On the sides AB and BC of $Δ l e$ ABC, squares are drawn with centres D and E such that points C and D lie on the same side of line AB and the points A and E lie on either side of line BC. A, B, C are represented by complex numbers $1, ω, ω^{2}$ respectively (where $ω = \frac{- 1 + i \sqrt{3}}{2}$ ). Which of the following statement(s) is (are) CORRECT?

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length of DE is $\frac{\sqrt{3}}{2}$

length of AE is $\frac{3 + \sqrt{3}}{2}$

Area of triangle ADE is $\frac{3}{8} s q . u n i t s$

Acute angle between AC and DE is equal to $\frac{π}{4}$

answer is A, C, D.

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Detailed Solution

ABPQ is a square with centre D
BCSR is a square with centre E
Apply rotation at B
$\frac{z_{P} - z_{B}}{z_{A} - z_{B}} = \frac{| z_{P} - z_{B} |}{| z_{A} - z_{B} |} e^{i (- \frac{π}{2})} \frac{z_{P} - ω}{1 - ω} = - i \Rightarrow z_{P} = ω + i (ω - 1) \frac{z_{P} - ω}{1 - ω} = - i \Rightarrow z_{P} = ω + i (ω - 1) z_{P} \to (\frac{- 1 + i \sqrt{3}}{2} + i (\frac{- 1 + i \sqrt{3}}{2} - 1)) = (\frac{- 1}{2} - \frac{\sqrt{3}}{2} + i (\frac{\sqrt{3} - 3}{2}))$
Apply rotation at point B

$\frac{z_{R} - z_{B}}{z_{C} - z_{B}} = \frac{| z_{R} - z_{B} |}{| z_{C} - z_{B} |} e^{i (\frac{- π}{2})} \Rightarrow \frac{z_{R} - ω}{ω^{2} - ω} = - i \Rightarrow z_{R} = ω - i (ω^{2} - ω) = (\frac{- 1}{2} - \sqrt{3} + i \frac{\sqrt{3}}{2})$

$z_{D} = \frac{z_{A} + z_{P}}{2} = \frac{\frac{- 1}{2} - \frac{i \sqrt{3}}{2} + i (\frac{\sqrt{3} - 2}{2})}{2} = \frac{1 - \sqrt{3}}{4} + i (\frac{\sqrt{3} - 3}{4}) z_{E} = \frac{z_{C} + z_{R}}{2} = \frac{\frac{1}{2} - \frac{\sqrt{3}}{2} + (\frac{- 1}{2} - \sqrt{3} + \frac{i \sqrt{3}}{2})}{2} = \frac{- 1 - \sqrt{3}}{2} D E = | z_{D} - z_{E} | = \sqrt{\frac{3}{2}} A E = | z_{A} - z_{E} | = \frac{3 + \sqrt{3}}{2} \bar{A C} = (\frac{- 3}{2}, \frac{- \sqrt{3}}{2}) \bar{D E} = (\frac{- 3 - \sqrt{3}}{4}, \frac{3 - \sqrt{3}}{- 4}) \cos θ = \frac{\bar{A C} . \bar{D E}}{| \bar{A C} | | \bar{D E} |} = \frac{1}{\sqrt{2}} \Rightarrow θ = 45^{0} = \frac{π}{4} A (1, 0) D (\frac{1 - \sqrt{3}}{4}, \frac{\sqrt{3} - 3}{4}) E (\frac{- 1 - \sqrt{3}}{2}, 0) A r e a o f Δ l e A D E = \frac{1}{2} \times (\frac{3 + \sqrt{3}}{2}) (\frac{3 - \sqrt{3}}{4}) = \frac{6}{16} = \frac{3}{8}$