ft.\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\right\}\) If \(P, Q\) and \(R\) are points,…
ft.\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\right\}\)
If \(P, Q\) and \(R\) are points, respectively representing the complex numbers \(z, z e^{\frac{i \pi}{3}}\) and \(z\left(1+e^{\frac{i \pi}{3}}\right)\) in argand plane, then the area of the triangle \(P Q R\), is
\(\sqrt{3}|z|^2\)
\(\frac{\sqrt{3}}{2}|z|^2\)
\(\frac{\sqrt{3}}{4}|z|^2\)
\(2 \sqrt{3}|z|^2\)
Solution
Given that,
\(\begin{aligned}
P Q & =\left|z e^{i \pi / 3}-z\right|=|z|\left|e^{i \pi / 3}-1\right| \\
& =|z|\left|\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}-1\right| \\
& =|z|\left|-2 \sin ^2 \frac{\pi}{6}+2 i \sin \frac{\pi}{6} \cos \frac{\pi}{6}\right| \\
& =|z|\left|2 \sin \frac{\pi}{6}\right|\left|\sin \frac{\pi}{6}-i \cos \frac{\pi}{6}\right|=|z| \cdot\left|2 \times \frac{1}{2}\right||1|
\end{aligned}\)
\(P Q=|z|\)
Now,
\(\begin{aligned}
& Q R=\left[z\left(1+e^{i \pi / 3}\right)-2 e^{i \pi / 3}\right] \\
& Q R=|z|
\end{aligned}\)
Similarly, \(P R=|z|\)
so, \(P Q=Q R=P R\)
\(\therefore \triangle P Q R\) is equilateral triangle with side length \(\geq 1\).
Now, area of \(\triangle P Q R=\frac{\sqrt{3}}{4}|z|^2\)