Geometrical Applications of Complex Numbers












1












$begingroup$



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











$endgroup$








  • 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 $frac{text{z1}+text{z3}}{text{z2}-text{z3}}=e^{-ifrac{pi }{3}}$ 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 $z1,z2,z3$...
    $endgroup$
    – 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


















1












$begingroup$



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











$endgroup$








  • 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 $frac{text{z1}+text{z3}}{text{z2}-text{z3}}=e^{-ifrac{pi }{3}}$ 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 $z1,z2,z3$...
    $endgroup$
    – 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
















1












1








1


2



$begingroup$



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











$endgroup$





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 '18 at 11:13









user376343

3,6783827




3,6783827










asked Mar 25 '16 at 6:17









Better WorldBetter World

10038




10038








  • 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 $frac{text{z1}+text{z3}}{text{z2}-text{z3}}=e^{-ifrac{pi }{3}}$ 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 $z1,z2,z3$...
    $endgroup$
    – 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
















  • 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 $frac{text{z1}+text{z3}}{text{z2}-text{z3}}=e^{-ifrac{pi }{3}}$ 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 $z1,z2,z3$...
    $endgroup$
    – 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










1




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




$begingroup$
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






$begingroup$
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 $frac{text{z1}+text{z3}}{text{z2}-text{z3}}=e^{-ifrac{pi }{3}}$ 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 $z1,z2,z3$...
$endgroup$
– Eddy Khemiri
Mar 25 '16 at 10:36




$begingroup$
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 $z1,z2,z3$...
$endgroup$
– 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






$begingroup$
@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






$begingroup$
@EddyKhemiri Edited, thanks for informing me.
$endgroup$
– 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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    $begingroup$

    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









    $endgroup$


















      1












      $begingroup$

      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









      $endgroup$
















        1












        1








        1





        $begingroup$

        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









        $endgroup$



        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









        user21820user21820

        38.9k543153




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