Complex Numbers – JEE Main MathematicsPractice Questions & PYQs

Representing complex numbers $z$ and $w$ in polar form as $z = r e^{i heta}$ and $w = r e^{i ext{arg}(w)}$ allows the use of the property $|z| = |w| = r$. Given the condition $heta + ext{arg}(w) = ext{\pi }$, we substitute $heta = ext{\pi } - ext{arg}(w)$ to rewrite $z$ as $r e^{i( ext{\pi } - ext{arg}(w))}$, which simplifies to $r e^{i ext{\pi }} e^{-i ext{arg}(w)}$. Since $e^{i ext{\pi }} = -1$ and the conjugate ar{w} equals $r e^{-i ext{arg}(w)}$, the expression becomes z = -ar{w}.

Starting at the complex number $2+i$, moving 1 unit east adds 1 to the real part, and moving 2 units north adds $2i$ to the imaginary part, bringing the position to $3+3i$. A movement of $2\sqrt{2}$ units in the south-west direction corresponds to a vector displacement of $2\sqrt{2}(\cos(\frac{5\pi}{4}) + i\sin(\frac{5\pi}{4}))$, which simplifies to $-2-2i$. Adding this displacement vector to the current position $3+3i$ yields the final position $1+i$.

For a quadratic equation $x^{2-}\sqrt{6}x+3=0$, the roots are given by $x = \frac{\sqrt{6}\pm i\sqrt{6}}{2} = \sqrt{3}e^{\pm i\pi/4}$. Expanding the expression $\frac{\alpha^{99}}{\beta}+\alpha^{98}$ as $\alpha^{98}(\frac{\alpha}{\beta}+1)$, we substitute the roots to find $\frac{\alpha}{\beta} = e^{i\pi/2}=i$, reducing the term to $\alpha^{98}(i+1)$. Evaluating $\alpha^{98} = (\sqrt{3})^{98}e^{i49\pi/2} = 3^{49}i$, the expression becomes $3^{49}i(i+1) = 3^{49}(-1+i)$, which matches $3^n(a+ib)$ with $n=49, a=-1, b=1$, summing to 49.

The intersection $P \cap Q$ defines a geometric region where $z_1$ and $z_2$ are identified as the points providing the maximum and minimum distance from the coordinate $(3, -2)$. Solving for these points involves finding the intersection of the circle $(x+2)^{2+}(y-3)^2 \le 1$ and the line $x-y \le -4$. Computing the squared moduli $|z_1|^2$ and $2|z_2|^2$ based on these coordinates leads to a result of $\alpha + \beta\sqrt{2}$, where the sum of the integers $\alpha$ and $\beta$ is 8.

The value of \alpha is determined by solving |1-i|^x = 2^x, which simplifies to (\sqrt{2})^x = 2^x, yielding the single solution x=0, thus \alpha=1. For z, evaluating the expression involving complex powers and conjugate pairs reveals z = 2\pi i, where the modulus |z| = 2\pi and \arg(z) = \frac{\pi}{2}, making \beta = 4. The distance from the point (1, 4) to the line 4x - 3y = 7 is then found using the standard perpendicular distance formula, resulting in a value of 3.

The condition $|z-2| \leq 1$ defines a circular disk of radius 1 centered at $(2,0)$, while $z(1+i)+\overline{z}(1-i) \leq 2$ simplifies to the linear inequality $x-y \leq 1$ in the Cartesian plane. The set $S$ is the intersection of this disk and the region $x-y \leq 1$. The distances $|z-4i|$ reach their minimum and maximum at specific points on the boundary of $S$. Calculation of these extrema for $|z_1|^2$ and $|z_2|^2$ reveals that $5(|z_1|^{2+}|z_2|^2) = 30-4\sqrt{5}$, and identifying the integer components yields $\alpha=30$ and $\beta=-4$, summing to 26.