Geometrical Applications of Complex Numbers












1















The complex numbers $z_1,z_2,z_3$ satisfying $$dfrac{z_1+z_3}{z_2-z_3}=dfrac{1-isqrt{3}}{2}$$ 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, $$argleft(dfrac{z_1+z_3}{z_2-z_3}right)=-pi/3$$ and that $$|z_1+z_3|=|z_2-z_3|$$
$$$$
Could somebody please show me how to solve this problem? Many thanks!






calculus complex-analysis algebra-precalculus geometry complex-numbers







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  • 1




    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 $frac{text{z1}+text{z3}}{text{z2}-text{z3}}=e^{-ifrac{pi }{3}}$ 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















The complex numbers $z_1,z_2,z_3$ satisfying $$dfrac{z_1+z_3}{z_2-z_3}=dfrac{1-isqrt{3}}{2}$$ 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, $$argleft(dfrac{z_1+z_3}{z_2-z_3}right)=-pi/3$$ and that $$|z_1+z_3|=|z_2-z_3|$$
$$$$
Could somebody please show me how to solve this problem? Many thanks!






calculus complex-analysis algebra-precalculus geometry complex-numbers







share|cite|improve this question




















  • 1




    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 $frac{text{z1}+text{z3}}{text{z2}-text{z3}}=e^{-ifrac{pi }{3}}$ 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












1








1


2






The complex numbers $z_1,z_2,z_3$ satisfying $$dfrac{z_1+z_3}{z_2-z_3}=dfrac{1-isqrt{3}}{2}$$ 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, $$argleft(dfrac{z_1+z_3}{z_2-z_3}right)=-pi/3$$ and that $$|z_1+z_3|=|z_2-z_3|$$
$$$$
Could somebody please show me how to solve this problem? Many thanks!






calculus complex-analysis algebra-precalculus geometry complex-numbers







share|cite|improve this question
















The complex numbers $z_1,z_2,z_3$ satisfying $$dfrac{z_1+z_3}{z_2-z_3}=dfrac{1-isqrt{3}}{2}$$ 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, $$argleft(dfrac{z_1+z_3}{z_2-z_3}right)=-pi/3$$ and that $$|z_1+z_3|=|z_2-z_3|$$
$$$$
Could somebody please show me how to solve this problem? Many thanks!






calculus complex-analysis algebra-precalculus geometry complex-numbers




calculus complex-analysis algebra-precalculus geometry complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 at 11:13









user376343

2,7782822




2,7782822










asked Mar 25 '16 at 6:17









Better World

10018




10018








  • 1




    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 $frac{text{z1}+text{z3}}{text{z2}-text{z3}}=e^{-ifrac{pi }{3}}$ 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




    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 $frac{text{z1}+text{z3}}{text{z2}-text{z3}}=e^{-ifrac{pi }{3}}$ 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




1




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




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






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 $frac{text{z1}+text{z3}}{text{z2}-text{z3}}=e^{-ifrac{pi }{3}}$ 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




You should edit your post: in fact it's $frac{text{z1}+text{z3}}{text{z2}-text{z3}}=e^{-ifrac{pi }{3}}$ 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






@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






@EddyKhemiri Edited, thanks for informing me.
– Better World
Mar 25 '16 at 11:03












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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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    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.






    share|cite|improve this answer


























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      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.






      share|cite|improve this answer
























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        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.






        share|cite|improve this 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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 25 '16 at 9:09









        user21820

        38.6k542153




        38.6k542153






























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