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

Question

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 $\sqrt{3}|{z}_{2}|=|{z}_{1}|$ and $arg({z}_{2})=arg({z}_{1})+\frac{\pi }{6}$. Then

Accepted Answer

${z}_{1}=\sqrt{3}+2\sqrt{2}i\, \&\frac{|{z}_{2}|}{|{z}_{1}|}=\frac{1}{\sqrt{3}}$

given $arg(\frac{{z}_{2}}{{z}_{1}})=\frac{\pi }{6}$

$\begin{gathered} {z}_{2}=\frac{|{z}_{2}|}{|{z}_{1}|}\cdot {z}_{1}{e}^{i(\frac{\pi }{6})} \ {z}_{2}=\frac{1}{\sqrt{3}}\cdot \frac{(\sqrt{3}+2\sqrt{2}i)(\sqrt{3}+i)}{2} \ {z}_{2}=\frac{(3-2\sqrt{2})+i(2\sqrt{6}+\sqrt{3})}{2\sqrt{3}} \end{gathered}$

Now,

${z}_{1}-{z}_{2}=\frac{(3+2\sqrt{2})+i(2\sqrt{6}-\sqrt{3})}{2\sqrt{3}}$

$|{z}_{1}-{z}_{2}|=|{z}_{2}|\Rightarrow \DeltaABO$ is isosceles with angles $\frac{\pi }{6},\frac{\pi }{6}\&\frac{2\pi }{3}$

Solutions · 1