Let \(O\) be the origin, the point \(A\) be \({z}_{1}=\sqrt{3}+2\sqrt{2}i\), the point \(B({z}_{2})\) be such that \(\sq…

\(\left|{z}_{1}\right|=\sqrt{3+8}=\sqrt{11}\)

Then, \(\left|{z}_{2}\right|=\sqrt{11}/\sqrt{3}=\sqrt{11/3}\)

In \(△ABO,\) (where is the origin),

sides are \(OA=\sqrt{11}\)

and \(OB=\sqrt{11/3}\)

The angle

\(∠AOB=\left|\text{arg}\left({z}_{2}\right)−\text{arg}\left({z}_{1}\right)\right|=\frac{\pi }{6}\)

Using cosine rule for

\(A{B}^{2}=O{A}^{2}+O{B}^{2}−2\left(OA\right)\left(OB\right)\)

\(\text{cos}\left(\frac{\pi }{6}\right)=11+\frac{11}{3}−2\left(\sqrt{11}\right)\left(\sqrt{\frac{11}{3}}\right)\left(\frac{\sqrt{3}}{2}\right)\)

\(=11+\frac{11}{3}−11=\frac{11}{3}\)

Thus, \(AB=\sqrt{11/3}=OB\)

The triangle is isosceles.

Check for obtuse angle:

\(O{A}^{2}=11,\)

\(O{B}^{2}+A{B}^{2}=\frac{11}{3}+\frac{11}{3}=\frac{22}{3}\approx 7.33\)

\(∵O{A}^{2}>O{B}^{2}+A{B}^{2}\), the angle at B is obtuse.

ABO is an obtuse angled isosceles triangle.