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Triangle formed by rotating a complex number: area, side or argument

This type of question, asked 5 times: 2021 Paper 7, 2021 Paper 10, 26 Jul 2022, 11 Apr 2023, 28 Jan 2025. All 5 questions below, oldest first.

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  1. Question 1

    2021 Paper 7

    73
    The area of the triangle with vertices A(z),B(iz)A(z),B(iz) and C(z+iz)C(z + iz) is :

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  2. Question 2

    2021 Paper 10

    70
    Let a complex number be w=1−3iw = 1 -\sqrt{3}i. Let another complex number zz be such that ∣zw∣=1|zw| = 1 and arg(z)−arg(w)=π2arg(z) - arg(w) =\frac{\pi}{2}. Then the area of the triangle with vertices origin, zz and ww is equal to:

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  3. Question 3

    26 Jul 2022

    62
    Let OO be the origin and AA be the point z1=1+2iz_{1}= 1 + 2i. If BB is the point z2,Re(z2)<0z_{2},Re\left( z_{2} \right)< 0, such that OABOAB is a right- angled isosceles triangle with OBOB as hypotenuse, then which of the following is NOT true?

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  4. Question 4

    11 Apr 2023

    68
    Let w1w_{1} be the point obtained by the rotation of z1=5+4iz_{1}= 5 + 4i about the origin through a right angle in the anticlockwise direction, and w2w_{2} be the point obtained by the rotation of z2=3+5iz_{2}= 3 + 5i about the origin through a right angle in the clockwise direction. Then the principal argument of w1−w2w_{1}-w_{2} is equal to

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  5. Question 5

    28 Jan 2025

    68
    Let OO be the origin, the point AA be z1=3+22iz_{1}=\sqrt{3}+ 2\sqrt{2}i, the point B(z2)B\left( z_{2} \right) be such that 3∣z2∣=∣z1∣\sqrt{3}\left| z_{2} \right|=\left| z_{1} \right| and arg(z2)=arg(z1)+π6arg\left( z_{2} \right)= arg\left( z_{1} \right)+\frac{\pi}{6}. Then

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