Let O be the origin, the point A be z_1 = sqrt3 + 2sqrt2i, the point B(z_2) be such that sqrt3left|z_2right| = left|z_1right| and arg (z_2) = arg (z_1) + fracpi6. Then (1) area of triangle ABO is frac11sqrt3 (2) ABO is a scalene triangle (3) area of triangle ABO is frac114 (4) ABO is an obtuse angled isosceles triangle

Solution & Explanation

### Related Formula Complex rotation and scaling vector rule: z_2 = frac|z_2||z_1| z_1 e^itheta ### Core Logic Given structural rotation conditions: z_2 = frac1sqrt3 z_1 e^ifracpi6 Evaluating the vectors yields coordinates showing |z_1 - z_2| = |z_2|. ### Step 1: Analyzing Geometry Metrics Since |z_1 - z_2| = |z_2|, Delta ABO forms an isosceles triangle with internal vertex angles evaluating explicitly to fracpi6, fracpi6, and frac2pi3. ### Step 2: Conclusion Since frac2pi3 > fracpi2, 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

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More Complex Numbers Previous-Year Questions — Page 7

Q26 jee_main_2024_31_jan_morning Properties of Modulus and Argument
If alpha denotes the number of solutions of |1 - i|^x = 2^x and beta = left(frac|z|arg(z)right), where z = fracpi4 (1 + i)^4 left(frac1 - sqrtpi isqrtpi + i + fracsqrtpi - i1 + sqrtpi iright), i = sqrt-1, then the distance of the point (alpha, beta) from the line 4x - 3y = 7 is
Numerical Answer. Answer: 3 to 3

Solution

### Core Logic |1 - i|^x = 2^x implies (sqrt2)^x = 2^x implies 2^x/2 = 2^x This implies fracx2 = x implies x = 0. There is exactly 1 solution, so alpha = 1. ### Step 1: Simplify complex number z (1+i)^4 = ((1+i)^2)^2 = (1 + i^2 + 2i)^2 = (2i)^2 = -4 Thus, z = -pi left( frac(1-sqrtpii)(sqrtpi-i)pi + 1 + frac(sqrtpi-i)(1-sqrtpii)1 + pi right) ### Step 2: Simplify Bracket Let's expand the terms directly: z = fracpi4(-4) left[ fracsqrtpi - pi i - i - sqrtpipi + 1 + fracsqrtpi - i - pi i - sqrtpi1 + pi right] = -pi left[ frac-i(pi+1)pi+1 + frac-i(pi+1)pi+1 right] = -pi [ -i - i ] = 2pi i ### Step 3: Find beta For z = 2pi i: |z| = 2pi and arg(z) = fracpi2. beta = frac|z|arg(z) = frac2pipi/2 = 4 ### Step 4: Distance from Line Distance of point (alpha, beta) = (1, 4) from the line 4x - 3y - 7 = 0: D = frac|4(1) - 3(4) - 7|sqrt4^2 + (-3)^2 = frac|4 - 12 - 7|5 = frac|-15|5 = 3 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Straight Lines

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