Let O be the origin, the point A be z_1=√3+2√2 i, and the point B(z_2) be such that |z_2||…

Question

Let $O$ be the origin, the point $A$ be $z_1=\sqrt3+2\sqrt2 i$, and the point $B(z_2)$ be such that $\dfrac{|z_2|}{|z_1|}=\dfrac{1}{\sqrt3}$ and $\arg\left(\dfrac{z_2}{z_1}\right)=\dfrac{\pi}{6}$. Then:

Accepted Answer

$$z_2 = \frac{|z_2|}{|z_1|}\cdot z_1 e^{i\pi/6} = \frac{1}{\sqrt3}\cdot\frac{(\sqrt3+2\sqrt2 i)(\sqrt3+i)}{2}=\frac{(3-2\sqrt2)+i(2\sqrt6+\sqrt3)}{2\sqrt3}.$$

Computing $z_1-z_2$:
$$z_1-z_2=\frac{(3+2\sqrt2)+i(2\sqrt6-\sqrt3)}{2\sqrt3}.$$

Careful computation shows $|z_1-z_2|=|z_2|$. Hence triangle $ABO$ is isosceles. Since $\arg(z_2/z_1)=\pi/6$ is the angle at $O$, and the side relation forces the other two base angles to each equal $\pi/6$, the third angle at $O$ (or at $A$, per the geometry) is $2\pi/3$ --- an obtuse angle. So $ABO$ is an obtuse-angled isosceles triangle.

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