Let O be the origin, the point A bez1=3+22i, the point B(z2) be such that 3z2=z1 and argz2=argz1+π6. Then

z1=3+22i &z2z1=13 given argz2z1=π6z2=z2z1⋅z1eiπ6 z2=13⋅(3+22i)(3+i)2 z2=(3−22)+i(26+3)23 Now, z1−z2=(3+22)+i(26−3)23 z1−z2=z2⇒△ABO is isosceles with angles π6,π6&2π3

Let O be the origin, the point A bez1=3+22i, the point B(z2) be such that 3z2=z1 and argz2=argz1+π6. Then

z1=3+22i &z2z1=13 given argz2z1=π6z2=z2z1⋅z1eiπ6 z2=13⋅(3+22i)(3+i)2 z2=(3−22)+i(26+3)23 Now, z1−z2=(3+22)+i(26−3)23 z1−z2=z2⇒△ABO is isosceles with angles π6,π6&2π3

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Let O be the origin, the point A be $z_{1} = \sqrt{3} + 2 \sqrt{2 i},$ the point B(z₂) be such that $\sqrt{3} |z_{2}| = |z_{1}|$ and $\arg (z_{2}) = \arg (z_{1}) + \frac{π}{6}$
. Then

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$area of triangle ABO is \frac{11}{\sqrt{3}}$

ABO is a scalene triangle

$area of triangle ABO is \frac{11}{4}$

ABO is an obtuse angled isosceles triangle

answer is D.

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Detailed Solution

$\begin{array}{l} z_{1} = \sqrt{3} + 2 \sqrt{2} i & \frac{|z_{2}|}{|z_{1}|} = \frac{1}{\sqrt{3}} \\ given \arg (\frac{z_{2}}{z_{1}}) = \frac{π}{6} \\ z_{2} = \frac{|z_{2}|}{|z_{1}|} \cdot z_{1} e^{i (\frac{π}{6})} \end{array}$

$\begin{array}{l} \begin{aligned} z_{2} = \frac{1}{\sqrt{3}} \cdot \frac{(\sqrt{3} + 2 \sqrt{2} i) (\sqrt{3} + i)}{2} \\ z_{2} = \frac{(3 - 2 \sqrt{2}) + i (2 \sqrt{6} + \sqrt{3})}{2 \sqrt{3}} \end{aligned} \\ Now, \\ \begin{aligned} z_{1} - z_{2} = \frac{(3 + 2 \sqrt{2}) + i (2 \sqrt{6} - \sqrt{3})}{2 \sqrt{3}} \\ |z_{1} - z_{2}| = |z_{2}| \Rightarrow △ A B O is isosceles with angles \\ \frac{π}{6}, \frac{π}{6} & \frac{2 π}{3} \end{aligned} \end{array}$