Complex Numbers – JEE Main MathematicsPractice Questions & PYQs

Because the coefficients of the quadratic equation are real, complex roots must occur in conjugate pairs. Since $1-2i$ is a root, the other must be $1+2i$. Applying Vieta's formulas, the sum of these roots is $(1-2i)+(1+2i)=2$, which equals $-\alpha$, implying $\alpha=-2$. The product of the roots is $(1-2i)(1+2i)=1^{2-}(2i)^2=5$, which equals $\beta$, so calculating $\alpha-\beta$ yields $-2-5=-7$.

A complex number $\frac{z-z_1}{z-z_2}$ is purely imaginary if $z$ lies on a circle with diameter endpoints $z_1=i$ and $z_2=1$. This circle has center $(\frac{1}{2}, \frac{1}{2})$ and radius $\frac{1}{\sqrt{2}}$, so the minimum distance to $3+3i$ is the distance between $(\frac{1}{2}, \frac{1}{2})$ and $(3,3)$ minus the radius, calculated as $\frac{5}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 2\sqrt{2}$.

The condition |zar{\omega}|=1 and $\arg(z)-\arg(\omega)=\frac{3\pi}{2}$ implies that the product $\overline{z}\omega$ is equivalent to $e^{i(\arg(\omega) - \arg(z))}$, which simplifies to $e^{-i(3\pi/2)} = i$. Substituting this value into the expression yields $\frac{1-2i}{1+3i}$, and multiplying the numerator and denominator by the complex conjugate $1-3i$ simplifies the fraction to $-\frac{1}{2}(1+i)$. Calculating the principal argument of this complex number, which lies in the third quadrant, results in $-\frac{3\pi}{4}$.

The equation of a circle is derived from the standard form $x+iy$ by substituting into the expression $ext{Re}(z^2) + 2 ext{Im}(z)^2 + 2 ext{Re}(z) = 0$, which simplifies to $(x+1)^2 + y^2 = 1$, identifying a circle centered at $(-1,0)$. The parabola $x^2 - 6x - y + 13 = 0$ is rewritten as $(x-3)^2 = y-4$, locating its vertex at $(3,4)$. The line passing through $(-1,0)$ and $(3,4)$ has the equation y - 0 = rac{4-0}{3 - (-1)}(x+1), which simplifies to $y = x+1$ and gives a y-intercept of $1$.

The geometry of complex regions involves identifying the intersection of a disk and a half-plane defined by the given inequalities. $S_1$ describes a disk centered at $(2,0)$ with radius $1$, while $S_2$ simplifies to the half-plane $x - y ext{ extgreater}= 2$. By expressing any point on the circle boundary as $z = 2 + ext{cos} heta + i ext{sin} heta$, the objective function $|z - 2.5|^2$ becomes $1.25 - ext{cos} heta$ for -rac{3 ext{\pi}}{4} ext{ extless}= heta ext{ extless}= rac{ ext{\pi}}{4}. Maximizing this distance occurs at heta = -rac{3 ext{\pi}}{4}, yielding a final value of rac{5 + 2 ext{sqrt}(2)}{4}.

When the ratio of two complex numbers is real, their argument must be a multiple of $\pi$, implying that the angle formed by these numbers is $0$ or $\pi$. This condition dictates that the points represented by $z$, $i$, and $-2i$ must be collinear in the complex plane. Consequently, the set of all such points $z$ forms a straight line passing through the points $i$ and $-2i$.

Since the term $\alpha|z-1|$ is real, the imaginary part of $z$ must be $-2$, so let $z=x-2i$. Substituting this into the equation results in $x+\alpha\sqrt{(x-1)^{2+}4}=0$, which can be rearranged as $\alpha^2 = \frac{x^2}{(x-1)^{2+}4}$. By analyzing the function $f(x) = \frac{x^2}{x^{2-}2x+5}$, we find that $\alpha^2$ ranges between $0$ and $\frac{5}{4}$. Consequently, $p$ and $q$ correspond to $-\frac{\sqrt{5}}{2}$ and $\frac{\sqrt{5}}{2}$, making $4(p^{2+}q^2) = 4(\frac{5}{4} + \frac{5}{4}) = 10$.

The real part of a complex number is determined by multiplying the numerator and denominator by the conjugate of the denominator, effectively clearing the imaginary component from the divisor. Applying this to z = rac{3+2i heta}{1-3i heta} results in a real part of rac{3 - 6 heta^2}{1+9 heta^2}, where $heta$ represents $ext{cos} heta$. Setting this expression to zero gives $3 - 6 ext{cos}^2 heta = 0$, which simplifies to ext{cos}^2 heta = rac{1}{2} and indicates heta = rac{ ext{\pi}}{4}. Finally, calculating $ext{sin}^2(3 heta) + ext{cos}^2 heta$ at rac{ ext{\pi}}{4} yields ext{sin}^2rac{3 ext{\pi}}{4} + ext{cos}^2rac{ ext{\pi}}{4} = rac{1}{2} + rac{1}{2} = 1.

Using the property $z = e^{-i\frac{\pi}{3}}$, the term $(z^r + \frac{1}{z^r})^3$ simplifies to $(2 \cos(\frac{r\pi}{3}))^3 = 8 \cos^3(\frac{r\pi}{3})$. The sum $8 \sum_{r=1}^{21} \cos^3(\frac{r\pi}{3})$ involves periodic values that cycle every six terms, where the sum of each cycle of six terms is zero. After dividing $21$ by $6$, we are left with the terms for $r=19, 20,$ and $21$, which correspond to the values at $r=1, 2,$ and $3$. These values are $\cos^3(\frac{\pi}{3}) = \frac{1}{8}$, $\cos^3(\frac{2\pi}{3}) = -\frac{1}{8}$, and $\cos^3(\pi) = -1$, yielding a final sum of $21 + 8(\frac{1}{8} - \frac{1}{8} - 1) = 21 - 8 = 13$.

The region is bounded by the circle $(x+5)^{2+}y^2 \le 16$ and the half-plane $x-y+5 \ge 0$, which is defined by the line $y=x+5$. The maximum value of $|z+1|^2$ occurs at the intersection points of the circle and the line, specifically $A(-2\sqrt{2}-5, -2\sqrt{2})$ and $B(2\sqrt{2}-5, 2\sqrt{2})$. Calculating the squared distance from the point $(-1, 0)$ to $A$ yields $(-2\sqrt{2}-5 - (-1))^2 + (-2\sqrt{2}-0)^2 = (-2\sqrt{2}-4)^2 + 8$, which simplifies to $32+16\sqrt{2}$. Given this matches $\alpha+\beta\sqrt{2}$, we find $\alpha=32$ and $\beta=16$, resulting in $\alpha+\beta=48$.

The requirement that the product of $A^n$ and any arbitrary matrix $X$ must equal $X$ implies that $A^n$ acts as the identity matrix $I$. To determine the valid values for $n$, we compute the powers of

. Squaring this matrix yields

, and squaring that result produces $A^4 = i^2 I = -I$. Since $A^4 = -I$, we find that $A^8 = (-I)^2 = I$, which means $n$ must be a multiple of 8. Identifying the two-digit multiples of 8 provides the set $\{16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96\}$, and counting these values confirms there are 11 such numbers.

The number of complex solutions is found by expanding $z^2 + 3 ext{conj}(z) = 0$ into its real and imaginary parts, resulting in the system $x^2 - y^2 + 3x = 0$ and $y(2x-3) = 0$. Solving these equations yields four distinct coordinate solutions, meaning $n=4$. The infinite geometric series defined by ext{\sum}_{k=0}^{ ext{infinity}} (rac{1}{n})^k then converges to rac{1}{1 - 1/4}, which equals rac{4}{3}.

Because the base of the logarithm is $\frac{1}{\sqrt{2}}$, which is less than $1$, the inequality $\log_{\frac{1}{\sqrt{2}}}(f(|z|)) \le 2$ implies that the argument must satisfy $f(|z|) \ge (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$. Setting $f(|z|) = \frac{|z|+11}{(|z|-1)^2}$, we obtain $\frac{|z|+11}{(|z|-1)^2} \ge \frac{1}{2}$, which leads to the quadratic inequality $|z|^2 - 4|z| - 21 \le 0$. Factoring this inequality as $(|z|-7)(|z|+3) \le 0$ shows that $|z|$ lies between $0$ and $7$, making the largest possible value $7$.

Utilizing De Moivre's Theorem, one can express $(\sqrt{3}+i)$ in polar form as $2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6})) = 2e^{i\frac{\pi}{6}}$. Raising this to the power of $100$ gives $2^{100}e^{i\frac{100\pi}{6}} = 2^{100}e^{i\frac{50\pi}{3}}$, which simplifies to $2^{100}e^{i\frac{2\pi}{3}}$ due to the periodicity of $e^{i\theta}$. Expanding this yields $2^{100}(\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3})) = 2^{100}(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 2^{99}(-1 + i\sqrt{3})$. Comparing this with $2^{99}(p+iq)$, we find $p=-1$ and $q=\sqrt{3}$, making the required quadratic equation $(x+1)(x-\sqrt{3}) = x^2 - (\sqrt{3}-1)x - \sqrt{3} = 0$.

The vertices of the triangle in the Argand plane are represented by $z$, $iz$, and $z+iz$. Because the vector $iz$ is simply $z$ rotated by $90^{\circ}$, the vectors $z$ and $iz$ are orthogonal and share the same magnitude $|z|$. These two vectors define the sides of a rectangle, and the triangle with vertices $0, z,$ and $z+iz$ has an area equal to half of this rectangle, given by $\frac{1}{2}|z||iz|$, which simplifies to $\frac{1}{2}|z|^{2}$.

The standard equation of a circle in the complex plane is given by $|z|^2 + \overline{\gamma}z + \gamma\bar{z} + c = 0$, where the radius squared is $r^2 = |\gamma|^2 - c$. Rewriting the provided equation by dividing by $a$ identifies the complex parameter $\gamma = \frac{\alpha}{a}$ and the constant term $c = \frac{d}{a}$. For this expression to represent a circle, the radius squared must be strictly positive, necessitating that $|\frac{\alpha}{a}|^2 - \frac{d}{a} > 0$, which simplifies to $|\alpha|^2 - ad > 0$ given $a \in \mathbb{R} - \{0\}$.

The intersection of sets in the complex plane is determined by the simultaneous overlap of the regions defined by each condition. The set $S_1$ represents a circular disk, while $S_2$ and $S_3$ define regions bounded by half-planes. Because these inequalities describe overlapping continuous geometric areas rather than isolated points or lines, the resulting intersection $S_1 \cap S_2 \cap S_3$ forms a region that contains infinitely many elements.

The vertices of the square $z=x+iy$ and its transformations form sides of length $4$ in the Argand plane. The vertical side length is determined by $|z-\bar{z}| = |2iy| = 2|y| = 4$, implying $|y|=2$, and the horizontal side is determined by the distance $|2\operatorname{Re}(z)| = 2|x| = 4$, implying $|x|=2$. The magnitude $|z|$ is calculated as $\sqrt{x^{2+}y^2} = \sqrt{2^{2+}2^2} = \sqrt{8} = 2\sqrt{2}$.

By substituting $\sin(\frac{2\pi}{9})$ and $\cos(\frac{2\pi}{9})$ with their complementary angle equivalents, the expression can be written in terms of $\cos(\frac{5\pi}{18})$ and $\sin(\frac{5\pi}{18})$. Applying trigonometric identities simplifies the numerator to $2\cos(\frac{5\pi}{36})e^{i5\pi/36}$ and the denominator to $2\cos(\frac{5\pi}{36})e^{-i5\pi/36}$. Raising the ratio $e^{i5\pi/18}$ to the power of $3$ yields $e^{i5\pi/6}$, which evaluates to $-\frac{\sqrt{3}}{2} + \frac{i}{2}$, or $-\frac{1}{2}(\sqrt{3}-i)$.

The complex value $u$ is defined by $\frac{2z+i}{z-ki}$, and the condition $\operatorname{Re}(u)+\operatorname{Im}(u)=1$ describes a geometric curve. By setting $z=x+iy$ and analyzing the equation on the $y$-axis where $x=0$, we derive the quadratic $y^{2+}y-k-k^2=0$, where $y_1$ and $y_2$ are the intersections $P$ and $Q$. Given the distance $PQ=5$ between roots $y_1$ and $y_2$, the relation $|y_1-y_2| = \sqrt{(y_1+y_2)^2 - 4y_1y_2} = \sqrt{(-1)^2 + 4(k+k^2)} = 2k+1$ leads to the equation $2k+1=5$, confirming $k=2$.