Complex Numbers – JEE Main MathematicsPractice Questions & PYQs

Substituting z=3+iy, where y is a real variable, allows the expression z-conj(z)+zconj(z) to be written as 2iy+9+y^2, while the denominator 2-3z+5conj(z) becomes 8-8iy. The real part of the resulting complex number simplifies to (1/8)((10/(1+y^2))-1). Since y ranges over all real values, (1+y^2) ranges over $[1, infinity), meaning 10/(1+y^2) ranges over (0, 10]$, which maps the real part to the interval (-1/8, 9/8]. Thus, alpha=-1/8 and beta=9/8, resulting in 24(9/8 - (-1/8)) = 30.

Rotating $z_1 = 5+4i$ by $90^\circ$ anticlockwise is achieved by multiplying by $i$, resulting in $w_1 = (5+4i)i = -4+5i$. Rotating $z_2 = 3+5i$ by $90^\circ$ clockwise corresponds to multiplying by $-i$, yielding $w_2 = (3+5i)(-i) = 5-3i$. The difference $w_1 - w_2 = -9+8i$ is a complex number in the second quadrant. The principal argument is calculated as $\pi - \tan^{-1}(\frac{8}{9})$.

To divide complex numbers, use the exponential form of the denominator where $\cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = e^{i\pi/3}$. The expression $z = \frac{i-1}{e^{i\pi/3}}$ simplifies to $(i-1)e^{-i\pi/3}$. Converting $i-1$ to polar form gives $\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})$, and multiplying by $e^{-i\pi/3}$ involves adding the arguments: $\frac{3\pi}{4} - \frac{\pi}{3} = \frac{5\pi}{12}$. Thus, $z = \sqrt{2}(\cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12})$.

The collinearity condition implies that the complex number $z_2$ can be expressed as $z_0 + k(z_1 - z_0)$ for some real $k$. Given the relation $|z_1 - z_0| |z_2 - z_0| = 1$ and $|z_1 - z_0| = |(1+i) - \frac{1}{2}(1+3i)| = |\frac{1}{2} - \frac{1}{2}i| = \frac{1}{\sqrt{2}}$, we find that $|z_2 - z_0| = \sqrt{2}$. Since $z_2$ lies on the line passing through $z_0$ and $z_1$, its coordinates are determined by translating $z_0$ by $\pm \sqrt{2}$ in the direction of the vector $z_1 - z_0 = (\frac{1}{2}, -\frac{1}{2})$, which yields the two possible positions $(-\frac{1}{2}, \frac{5}{2})$ and $(\frac{3}{2}, \frac{1}{2})$. Calculating the squared modulus $|z_2|^2$ for these positions gives $\frac{26}{4}$ and $\frac{5}{2}$, from which the smaller value is $\frac{5}{2}$.

The condition $\operatorname{Re}(a+\overline{z}) > \operatorname{Im}(\overline{a}+z)$ is equivalent to $x_1+x > y-y_1$, where $z=x+iy$ and $a=x_1+iy_1$. This inequality defines a half-plane in the complex plane, which does not contain all real numbers since the condition imposes a constraint on the relationship between the real and imaginary parts of $z$. Testing the condition with specific values for $a$ and $z$ demonstrates that neither (S1) nor (S2) holds for all real numbers, as the inequality boundaries exclude parts of the real axis. Consequently, both statements are false.

A complex fraction $\frac{z_1}{z_2}$ is real if and only if its numerator and denominator are proportional, which implies the imaginary part of $z_1 \overline{z_2}$ must be zero. For $z_1 = 2z-3i$ and $z_2 = 4z+2i$, writing $z=x+iy$ and equating the imaginary part of the product to zero leads to $x(2y+5) = 0$, implying $x=0$ (excluding the point where the denominator is zero). Because the denominator $4z+2i$ is zero at $z = -\frac{1}{2}i$, the point $(x,y) = (0, -1/2)$ is excluded from the set, making the statement that this point is included incorrect.

The equation simplifies by setting $z = x+iy$ and recognizing $\alpha z - \overline{\alpha}\overline{z} = 2i \operatorname{Im}(\alpha z)$. Given $\alpha = 8-14i$, the numerator becomes $2i(8y-14x)$ and the denominator is $z^2 - \overline{z}^2 = 4ixy - 112i$, which simplifies the locus to $(x-4)(y+7)=0$. Intersecting the lines $x=4$ and $y=-7$ with the circle $|z+3i|=4$, we find the intersection points $4-3i$ and $0-7i$. Summing $\operatorname{Re} z - \operatorname{Im} z$ for these points yields $(4 - (-3)) + (0 - (-7)) = 7+7=14$.

For a complex number $z$ on the circle $|z|=4$, writing $z=4e^{i\theta}$ allows the expression $z+\frac{1}{z}$ to be represented as $4(\cos\theta+i\sin\theta) + \frac{1}{4}(\cos\theta-i\sin\theta)$. This sum simplifies to $(\frac{17}{4}\cos\theta) + i(\frac{15}{4}\sin\theta)$, which describes an ellipse with a semi-major axis of $4.25$ and a semi-minor axis of $3.75$. Since the ellipse spans values outside the circle $|z|=4$ along the real axis but remains inside along the imaginary axis, the ellipse and circle must intersect at four points.

Given $\left|\frac{1+ai}{b+i}\right|=1$, the modulus property implies $|1+ai|=|b+i|$, leading to $1+a^2=b^{2+}1$ or $a^2=b^2$, which forces $b=-a$ since $ab<0$. Substituting $z=a-ia$ into the locus $|z-1|=2|z|$ yields $(a-1)^{2+}a^2=4(2a^2)$, which simplifies to the quadratic $6a^{2+}2a-1=0$. Solving for $a$ gives $a=\frac{-1 \pm \sqrt{7}}{6}$. Choosing the root $a=\frac{-1-\sqrt{7}}{6}$ where $[a]=-1$, the expression $\frac{1+[a]}{4b}$ becomes $\frac{1-1}{4b}=0$.

The real part Re(omega)=0 describes a circle with center (1,2) and radius 1, which identifies the coefficients as k_1=-2, k_2=4, and lambda=4. Substituting these into the imaginary part Im(omega)=0 gives the equation 4x-2y+4=0, or 2x-y+2=0, for the intersecting line. The perpendicular distance from the center (1,2) to this line is |2(1)-2+2|/sqrt(2^{2+}(-1)^2) = 2/sqrt(5), and the squared chord length is 4(r^{2-}d^2) = 4(1-4/5) = 4/5, making 30(AB)^2=24.

For the equation $|z+2|=z+4(1+i)$, since the modulus is a real value, the imaginary part of the right side must be zero, implying $\beta+4=0$ or $\beta=-4$. Substituting this back into the equation $|(\alpha+2)+i\beta|=(\alpha+4)+i(\beta+4)$ simplifies to $\sqrt{(\alpha+2)^{2+}16}=\alpha+4$. Squaring both sides yields $(\alpha+2)^{2+}16=(\alpha+4)^2$, which simplifies to $4\alpha=4$, giving $\alpha=1$. The roots of the quadratic equation are the sum $\alpha+\beta=-3$ and product $\alpha\beta=-4$, resulting in the quadratic $x^{2-}(-3-4)x+(-3)(-4)=0$, which simplifies to $x^{2+}7x+12=0$.

Expressing the condition $\left|\frac{z-2}{z-3}\right|=2$ in Cartesian coordinates $z=x+iy$ leads to the equation $(x-2)^{2+}y^2=4((x-3)^{2+}y^2)$. Expanding the terms yields $x^{2-}4x+4+y^2=4(x^{2-}6x+9+y^2)$, which simplifies to $3x^{2+}3y^{2-}20x+32=0$. Dividing by $3$ gives the circle equation $x^{2+}y^{2-}\frac{20}{3}x+\frac{32}{3}=0$, identifying the center $(\alpha, \beta) = (\frac{10}{3}, 0)$ and radius $\gamma = \sqrt{(\frac{10}{3})^2 - \frac{32}{3}} = \frac{2}{3}$. The value $3(\alpha+\beta+\gamma) = 3(\frac{10}{3}+0+\frac{2}{3}) = 12$.

The expression can be simplified by identifying the modulus of the term $w = \sin(2\pi/9) + i\cos(2\pi/9)$ as 1, which implies $\overline{w} = 1/w$. Substituting this into the given expression yields $(\frac{1+w}{1+1/w})^3 = w^3$. Using polar form, $w = i(\cos(2\pi/9) - i\sin(2\pi/9)) = i e^{-i(2\pi/9)}$, so $w^3 = i^3 e^{-i(2\pi/3)} = -i(\cos(2\pi/3) - i\sin(2\pi/3))$, which evaluates to $-\frac{1}{2}(\sqrt{3}-i)$.

The expression $|\frac{z-2i}{z+i}|=2$ can be expanded by squaring both sides, resulting in $(z-2i)(\overline{z}+2i) = 4(z+i)(\overline{z}-i)$. Expanding these products leads to $z\overline{z} + 2i(z-\overline{z}) + 4 = 4z\overline{z} - 4i(z-\overline{z}) + 4$. Setting $z=x+iy$ reduces this to $3(x^{2+}y^2) + 12y = 0$, or $x^2 + (y+2)^2 = 4$, which is a circle with radius 2 centered at $(0,-2)$.

For complex numbers $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$, the condition $\operatorname{Re}(z_1+z_2)=0$ implies $x_1+x_2=0$, meaning $x_2=-x_1$. Substituting this into $\operatorname{Re}(z_1 z_2) = x_1 x_2 - y_1 y_2 = 0$ gives $-x_1^2 = y_1 y_2$, which indicates that $y_1$ and $y_2$ must have opposite signs. Thus, one of the imaginary parts is positive and the other is negative, which corresponds to options B and C.

The expression $\frac{1+2i\sin\theta}{1-i\sin\theta}$ is purely imaginary when the real part of its product with the conjugate of the denominator is zero. Multiplying the numerator by the conjugate $1+i\sin\theta$ gives a real part of $1-2\sin^2\theta$, which must equal zero. Solving $1-2\sin^2\theta=0$ gives $\sin^2\theta=\frac{1}{2}$, yielding $\theta$ values of $\frac{\pi}{4}$, $\frac{3\pi}{4}$, $\frac{5\pi}{4}$, and $\frac{7\pi}{4}$ within the interval $(0, 2\pi)$. The sum of these four angles is $\frac{16\pi}{4}=4\pi$.

Simplifying $z_1 = \frac{1+i \overline{z}}{\overline{z}(1-z) + \frac{1}{z}}$ with $z=1+i$ and $\overline{z}=1-i$ requires calculating the components separately. The numerator evaluates to $1+i(1-i) = 2+i$, while the denominator simplifies to $(1-i)(-i) + \frac{1-i}{2} = -0.5-1.5i$. Dividing $2+i$ by $-0.5-1.5i$ is equivalent to $\frac{4+2i}{-1-3i}$, which upon rationalization yields $-1+i$. Since the argument of $-1+i$ is $\frac{3\pi}{4}$, the final result is $\frac{12}{\pi} \times \frac{3\pi}{4} = 9$.

The expression $|z-\alpha|^{2+}|z-\beta|^2=2\lambda$ represents a circle centered at the midpoint $c=\frac{\alpha+\beta}{2}$. Expanding the moduli gives $2|z-c|^{2+}\frac{1}{2}|\alpha-\beta|^2=2\lambda$, which rearranges to $|z-c|^2=\lambda-\frac{1}{4}|\alpha-\beta|^2$. Comparing this to the standard circle form $|z-c|^2=r^2$ where $r^2=\lambda-1$, we identify that $\frac{1}{4}|\alpha-\beta|^2=1$. Therefore, $|\alpha-\beta|^2=4$, which means $|\alpha-\beta|=2$.

A complex number is purely imaginary if its real part is zero. For the expression $\frac{2z-3i}{2z+i}$, setting the real part of the product of the numerator and the conjugate of the denominator to zero yields $4x^{2+}4y^{2-}4y-3=0$. Given the constraint $x+y^2=0$, we substitute $x=-y^2$ into this equation, which results in $4(-y^2)^{2+}4y^{2-}4y-3=0$. Simplifying this leads to $4y^{4+}4y^{2-}4y=3$, and dividing by four reveals $y^{4+}y^{2-}y=\frac{3}{4}$.

The distance formula in the complex plane states that $|z-z_0|^2 = (x-x_0)^2 + (y-y_0)^2$. Substituting $z_1=2+3i$ and $z_2=3+4i$ into the equation $|z-z_1|^2 - |z-z_2|^2 = |z_1-z_2|^2$ results in $((x-2)^2 + (y-3)^2) - ((x-3)^2 + (y-4)^2) = |(2-3) + (3-4)i|^2$. Simplifying the quadratic terms reveals a linear equation $2x+2y=14$, which represents a straight line with intercepts at $x=7$ and $y=7$, summing to 14.