Complex Numbers – JEE Main MathematicsPractice Questions & PYQs

We can parameterize $z+2$ as $\cos \theta + i \sin \theta$ because $|z+2|=1$. Substituting this into the expression $\frac{z+1}{z+2}$, we get $1 - \frac{1}{z+2} = 1 - (\cos \theta - i \sin \theta) = (1-\cos \theta) + i \sin \theta$. The given imaginary part $\sin \theta = \frac{1}{5}$ allows us to calculate $\cos \theta = \pm \sqrt{1 - (\frac{1}{5})^2} = \pm \frac{\sqrt{24}}{5} = \pm \frac{2\sqrt{6}}{5}$. Thus, $|\operatorname{Re}(\overline{z+2})| = |\cos \theta| = \frac{2\sqrt{6}}{5}$.

A complex number $Z$ is purely imaginary if $Z = -\overline{Z}$, or equivalently, if its real part is zero. Applying this condition to $\frac{1+i \cos \theta}{1-2 i \cos \theta}$ by multiplying the numerator and denominator by the conjugate of the denominator, $(1+2 i \cos \theta)$, we obtain $\frac{1 + 3i \cos \theta - 2 \cos^2 \theta}{1 + 4 \cos^2 \theta}$. Setting the real part, $\frac{1-2 \cos^2 \theta}{1+4 \cos^2 \theta}$, to zero yields $\cos^2 \theta = \frac{1}{2}$, leading to $\cos \theta = \pm \frac{1}{\sqrt{2}}$. In the interval $[-\pi, 2\pi]$, this gives solutions $\theta = \pm \frac{\pi}{4}, \pm \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$, which sum to $3\pi$.

The sum of cubes identity z_1^3 + z_2^3 = (z_1 + z_2)^3 - 3z_1 z_2 (z_1 + z_2) serves as the starting point for finding the product z_1 z_2. Given z_1 + z_2 = 5 and z_1^3 + z_2^3 = 20 + 15i, one can substitute these values to obtain z_1 z_2 = 7 - i. Next, compute z_1^2 + z_2^2 = (z_1 + z_2)^2 - 2z_1 z_2 and use the relation (z_1^2 + z_2^2)^2 = z_1^4 + z_2^4 + 2(z_1 z_2)^2 to find z_1^4 + z_2^4. The final modulus |z_1^4 + z_2^4| calculates to 75.

Given that $\alpha$ and $\frac{1}{\bar{\alpha}}$ lie on circles centered at $z_0=1+i$ with squared radii 4 and 16, we apply the squared modulus condition $(\alpha-z_0)(\bar{\alpha}-\bar{z}_0)=4$ to obtain $|\alpha|^2 - (\alpha\bar{z}_0+\bar{\alpha}z_0) = 2$ using $|z_0|^2=2$. The second condition leads to $1-(\bar{\alpha}z_0+\alpha\bar{z}_0)+2|\alpha|^2=16|\alpha|^2$. By isolating the term $(\alpha\bar{z}_0+\bar{\alpha}z_0)$ and substituting it, we solve $|\alpha|^{2-}2=1-14|\alpha|^2$, giving $15|\alpha|^2=3$, which implies $100|\alpha|^2=20$.

The region is defined by the disk $|z-1| \le 2$ centered at $(1, 0)$ with radius 2, restricted to the upper half-plane. The linear constraint $(z+\overline{z})+i(z-\overline{z}) \le 2$ simplifies to $2x - 2y \le 2$, or $y \ge x-1$. This line intersects the semicircle, and the resulting area is obtained by calculating the area of the semicircle and subtracting the area of the circular segment excluded by the linear inequality, which results in a final area of $\frac{3 \pi}{2}$.

Given the quadratic x^{2+}x+1=0, its roots are the cube roots of unity, alpha=omega and alpha=omega^2, which satisfy 1+alpha+alpha^2=0. By taking alpha=omega, the expression simplifies using the property 1+alpha = -alpha^2, resulting in (1+alpha)^7 = (-alpha^2)^7 = -alpha^14 = -alpha^2 = 1+alpha. Comparing this to A+B(alpha)+C(alpha^2) yields A=1, B=1, and C=0, and calculating the final expression gives 5(3(1)-2(1)-0) = 5.

The roots of the polynomial z^3 + 2z^2 + 2z + 1 = 0 are identified by factoring it as (z+1)(z^2 + z + 1) = 0, giving solutions z = -1, z = \omega, and z = \omega^2, where \omega is a complex cube root of unity. Testing these in the equation z^{1985} + z^{100} + 1 = 0 shows that z = -1 is not a solution, whereas the complex roots \omega and \omega^2 satisfy the equation due to the property 1 + \omega + \omega^2 = 0. Thus, there are two common roots shared by the two equations.

To determine $m$, we find the number of integer pairs $(a, b)$ satisfying $a+5b=42$ with $a, b \in N$, which range from $b=1$ ($a=37$) to $b=8$ ($a=2$), giving $m=8$. The sum $\sum_{n=1}^{8} (1-i^{n!})$ evaluates the terms for $n=1, 2, 3$ as $(1-i)$, $(1+1)$, and $(1+1)$ respectively, while for $n \ge 4$, $n!$ is a multiple of $4$, so $i^{n!}=1$ and $1-i^{n!} = 0$. Summing these yields $(1-i) + 2 + 2 = 5-i$, identifying $x=5$ and $y=-1$. Consequently, $m+x+y = 8 + 5 - 1 = 12$.

Expanding the equation $(\overline{z})^{2+}|z|=0$ using $z=x+iy$ yields $(x^{2-}y^{2-}2ixy) + \sqrt{x^{2+}y^2} = 0$. Equating the imaginary part to zero implies $-2xy=0$, so either $x=0$ or $y=0$. For $x=0$, the equation becomes $-y^{2+}|y|=0$, yielding non-zero solutions $z=i$ and $z=-i$, while the case $y=0$ leads to $x^{2+}|x|=0$ which has no non-zero real solutions. The sum of these roots is $\alpha=0$ and their product is $\beta=1$, resulting in $4(0^{2+}1^2)=4$.

The problem defines a set S as the intersection of a circle |z-1|=1 and a line derived from the given complex equation. Substituting z=x+iy into the equations yields (x-1)^2 + y^2 = 1 for the circle and (\sqrt{2}-1)x + y = \sqrt{2} for the line. Solving this system simultaneously identifies the intersection points z_1 and z_2 that minimize and maximize the modulus |z|. Calculating |\sqrt{2} z_1 - z_2|^2 using these coordinates results in a value of 2.

The modulus equation $\left| \frac{z_1-2z_2}{\frac{1}{2}-z_1\overline{z}_2} \right|^2 = 4$ can be expanded using the property $|w|^2 = w\overline{w}$ to $(z_1-2z_2)(\overline{z}_1-2\overline{z}_2) = 4(\frac{1}{2}-z_1\overline{z}_2)(\frac{1}{2}-\overline{z}_1z_2)$. Expanding both sides results in $|z_1|^2 - 2z_1\overline{z}_2 - 2\overline{z}_1z_2 + 4|z_2|^2 = 1 - 2\overline{z}_1z_2 - 2z_1\overline{z}_2 + 4|z_1|^2|z_2|^2$. Canceling common terms leads to $|z_1|^2 + 4|z_2|^2 - 1 - 4|z_1|^2|z_2|^2 = 0$, which factors as $(|z_1|^{2-}1)(1-4|z_2|^2)=0$, implying either $|z_1|=1$ or $|z_2|=\frac{1}{2}$.

Since $\alpha$ is a root of $x^{2-}x+2=0$, it satisfies the relation $\alpha^2=\alpha-2$. We express higher powers of the root as $\alpha^4=(\alpha-2)^2 = -3\alpha+2$ and $\alpha^6=\alpha^2(-3\alpha+2) = 5\alpha+2$, with a similar reduction for $\beta^4 = -3\beta+2$. Substituting these into the expression $\alpha^{6+}\alpha^{4+}\beta^{4-}5\alpha^2$ yields $(5\alpha+2)+(-3\alpha+2)+(-3\beta+2)-5(\alpha-2)$, which simplifies to $-3(\alpha+\beta)+16$. Using the root property $\alpha+\beta=1$, the final value is 13.

The polar representation of a complex number $z=x+iy$ is defined by its modulus $r=\sqrt{x^{2+}y^2}$ and argument $\theta=\tan^{-1}(y/x)$. Given $z=2-i(2\tan \frac{5\pi}{8})$, we utilize the trigonometric identity $\tan \frac{5\pi}{8} = -\tan \frac{3\pi}{8}$ to express the components as $x=2$ and $y=2\tan \frac{3\pi}{8}$. Calculating the modulus yields $r=\sqrt{4 + 4\tan^2 \frac{3\pi}{8}} = 2\sec \frac{3\pi}{8}$, and the argument is found by identifying $\theta = \tan^{-1}(\tan \frac{3\pi}{8}) = \frac{3\pi}{8}$.

The expression $|z - (-1.5 - 2i)|$ represents the distance between a complex number $z$ and the fixed point $w = -1.5 - 2i$. Since the modulus of this point is $\sqrt{(-1.5)^2 + (-2)^2} = \sqrt{2.25+4} = 2.5$, which is greater than 1, the point $w$ lies within the region defined by $|z| \ge 1$. Because the point is contained within the set, the minimum distance from any $z$ in that set to $w$ is zero.

The condition $|z-1| \leq 1$ restricts $z$ to a disk centered at $(1,0)$ with radius $1$, while $|z-5| \leq |z-5i|$ simplifies to the half-plane $x \geq y$. Checking the integer coordinates $(x,y)$ within the disk, the valid points are $(0,0), (1,-1), (1,0), (1,1),$ and $(2,0)$. Calculating the sum of the squared moduli $|z|^2 = x^{2+}y^2$ for these five points yields $0 + 2 + 1 + 2 + 4 = 9$.

The condition that the real part of the complex fraction $\frac{z-2i}{z+2i}$ is zero requires the numerator and denominator to satisfy a relationship equivalent to $|z|=2$, placing $z$ on a circle of radius $2$ centered at the origin. The maximum value of $|z-(6+8i)|$ represents the greatest distance from a point on this circle to the point $(6,8)$. This is calculated by finding the distance from the origin to $(6,8)$, which is $\sqrt{6^{2+}8^2} = 10$, and adding the radius of the circle, resulting in $10 + 2 = 12$.

The region $S_1$ is a disk of radius $5$ centered at the origin, while $S_3$ restricts the domain to $x \ge 0$. The condition for $S_2$ simplifies to $\sqrt{3}x+y \ge 0$, which defines a half-plane bounded by the line $y = -\sqrt{3}x$. The intersection $S_1 \cap S_2 \cap S_3$ corresponds to a sector of the circle with an angle spanning from $-60^\circ$ to $90^\circ$, totaling $150^\circ$ or $\frac{5\pi}{6}$ radians. The area of this sector is $\frac{1}{2} r^2 \theta = \frac{1}{2} (25) (\frac{5\pi}{6}) = \frac{125\pi}{12}$.

Given the equation z^2 = -i\bar{z}, one can apply the modulus operation to both sides to relate |z^2| and |\bar{z}|. Using the properties that |z^2| = |z|^2 and |i\bar{z}| = |i||\bar{z}| = |z|, the equation transforms into |z|^2 = |z|. Since xy \neq 0 implies z \neq 0, dividing by |z| gives |z| = 1, which consequently means |z^2| = |z|^2 = 1.

The condition |z-i|=|z+i|=|z-1| defines the set of points in the complex plane equidistant from three specific points: (0,1), (0,-1), and (1,0). Geometrically, this point corresponds to the circumcenter of the triangle formed by these three vertices. Since any three non-collinear points determine a unique circumcenter, the set S contains exactly one element, so n(S)=1.

Using de Moivre's theorem, we convert the base $1-\sqrt{3}i$ into polar form, yielding $2(\cos(\frac{-\pi}{3}) + i\sin(\frac{-\pi}{3})) = 2e^{-i\pi/3}$. Raising this to the power of 200, we calculate $2^{200}e^{-i 200\pi/3}$, which simplifies to $2^{200}e^{-i(66\pi + 2\pi/3)} = 2^{200}e^{-i2\pi/3} = 2^{200}(-\frac{1}{2} - i\frac{\sqrt{3}}{2}) = 2^{199}(-1 - i\sqrt{3})$. Comparing this result to $2^{199}(p+iq)$, we find $p=-1$ and $q=-\sqrt{3}$. Substituting these values into the given roots gives $\alpha = p+q+q^2 = -1-\sqrt{3}+3 = 2-\sqrt{3}$ and $\beta = p-q+q^2 = -1+\sqrt{3}+3 = 2+\sqrt{3}$. The quadratic equation with these roots is $x^2 - (\alpha+\beta)x + \alpha\beta = x^2 - 4x + 1 = 0$.