}. This minimum value of rac{3}{2} falls within the interval $(1,2)$.

Utilizing the identity ar{z} = rac{1}{z} for any complex number with unit modulus, the expression rac{1+z}{1+ar{z}} simplifies significantly. Substituting the identity into the denominator yields rac{1+z}{1+1/z}, which simplifies to rac{1+z}{(z+1)/z} = z. Since the argument of the result is equivalent to the argument of $z$, the value is $heta$.