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!

  • 1
    $\begingroup$ 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 $60^\circ$. $\endgroup$ – grand_chat Mar 25 '16 at 7:04
  • $\begingroup$ 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. $\endgroup$ – mea43 Mar 25 '16 at 7:10
  • $\begingroup$ 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 $z1,z2,z3$... $\endgroup$ – Eddy Khemiri Mar 25 '16 at 10:36
  • $\begingroup$ @mea43 Sir I tried for very long, but couldn't reach anywhere. Could you please help me? $\endgroup$ – Better World Mar 25 '16 at 11:01
  • $\begingroup$ @EddyKhemiri Edited, thanks for informing me. $\endgroup$ – Better World Mar 25 '16 at 11:03

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.


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