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Q.
If centre of a regular hexagon is at origin and one of the vertex on argand diagram is $1 + 2i$, then its perimeter is
Complex Numbers and Quadratic Equations
Solution:
Let the vertices be $z_{0}, z_{1}, \ldots, z_{5}$ w.r.t. centre $O$ at origin and $\left|z_{0}\right|=\sqrt{5}$.
$\Rightarrow A _{0} A _{1}=\left|z_{1}-z_{0}\right|=\left|z_{0} e^{i \theta}-z_{0}\right|=\left|z_{0}\right||\cos \theta+i \sin \theta-1|$
$=\sqrt{5} \sqrt{(\cos \theta-1)^{2}+\sin ^{2} \theta}=\sqrt{5} \sqrt{2(1-\cos \theta)}=\sqrt{5} 2 \sin (\theta / 2)$
$\Rightarrow A _{0} A _{1}=\sqrt{5} .2 \sin \left(\frac{\pi}{6}\right)=\sqrt{5}\left(\because \quad \theta=\frac{2 \pi}{6}=\frac{\pi}{3}\right)$
Similarly, $A _{1} A _{2}= A _{2} A _{3}= A _{3} A _{4}= A _{4} A _{5}+ A _{5} A _{0}=6 \sqrt{5}$
Hence the perimeter of, regular polygon is
$= A _{0} A _{1}+ A _{1} A _{2}+ A _{2} A _{3}+ A _{3} A _{4}+ A _{4} A _{5}+ A _{5} A _{0}=6 \sqrt{5}$
