### Related Formula For any equilateral triangle with vertices z_1, z_2, z_3$z_1, z_2, z_3$: z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1 quad text--- (1)$z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1 \quad \text{--- (1)}$ Centroid equation: z_0 = fracz_1 + z_2 + z_33 implies z_1 + z_2 + z_3 = 3z_0$z_0 = \frac{z_1 + z_2 + z_3}{3} \implies z_1 + z_2 + z_3 = 3z_0$ ### Core Logic Let's expand the target sum: sum_k=1^3 (z_k - z_0)^2 = (z_1 - z_0)^2 + (z_2 - z_0)^2 + (z_3 - z_0)^2$\sum_{k=1}^{3} (z_k - z_0)^2 = (z_1 - z_0)^2 + (z_2 - z_0)^2 + (z_3 - z_0)^2$ ### Step 1: Expansion and Algebraic Grouping Expanding each quadratic term: = (z_1^2 + z_2^2 + z_3^2) - 2z_0(z_1 + z_2 + z_3) + 3z_0^2$= (z_1^2 + z_2^2 + z_3^2) - 2z_0(z_1 + z_2 + z_3) + 3z_0^2$ Substitute z_1 + z_2 + z_3 = 3z_0$z_1 + z_2 + z_3 = 3z_0$: = (z_1^2 + z_2^2 + z_3^2) - 2z_0(3z_0) + 3z_0^2$= (z_1^2 + z_2^2 + z_3^2) - 2z_0(3z_0) + 3z_0^2$ = z_1^2 + z_2^2 + z_3^2 - 3z_0^2$= z_1^2 + z_2^2 + z_3^2 - 3z_0^2$ ### Step 2: Resolving using Equilateral Condition Substitute 3z_0^2 = 3 left(fracz_1+z_2+z_33right)^2 = frac(z_1+z_2+z_3)^23$3z_0^2 = 3 \left(\frac{z_1+z_2+z_3}{3}\right)^2 = \frac{(z_1+z_2+z_3)^2}{3}$: = (z_1^2 + z_2^2 + z_3^2) - fracz_1^2 + z_2^2 + z_3^2 + 2(z_1z_2 + z_2z_3 + z_3z_1)3$= (z_1^2 + z_2^2 + z_3^2) - \frac{z_1^2 + z_2^2 + z_3^2 + 2(z_1z_2 + z_2z_3 + z_3z_1)}{3}$ = frac2(z_1^2 + z_2^2 + z_3^2) - 2(z_1z_2 + z_2z_3 + z_3z_1)3$= \frac{2(z_1^2 + z_2^2 + z_3^2) - 2(z_1z_2 + z_2z_3 + z_3z_1)}{3}$ = frac23(z_1^2 + z_2^2 + z_3^2 - (z_1 z_2 + z_2 z_3 + z_3 z_1))$= \frac{2}{3}(z_1^2 + z_2^2 + z_3^2 - (z_1 z_2 + z_2 z_3 + z_3 z_1))$ Using condition (1) for equilateral triangles, the terms inside the parentheses equal 0$0$. Thus: = 0$= 0$ ### Pattern Recognition This is a standard invariant of equilateral triangles. Any translation to center of mass coordinates leaves the shape invariant, making the sum of squares of coordinate vectors relative to the centroid equal to zero. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers

### Related Formula A complex number w$w$ is purely real if w = barw$w = \bar{w}$. A complex number w$w$ is purely imaginary if w + barw = 0$w + \bar{w} = 0$. ### Core Logic Let's evaluate statement **(S1)**: w = fracz - iz + i$w = \frac{z - i}{z + i}$ If w$w$ is purely real, then w = barw$w = \bar{w}$: fracz - iz + i = fracbarz + ibarz - i$\frac{z - i}{z + i} = \frac{\bar{z} + i}{\bar{z} - i}$ (z - i)(barz - i) = (z + i)(barz + i)$(z - i)(\bar{z} - i) = (z + i)(\bar{z} + i)$ |z|^2 - iz - ibarz - 1 = |z|^2 + iz + ibarz - 1$|z|^2 - iz - i\bar{z} - 1 = |z|^2 + iz + i\bar{z} - 1$ -i(z + barz) = i(z + barz) implies 2i(z + barz) = 0 implies z + barz = 0$-i(z + \bar{z}) = i(z + \bar{z}) \implies 2i(z + \bar{z}) = 0 \implies z + \bar{z} = 0$ Since z + barz = 2textRe(z) = 0$z + \bar{z} = 2\text{Re}(z) = 0$, z$z$ must lie on the imaginary axis (y-axis). Given the condition |z| = 1$|z| = 1$, the only points are z = i$z = i$ and z = -i$z = -i$. However, the domain excludes z = -i$z = -i$. Let's test z = i$z = i$: For z = i$z = i$, fraci - ii + i = 0$\frac{i - i}{i + i} = 0$, which is purely real. So it contains elements on the unit circle. But the condition z + barz = 0$z + \bar{z} = 0$ alongside |z|=1$|z|=1$ explicitly limits it to z=i$z=i$ only, which is one element, not two. Thus, (S1) is incorrect. ### Step 1: Evaluate Statement S2 Let's evaluate statement **(S2)**: u = fracz - 1z + 1$u = \frac{z - 1}{z + 1}$ If u$u$ is purely imaginary, then u + baru = 0$u + \bar{u} = 0$: fracz - 1z + 1 + fracbarz - 1barz + 1 = 0$\frac{z - 1}{z + 1} + \frac{\bar{z} - 1}{\bar{z} + 1} = 0$ frac(z - 1)(barz + 1) + (z + 1)(barz - 1)(z + 1)(barz + 1) = 0$\frac{(z - 1)(\bar{z} + 1) + (z + 1)(\bar{z} - 1)}{(z + 1)(\bar{z} + 1)} = 0$ (|z|^2 + z - barz - 1) + (|z|^2 - z + barz - 1) = 0$(|z|^2 + z - \bar{z} - 1) + (|z|^2 - z + \bar{z} - 1) = 0$ 2|z|^2 - 2 = 0 implies |z|^2 = 1 implies |z| = 1$2|z|^2 - 2 = 0 \implies |z|^2 = 1 \implies |z| = 1$ This condition holds true for ALL points on the unit circle |z| = 1$|z| = 1$ except z = -1$z = -1$ (which makes the denominator zero). Because there are infinitely many points on the unit circle, the set contains infinitely many elements. Thus, (S2) is correct. ### Pattern Recognition Geometric shortcut: The transformation w = fracz-1z+1$w = \frac{z-1}{z+1}$ maps the unit circle |z|=1$|z|=1$ directly onto the imaginary axis textRe(w)=0$\text{Re}(w)=0$. Hence, any point on the unit circle (except the pole at z=-1$z=-1$) satisfies the condition naturally. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers and Quadratic Equations

### Related Formula z + overlinez = 2operatornameRe(z)$z + \overline{z} = 2\operatorname{Re}(z)$ ### Core Logic Isolate the real fractional component block by conjugating the complex quotient matrix expression, then resolve the structural wave equations across bounds boundaries. ### Step 1: Expand Complex Real Operator frac2cos^2theta - 3sin^2thetacos^2theta + 9sin^2theta = -frac110$\frac{2\cos^2\theta - 3\sin^2\theta}{\cos^2\theta + 9\sin^2\theta} = -\frac{1}{10}$ 20cos^2theta - 30sin^2theta = -cos^2theta - 9sin^2theta$20\cos^2\theta - 30\sin^2\theta = -\cos^2\theta - 9\sin^2\theta$ ### Step 2: Factor Trigonometric Expressions 21cos^2theta - 21sin^2theta = 0 implies cos(2theta) = 0$21\cos^2\theta - 21\sin^2\theta = 0 \implies \cos(2\theta) = 0$ ### Step 3: Collect Domain Solutions and Evaluate Squares Since angular coordinate parameters scan [0, 2pi]$[0, 2\pi]$, multi frequency vectors trace out: 2theta = fracpi2, frac3pi2, frac5pi2, frac7pi2$2\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$ sum theta^2 = fracpi^216 + frac9pi^216 + frac25pi^216 + frac49pi^216 = frac84pi^216 = frac214pi^2$\sum \theta^2 = \frac{\pi^2}{16} + \frac{9\pi^2}{16} + \frac{25\pi^2}{16} + \frac{49\pi^2}{16} = \frac{84\pi^2}{16} = \frac{21}{4}\pi^2$ ### Pattern Recognition Transforming algebraic equations to clean forms like \cos(2\theta) = 0$\\cos(2\\theta) = 0$ guarantees evenly distributed coordinate solutions across standard periodicity ranges. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers Class 11 Mathematics: Trigonometric Functions

### Related Formula Properties of cube roots of unity: 1 + omega + omega^2 = 0, quad omega^3 = 1$1 + \omega + \omega^2 = 0, \quad \omega^3 = 1$ ### Core Logic Simplify the determinant by performing row operation R_1 to R_1 + R_2 + R_3$R_1 \to R_1 + R_2 + R_3$: Delta = beginvmatrix z + 1 + omega + omega^2 & z + 1 + omega + omega^2 & z + 1 + omega + omega^2 \\ omega & z + omega^2 & 1 \\ omega^2 & 1 & z + omega endvmatrix$\Delta = \begin{vmatrix} z + 1 + \omega + \omega^2 & z + 1 + \omega + \omega^2 & z + 1 + \omega + \omega^2 \\ \omega & z + \omega^2 & 1 \\ \omega^2 & 1 & z + \omega \end{vmatrix}$ Using 1 + omega + omega^2 = 0$1 + \omega + \omega^2 = 0$, the top row simplifies to vector [z, z, z]$[z, z, z]$. Factoring out z$z$: Delta = z cdot (z^2) = z^3$\Delta = z \cdot (z^2) = z^3$ Given modulus constraint |z^3| = 1 implies |z| = 1$|z^3| = 1 \implies |z| = 1$. The root solutions are: z in \1, omega, omega^2\$z \in \{1, \omega, \omega^2\}$ ### Step 1: Evaluate Geometric Magnitude Metric The condition left|fracz - az + bright| = 1 implies |z - a| = |z + b|$\left|\frac{z - a}{z + b}\right| = 1 \implies |z - a| = |z + b|$. This equation represents the perpendicular bisector of the segment connecting real coordinate points a$a$ and -b$-b$ on the complex plane. Since a$a$ and b$b$ are integers, the bisector is a vertical line: x = fraca - b2$x = \frac{a - b}{2}$. ### Step 2: Match Root Solutions and Count Pairs For z=1$z=1$, it must lie on the line: fraca-b2 = 1 implies a - b = 2$\frac{a-b}{2} = 1 \implies a - b = 2$. For z = omega, omega^2$z = \omega, \omega^2$, their real part is -frac12$-\frac{1}{2}$, so the line must be: fraca-b2 = -frac12 implies a - b = -1$\frac{a-b}{2} = -\frac{1}{2} \implies a - b = -1$. Counting integer pairs (a,b) in [-3, 3]^2$(a,b) \in [-3, 3]^2$ with a+b neq 0$a+b \neq 0$: From a - b = 2$a - b = 2$: valid pairs are (3,1), (1,-1), (0,-2), (-1,-3)$(3,1), (1,-1), (0,-2), (-1,-3)$. Note: (2,0)$(2,0)$ is valid, but a+b=2 neq 0$a+b=2 \neq 0$. Total = 5 pairs. From a - b = -1$a - b = -1$: valid pairs match another 5 configurations. Combining both groups gives a final count of 10 pairs. ### Pattern Recognition Using matrix summation properties (1+omega+omega^2=0$1+\omega+\omega^2=0$) helps simplify large complex variable equations quickly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers Class 12 Mathematics: Matrices and Determinants

### Related Formula Complex rotation and scaling vector rule: z_2 = frac|z_2||z_1| z_1 e^itheta$z_2 = \frac{|z_2|}{|z_1|} z_1 e^{i\theta}$ ### Core Logic Given structural rotation conditions: z_2 = frac1sqrt3 z_1 e^ifracpi6$z_2 = \frac{1}{\sqrt{3}} z_1 e^{i\frac{\pi}{6}}$ Evaluating the vectors yields coordinates showing |z_1 - z_2| = |z_2|$|z_1 - z_2| = |z_2|$. ### Step 1: Analyzing Geometry Metrics Since |z_1 - z_2| = |z_2|$|z_1 - z_2| = |z_2|$, Delta ABO$\Delta ABO$ forms an isosceles triangle with internal vertex angles evaluating explicitly to fracpi6, fracpi6$\frac{\pi}{6}, \frac{\pi}{6}$, and frac2pi3$\frac{2\pi}{3}$. ### Step 2: Conclusion Since frac2pi3 > fracpi2$\frac{2\pi}{3} > \frac{\pi}{2}$, the triangle is an obtuse-angled isosceles triangle. ### Pattern Recognition Complex argument shifts represent pure coordinate system rotations on the Argand plane diagram matrix. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers