Complex Numbers – JEE Main MathematicsPractice Questions & PYQs

For a quadratic equation with real coefficients, complex roots must be conjugate pairs, so the product of the roots $z$ and $\bar{z}$ is $45$ and the sum is $-b$. Substituting $z = x+iy$ into $|z+1|^2 = 40$ gives $(x+1)^2 + y^2 = 40$, which expands to $x^{2+}y^2 + 2x + 1 = 40$. Using the property that $x^{2+}y^2$ equals the product of the roots, we get $45 + 2x + 1 = 40$, leading to $x = -3$. Since the real part $x$ of the root -rac{b}{2} equals $-3$, we find $b = 6$, and the required value $b^{2-}b$ is $6^{2-}6 = 30$.

The condition $|z-z_1| = |z-z_2|$ represents the perpendicular bisector of the line segment joining $z_1$ and $z_2$ in the complex plane. Given $|z-i| = |z-(-2i)|$, the complex number $z$ must lie on the horizontal line ext{Im}(z) = -rac{1}{2}. By substituting z = x - rac{i}{2} into the magnitude equation |z| = rac{5}{2}, we obtain x^2 + rac{1}{4} = rac{25}{4}, which simplifies to $x^2 = 6$. The value of |z+3i| = |x + rac{5}{2}i| is then calculated as \sqrt{x^2 + rac{25}{4}} = \sqrt{6 + rac{25}{4}} = \frac{7}{2}.

Expressing the complex number $3 \pm 2\sqrt{-54}$ as $3 \pm 6i\sqrt{6}$ allows us to determine its square root, resulting in $\pm(3 \pm i\sqrt{6})$. Calculating the difference of these square roots yields values of $\pm 2i\sqrt{6}$ and $\pm 6$. The imaginary part of these results is $\pm 2\sqrt{6}$, so $-2\sqrt{6}$ is a possible value.

The locus is determined by substituting $z = x+iy$ into the expression and isolating the real part. Multiplying the numerator and denominator by the conjugate of the denominator, $2x - i(2y+1)$, results in a real part of $\frac{2x^2 + 2y^2 - 2x + y}{4x^2 + (2y+1)^2}$. Setting this equal to 1 leads to the equation $2x^2 + 2y^2 + 2x + 3y + 1 = 0$, which defines a circle. The radius of this circle is $\frac{\sqrt{5}}{4}$, which means the diameter is $\frac{\sqrt{5}}{2}$.

The term $\alpha = \frac{-1+i\sqrt{3}}{2}$ represents the primitive cube root of unity $\omega$, which follows the identity $\omega^{2+}\omega+1=0$ and $\omega^3=1$. Expanding $(2+\omega)^4$ via the binomial theorem yields $16 + 32\omega + 24\omega^2 + 8 + \omega$, and substituting $\omega^2 = -1-\omega$ reduces this expression to $24+33\omega -24-24\omega = 9\omega$. Matching this with $a+b\omega$ identifies $a=0$ and $b=9$, so their sum is $9$.

The equation $|Re(z)| + |Im(z)| = 4$ describes a square in the complex plane with vertices at $(\pm 4, 0)$ and $(0, \pm 4)$. The distance from the origin $|z|$ is minimized at the midpoints of the sides where $|x|=|y|=2$, yielding $|z| = \sqrt{2^{2+}2^2} = \sqrt{8}$, and maximized at the vertices where $|z|=4 = \sqrt{16}$. Since the range of $|z|$ is $[\sqrt{8}, 4]$, the value $\sqrt{7}$ is impossible.

Substituting $z=x+iy$ into the equation $z^2=i|z|^2$ transforms it to $(x^{2-}y^2)+i(2xy)=i(x^{2+}y^2)$. Comparing the real and imaginary parts, we must have $x^{2-}y^2=0$ and $2xy=x^{2+}y^2$. The first equation implies $x^2=y^2$, and substituting this into the second yields $2xy=2x^2$, which simplifies to $x(x-y)=0$, confirming that $z$ lies on the line $y=x$.

A complex number $\frac{z_1}{z_2}$ is real if the imaginary part of the fraction vanishes, which leads to the condition $3(-\cos\theta) - 4\sin\theta = 0$, implying $\tan\theta = -\frac{3}{4}$. We seek the argument of $\sin\theta + i\cos\theta$; for $\theta$ in the fourth quadrant where $\cos\theta = \frac{4}{5}$ and $\sin\theta = -\frac{3}{5}$, the complex number becomes $-\frac{3}{5} + i\frac{4}{5}$. The argument of this value is $\pi - \tan^{-1}\frac{4}{3}$.

The condition $\Re(z) = |z-1|$ defines a parabola $y^2 = 2x - 1$ with focus at $(1, 0)$ and the imaginary axis as the directrix. For any two points $z_1$ and $z_2$ on this parabola, the slope of the chord connecting them is given by $\frac{2}{y_1+y_2}$, where $y_1$ and $y_2$ are the imaginary parts of $z_1$ and $z_2$. Given the argument of $z_1-z_2$ is $\frac{\pi}{6}$, the slope is $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$, which implies $\frac{2}{y_1+y_2} = \frac{1}{\sqrt{3}}$, leading to $y_1+y_2 = 2\sqrt{3}$.

Simplifying the given expression involves properties of the cube root of unity $\omega = \frac{-1+i\sqrt{3}}{2}$. Recognizing the numerator as $2\omega$, the expression becomes $\left(\frac{2\omega}{1-i}\right)^{30} = \frac{2^{30}\omega^{30}}{(1-i)^{30}}$. Since $\omega^3 = 1$ and $(1-i)^2 = -2i$, raising $(1-i)$ to the power $30$ gives $(-2i)^{15} = -2^{15}i^{15} = 2^{15}i$. Thus, the expression results in $\frac{2^{30}}{2^{15}i} = \frac{2^{15}}{i} = -2^{15}i$.

Simplifying the bases first, $\frac{1+i}{1-i}$ reduces to $i$, and $\frac{1+i}{i-1}$ reduces to $-i$. The given equation becomes $i^{m/2} = 1$ and $(-i)^{n/3} = 1$, which requires $m/2$ to be a multiple of $4$ and $n/3$ to be a multiple of $4$. This gives the least values $m=8$ and $n=12$, so their greatest common divisor is $4$.

To evaluate the region, express the complex number $z$ as $x+iy$, where $|z| = \sqrt{x^{2+}y^2}$ and $\operatorname{Re}(z) = x$. The given inequality $\sqrt{x^{2+}y^2} - x \le 1$ rearranges to $\sqrt{x^{2+}y^2} \le 1+x$, and squaring both sides gives $x^{2+}y^2 \le (1+x)^2 = 1+2x+x^2$. This simplifies to $y^2 \le 2x+1$, which can be factored to $y^2 \le 2(x + \frac{1}{2})$.

The expansion of the cubic expression $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ allows for direct algebraic simplification. By computing $-\left(2+\frac{1}{3}i\right)^3$, we obtain $-(8 + 4i - \frac{2}{3} - \frac{1}{27}i)$, which simplifies to $-\frac{22}{3} - \frac{107}{27}i$. Expressing this with a common denominator of 27 results in $\frac{-198 - 107i}{27}$, implying $x = -198$ and $y = -107$, yielding $y - x = -107 - (-198) = 91$.

The product $\bar{z}w$ can be determined by expressing $z$ and $w$ in polar form as $z=r_1 e^{i\theta_1}$ and $w=r_2 e^{i\theta_2}$. Given $|zw|=r_1r_2=1$ and the angle condition $\theta_1 - \theta_2 = \pi/2$, the product $\bar{z}w$ becomes $r_1 e^{-i\theta_1} \cdot r_2 e^{i\theta_2} = (r_1 r_2) e^{i(\theta_2 - \theta_1)} = 1 \cdot e^{-i\pi/2}$, which evaluates to $-i$.

Let the magnitude of the complex number be $r=|z|$, allowing us to rewrite the equation as $z = 3-r+i$. Substituting this into the modulus definition $r = ext{\sqrt{\text{Re}(z)^2 + \text{Im}(z)^2}}$ gives $r = ext{\sqrt{(3-r)^{2+}1^2}}$. Squaring both sides produces the equation $r^2 = 9 - 6r + r^2 + 1$, which simplifies to $6r=10$, providing r = rac{5}{3}.

The set of points is defined by the modulus of the expression $z = \frac{\alpha+i}{\alpha-i}$. Evaluating this for any real $\alpha$, we find $|z| = \frac{|\alpha+i|}{|\alpha-i|} = \frac{\sqrt{\alpha^{2+}1}}{\sqrt{\alpha^{2+}(-1)^2}} = 1$. Since the modulus remains $1$ for all $\alpha$, these points form a circle centered at the origin with a radius of $1$.

To determine the relationship, rearrange the equation to express $z$ in terms of $\omega$, yielding $z = \frac{5\omega-5}{5\omega+3}$. The condition $|z|<1$ implies $|5\omega-5| < |5\omega+3|$, which geometrically requires the distance of $\omega$ from $1$ to be less than its distance from $-3/5$. Expanding the squared modulus inequality $25|\omega-1|^2 < 25|\omega+3/5|^2$ simplifies to the linear inequality $\operatorname{Re}(\omega) > 0.2$, which corresponds to $5\operatorname{Re}(\omega) > 1$.

The roots of the equation $x^{2}-2x+2=0$ are obtained using the quadratic formula as $1 ext{\pm} i$. Forming the ratio rac{ ext{\alpha}}{ ext{\beta}} = rac{1+i}{1-i} and multiplying the numerator and denominator by the conjugate $1+i$ yields $i$. Given the condition (rac{ ext{\alpha}}{ ext{\beta}})^{n}=1, we solve for the smallest integer $n$ such that $i^{n}=1$, which is 4.

Euler's formula $e^{i\theta} = \cos \theta + i \sin \theta$ simplifies complex powers, where $\frac{\sqrt{3}}{2} + \frac{i}{2} = e^{i\pi/6}$ and $\frac{\sqrt{3}}{2} - \frac{i}{2} = e^{-i\pi/6}$. Raising these to the fifth power gives $e^{i5\pi/6} + e^{-i5\pi/6}$, which simplifies via the identity $2\cos\theta$ to $2\cos\left(\frac{5\pi}{6}\right)$. Since $\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$, the total value is $z = -\sqrt{3}$, meaning the imaginary part $I(z)$ is $0$.

The complex number z=rac{ ext{\sqrt{3}}}{2}+rac{i}{2} is equivalent to $e^{i ext{\pi}/6}$, implying $z^6 = e^{i ext{\pi}} = -1$. By substituting $z^8 = z^6 ext{\cdot} z^2 = -z^2$ into the expression $(1+iz+z^{5}+iz^{8})^{9}$, the term $iz^8$ becomes $-iz^2$. This manipulation reveals that the sum inside the parentheses simplifies to $-1$, which raised to the power of 9 results in $-1$.