Complex Numbers And Quadratic Equations question 312
Question: Let the complex numbers $ z_1,z_2 $ and $ z_3 $ be the vertices of an equilateral triangle. Let $ z_0 $ be the circumcentre of the triangle, then $ z_1^{2}+z_2^{2}+z_3^{2}= $ [IIT 1981]
Options:
A) $ z_0^{2} $
B) $ -z_0^{2} $
C) $ 3z_0^{2} $
D) $ -3z_0^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ r $ be the circum radius of the equilateral triangle and $ \omega $ the cube root of unity. Let $ ABC $ be the equilateral triangle with $ z_1,z_2 $ and $ z_3 $ as its vertices $ A,B $ and C respectively with circumcentre $ {O}’(z_0) $ . The vectors $ {O}‘A,{O}‘B,{O}‘C $ are equal and parallel to $ O{A}’,O{B}’,O{C}’ $ respectively. Then the vectors $ \overrightarrow{O{A}’}=z_1-z_0=r{e^{i\theta }} $ $ \overrightarrow{O{B}’}=z_2-z_0=r{e^{( \theta +\frac{2\pi }{3} )}}=r\omega {e^{i\theta }} $ $ \overrightarrow{O{C}’}=z_3-z_0=r{e^{i( \theta +\frac{4\pi }{3} )}}=r{{\omega }^{2}}{e^{i\theta }} $
$ \therefore $ $ z_1=z_0+r{e^{i\theta }},z_2=z_0+r\omega {e^{i\theta }},z_3=z_0+r{{\omega }^{2}}{e^{i\theta }} $ Squaring and adding $ z_1^{2}+z_2^{2}+z_3^{2}=3z_0^{2}+2(1+\omega +{{\omega }^{2}})z_0r{e^{i\theta }} $ + $ (1+{{\omega }^{2}}+{{\omega }^{4}})r^{2}{e^{i2\theta }} $ $ =3z_{^{0}}^{2}, $ since $ 1+\omega +{{\omega }^{2}}=0=1+{{\omega }^{2}}+{{\omega }^{4}} $ Note: Students should remember this question as a formula.
BETA
