${z}_{1}=\sqrt{3}+2\sqrt{2}i&\sqrt{3}\left|{Z}_{1}\right|=\left|{Z}_{2}\right|\text{ and }\arg \left({z}_{2}\right)=\arg \left({z}_{1}\right)+\frac{\pi }{6}\text{ then area }$$\text{ of triangle with vertices }{z}_{1},{z}_{2}\text{ and origin. }$ (28 Jan, Shift I, Memory Based)

Question

YESPYQ · Accepted Answer

Correct answer: a) $\frac{11\sqrt{3}}{4}$. $\arg \left({z}_{2}\right)=\arg \left({z}_{1}\right)+\pi /6\ \arg \left({z}_{2}\right)-\arg \left({z}_{1}\right)=\pi /6\Rightarrow \arg \left(\frac{{z}_{1}}{{z}_{2}}\right)=\pi /6\ \text{ Area of }∆OAB=\frac{1}{2}\left|{z}_{1}\right|\left|{z}_{2}\right|\sin \pi /6\ =\frac{1}{2}|{z}_{1}|\sqrt{3}|{z}_{1}|\sin \frac{\pi }{6}=\frac{1}{2}|{z}_{1}{|}^{2}\sqrt{3}\frac{1}{2}=\frac{11\sqrt{3}}{4}$

\({z}_{1}=\sqrt{3}+2\sqrt{2}i&\sqrt{3}\left|{Z}_{1}\right|=\left|{Z}_{2}\right|\text{ and }\arg \left({z}_{2}\right)=\ar…

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