Question

The point $A,B,C$ depict the complex numbers $z_1,z_2,z_3$ respectively on a complex plane & the angle $B$ and $C$ of the triangle $ABC$ are each equal to $\frac{1}{2}(\pi -\alpha )$, then $4(z_3-z_1)(z_1-z_2){\sin }^2\frac{\alpha }{2}=$

• A. $(z_2-z_3{)}^2$
• B. $(z_2+z_3{)}^2$
• C. $(z_1-z_3{)}^2$
• D. $(z_1+z_3{)}^2$

Answer 1

The correct option is A $(z_2-z_3{)}^2$

$\frac{z_3-z_1}{z_2-z_1}=e^{i\alpha }$

$\Rightarrow \frac{z_3-z_1}{z_2-z_1}=\cos \alpha +i\sin \alpha$

$\Rightarrow \frac{z_3-z_1}{z_2-z_1}-1=\cos \alpha -1+i\sin \alpha$

$\Rightarrow \frac{z_3-z_2}{z_2-z_1}=-2{\sin }^2\frac{\alpha }{2}+2i\sin \frac{\alpha }{2}\cdot \cos \frac{\alpha }{2}$

$\Rightarrow \frac{z_3-z_2}{z_2-z_1}=2i\sin \frac{\alpha }{2}(\cos \frac{\alpha }{2}+i\sin \frac{\alpha }{2})$

Squaring both sides
$\frac{(z_3-z_2{)}^2}{(z_2-z_1{)}^2}=-4{\sin }^2\frac{\alpha }{2}(\cos \alpha +i\sin \alpha )$

$\frac{(z_3-z_2{)}^2}{(z_2-z_1{)}^2}=-4{\sin }^2\frac{\alpha }{2}\left(\frac{z_3-z_1}{z_2-z_1}\right)$

$\Rightarrow (z_3-z_2{)}^2=4{\sin }^2\frac{\alpha }{2}(z_3-z_1\left)\right(z_1-z_2)$