# Geometrical Applications of Complex Numbers

The complex numbers $$z_1,z_2,z_3$$ satisfying $$\dfrac{z_1+z_3}{z_2-z_3}=\dfrac{1-i\sqrt{3}}{2}$$ are the vertices of a triangle which is:

a) of area $$0$$

b) equilateral

c) right angled and isosceles

d) obtuse angled



All I got was that from the Rotation Theorem, $$\arg\left(\dfrac{z_1+z_3}{z_2-z_3}\right)=-\pi/3$$ and that $$|z_1+z_3|=|z_2-z_3|$$  Could somebody please show me how to solve this problem? Many thanks!

• If the numerator were $z_1-z_3$ instead of $z_1+z_3$, then the answer would be (b), because the information you're given would imply that two of the sides have equal length and the angle between those two sides is $60^\circ$. – grand_chat Mar 25 '16 at 7:04
• Hint: Shift the origin to any of $z_1, z_2, z_3$ to see the properties of triangle, as angles and sides are invariant under translation of origin. – mea43 Mar 25 '16 at 7:10
• You should edit your post: in fact it's $\frac{\text{z1}+\text{z3}}{\text{z2}-\text{z3}}=e^{-i\frac{\pi }{3}}$ and $\arg \left(\frac{\text{z1}+\text{z3}}{\text{z2}-\text{z3}}\right)=-\frac{\pi }{3}$. As @user21820, I don't think you can know much more about $z1,z2,z3$... – Eddy Khemiri Mar 25 '16 at 10:36
• @mea43 Sir I tried for very long, but couldn't reach anywhere. Could you please help me? – Better World Mar 25 '16 at 11:01
• @EddyKhemiri Edited, thanks for informing me. – Better World Mar 25 '16 at 11:03

## 1 Answer

There is a problem with the problem. By what you have stated, we know that $(-z_1,z_2,z_3)$ are the vertices of an isosceles triangle with equal sides meeting at $120^\circ$. But then we don't know anything much about $(z_1,z_2,z_3)$, because as you can see translating $(-z_1,z_2,z_3)$ around makes $(z_1,z_2,z_3)$ change shape.