Complex Numbers – JEE Main MathematicsPractice Questions & PYQs

The modulus properties allow us to state $||z| - |1/z|| \le |z - 1/z| = 2$. Let $r = |z|$, then the inequality $|r - 1/r| \le 2$ must be satisfied. Expanding this into two cases, $r - 1/r \le 2$ implies $r^2 - 2r - 1 \le 0$, which yields $r \le 1+\sqrt{2}$ because $r > 0$. The other case, $r - 1/r \ge -2$, implies $r^2 + 2r - 1 \ge 0$, leading to $r \ge \sqrt{2}-1$. Thus, the maximum value for $|z|$ is $1+\sqrt{2}$.

The condition $\left|\frac{z+1}{z-1}\right|<1$ geometrically represents the open half-plane where $\text{Re}(z) < 0$. The equation $\arg\left(\frac{z-1}{z+1}\right)=\frac{2\pi}{3}$ defines an arc of a circle passing through $-1$ and $1$, where any point on this arc subtends an angle of $120^\circ$ at the chord. Intersecting this arc with the region $\text{Re}(z) < 0$ restricts the locus to a portion of the circle situated entirely in the second quadrant.

Representing complex numbers in polar form as $z=re^{i\theta}$ allows us to equate the magnitudes and arguments of the equation $\bar{z}=iz^{2}$, which simplifies to $re^{-i\theta} = r^{2}e^{i(2\theta+\pi/2)}$. Setting $r=1$ for the non-zero solutions leads to the condition $3\theta = -\pi/2 + 2k\pi$, producing the three vertices $(\frac{\sqrt{3}}{2}, -\frac{1}{2})$, $(-\frac{\sqrt{3}}{2}, -\frac{1}{2})$, and $(0, 1)$. These points form an isosceles triangle with a horizontal base of length $\sqrt{3}$ and a vertical altitude of $\frac{3}{2}$. Applying the area formula $\frac{1}{2} \times \text{base} \times \text{height}$ results in $\frac{1}{2} \times \sqrt{3} \times \frac{3}{2} = \frac{3\sqrt{3}}{4}$.

The numerator can be simplified using $(1+2i)(1-2i)=5$, transforming the expression to $25(1+2i)^6$ divided by the denominator $(3+2i)(4+6i)=26i$. Expanding $(1+2i)^6$ gives $117+44i$, which leads to the complex value $25(117+44i)/26i$. Multiplying by $i/i$ and separating the real part yields $550/13$.

The set of points satisfying $\left|z-(4+3i)\right|=2$ represents a circle centered at $(4,3)$ with radius 2, while $\left|z\right|+\left|z-4\right|=6$ defines an ellipse with foci at $(0,0)$ and $(4,0)$ and a major axis length of 6. Geometric analysis shows that the circle's center $(4,3)$ lies outside the ellipse. However, because the circle and the ellipse overlap such that the boundary of the circle intersects the ellipse curve, there are exactly 2 points of intersection.

For an isosceles right-angled triangle $OAB$ with hypotenuse $OB$, the vector $\vec{AB}$ is perpendicular to $\vec{OA}$ and equal in length, satisfying $z_2-z_1 = \pm i z_1$. The condition $\operatorname{Re}(z_2)<0$ selects $z_2=z_1(1+i)=-1+3i$. Checking the provided options reveals that $\arg z_2 = \pi - \tan^{-1} 3$, $\arg(z_1-2z_2) = -\tan^{-1}(4/3)$, and $|z_2|=\sqrt{10}$ are all true, but $|2z_1-z_2|=\sqrt{10}$, which contradicts the value $5$ stated in the fourth option.

The condition $|z|=2$ represents a circle of radius 2 centered at the origin, while $|z-i|=|z+5i|$ describes the perpendicular bisector of the points $(0,1)$ and $(0,-5)$, which is the horizontal line $y=-2$. Substituting $y=-2$ into the circle equation $x^{2+}y^2=4$ yields $x^{2+}4=4$, meaning $x=0$. The point $(0, -2)$ satisfies the line equation $x+2y+4=0$, confirming that this is the correct relation.

The condition $|z+i|=|z-1|$ defines the perpendicular bisector of the points $-i$ and $1$, resulting in the line $y=-x$. Applying the inequality $|z-1+i| \ge |z|$ with $y=-x$ leads to $(x-1)^2 + (-x+1)^2 \ge x^{2+}(-x)^2$, which simplifies to $2(x-1)^2 \ge 2x^2$, yielding $x \le 1/2$. Given $u=2x+iy \in S$, the conditions must hold for $u$, so $y=-2x$ and the inequality $2x \le 1/2$ implies $x \le 1/4$. Furthermore, the condition $|u| < 2$ implies $\sqrt{(2x)^{2+}y^2} < 2$, which becomes $\sqrt{8x^2} < 2$, or $|x| < 1/\sqrt{2}$. Combining these constraints gives the range $(-\frac{1}{\sqrt{2}}, \frac{1}{4}]$.

Expressing the complex numbers in polar form, $-1+i\sqrt{3}=2e^{i2\pi/3}$ and $1-i=\sqrt{2}e^{-i\pi/4}$, the first fraction evaluates to $\frac{2^{21}e^{i14\pi}}{2^{12}e^{-i6\pi}} = 512$. Similarly, $1+i\sqrt{3}=2e^{i\pi/3}$ and $1+i=\sqrt{2}e^{i\pi/4}$ give a second term of $\frac{2^{21}e^{i7\pi}}{2^{12}e^{i6\pi}} = -512$, resulting in $k=0$ and therefore $n=0$. The summation $\sum_{j=0}^{5}(j+5)^2 - \sum_{j=0}^{5}(j+5)$ simplifies to the sum of squares $25+36+49+64+81+100$ minus the sum $5+6+7+8+9+10$, yielding $355-45=310$.

For three complex numbers to form an equilateral triangle with the origin, they must satisfy the relation $z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1$, which simplifies to $z_1^2 + z_2^2 = z_1z_2$ when the third vertex is the origin. By adding $2z_1z_2$ to both sides, the identity becomes $(z_1+z_2)^2 = 3z_1z_2$. Substituting the sum $-(a)$ and product $12$ from the quadratic equation $z^{2+}az+12=0$ into this identity results in $(-a)^2 = 3(12)$, which confirms that $|a|=6$.

The condition $\arg(\frac{z-2}{z+2}) = \frac{\pi}{4}$ identifies $z$ as lying on an arc of a circle with center $(0,2)$ and radius $2\sqrt{2}$. The expression $|z-9\sqrt{2}-2i|^2$ is the square of the distance from $z$ to the fixed point $(9\sqrt{2}, 2)$, which is minimized when $z$ lies on the line segment joining the center to this point, resulting in a distance of $9\sqrt{2}-2\sqrt{2} = 7\sqrt{2}$ and a squared value of $98$.

The equation $|Z-3| = \text{Re}(Z)$ simplifies to $b^2 = 6a-9$, and the condition $\arg(Z_1-Z_2) = \frac{\pi}{4}$ implies $a_1-a_2 = b_1-b_2$. Substituting the former into the latter leads to $\frac{b_1^{2-}b_2^2}{6} = b_1-b_2$, which simplifies to the conclusion that the imaginary part $b_1+b_2 = 6$.

The condition $\left|\frac{z+i}{z-3i}\right|=1$ represents the perpendicular bisector of the points $-i$ and $3i$, which simplifies to the horizontal line $y=1$. Substituting $z=x+i$ into $w = z\overline{z}-2z+2$ allows us to express $w$ as $(x-1)^{2+}2-2i$. To minimize $Re(w)$, we set $x=1$, resulting in $w = 2-2i$, which in polar form is $2\sqrt{2}e^{-i\pi/4}$. For $w^n$ to be real, the imaginary part must be zero, meaning the argument $-n\pi/4$ must be an integer multiple of $\pi$, which identifies the minimum natural number $n=4$.

By simplifying the expression $\frac{(2i)^{n}}{(1-i)^{n-2}}$, it can be rewritten as $(\frac{2i}{1-i})^n (1-i)^2$. The term $\frac{2i}{1-i}$ simplifies to $\frac{2i(1+i)}{2} = i-1$, while $(1-i)^2$ equals $-2i$, resulting in the expression $(i-1)^n(-2i)$. Testing small integer values for $n$, for $n=2$, the value is $-4$; for $n=4$, it is $8i$; and for $n=6$, it is $(i-1)^6(-2i) = (-2i)^3(-2i) = (8i)(-2i) = 16$. Since $16$ is a positive integer, $6$ is the least positive integer satisfying the condition.

Representing $\arg(\frac{z-1}{z+1}) = \frac{\pi}{4}$ as $\arg(z-1) - \arg(z+1) = \frac{\pi}{4}$ and setting $z=x+iy$ leads to the equation $\tan^{-1}(\frac{y}{x-1}) - \tan^{-1}(\frac{y}{x+1}) = \frac{\pi}{4}$. Applying the tangent subtraction formula $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$, we get $\frac{\frac{y}{x-1} - \frac{y}{x+1}}{1 + \frac{y^2}{x^{2-}1}} = 1$. Simplifying this expression results in $\frac{2y}{x^{2+}y^{2-}1} = 1$, which rearranges to $x^2 + y^2 - 2y - 1 = 0$, or in standard form, $x^2 + (y-1)^2 = 2$. This represents a circle centered at $(0, 1)$ with a radius of $\sqrt{2}$.

Determining the number of solutions requires finding the intersection of a circle and specific regional constraints in the complex plane. $S_1$ is a circle centered at $(3,2)$ with radius $2 ext{sqrt}(2)$, $S_2$ restricts the region to $x ext{ extgreater}= 5$, and $S_3$ restricts points to $y ext{ extgreater}= 4$ or $y ext{ extless}= -4$. The circle intersects the boundary $x=5$ at points $(5,4)$ and $(5,0)$, but only $(5,4)$ satisfies the condition $|y| ext{ extgreater}= 4$, confirming exactly one common solution.

First, we evaluate the constant $\log_{\sqrt{2}} |5\sqrt{7}+9i|$. The modulus is $\sqrt{(5\sqrt{7})^2 + 9^2} = \sqrt{175+81} = \sqrt{256} = 16$, and since $\log_{\sqrt{2}} 16 = \log_{2^{1/2}} 2^4 = 8$, the inequality becomes $\exp(\dots) \ge 8$, or equivalently, $2^{\frac{(|z|+3)(|z|-1)}{|z|+1}} \ge 2^3$. This simplifies to $\frac{(|z|+3)(|z|-1)}{|z|+1} \ge 3$, which rearranges to $|z|^2 - |z| - 6 \ge 0$. Factoring this quadratic expression gives $(|z|-3)(|z|+2) \ge 0$, and since $|z| \ge 0$, the minimum value is $3$.

The expression $|3iz+6|$ simplifies to $3|z-2i|$, which represents three times the distance from $z$ to the point $2i$. As $z$ is constrained within the disk $|z-(2+2i)| ext{ extless{}} 1$ with radius $1$, the maximum value is found at the point on the boundary furthest from $2i$, which is $3+2i$, giving $a+b=3+2=5$.

The area of a triangle formed by the origin and two complex numbers $z$ and $w$ is determined by the formula $\frac{1}{2}|z||w|\sin(\theta)$, where $\theta$ is the angle between the two vectors. Given $|zw|=1$, the product of the magnitudes is $1$, and the condition $\arg(z)-\arg(w)=\frac{\pi}{2}$ indicates that the angle between them is $\frac{\pi}{2}$. Substituting these values into the area formula yields $\frac{1}{2}(1)\sin(\frac{\pi}{2}) = \frac{1}{2}$.

The center of the circle is the intersection of the two given normal lines. Substituting $z=x+iy$ into $(2-i)z=(2+i)\overline{z}$ yields the equation $x=2y$, and doing the same for $(2+i)z+(i-2)\overline{z}-4i=0$ simplifies to $x+2y=2$, which intersect at $(1, 0.5)$. The tangent line $iz+\overline{z}+1+i=0$ corresponds to $x-y+1=0$ in Cartesian coordinates. The radius is the perpendicular distance from $(1, 0.5)$ to this line, calculated as $\frac{|1-0.5+1|}{\sqrt{1^{2+}(-1)^2}}$, which equals $\frac{1.5}{\sqrt{2}}$ or $\frac{3}{2\sqrt{2}}$.