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If $z_1$,$z_2$,$z_3$ are the vertices of an equilateral triangle with circumcentre at $(1-2i)$.Find $z_2$ and $z_3$ if $z_3=1+i$

MY ATTEMPT:[Take $z_0$ as circumcentre]

For equilateral triangle:
$z_1+z_2+z_3=3z_0$---(1)

$(z_1)^2+(z_2)^2+(z_3)^2=z_1z_2+z_2z_3+z_1z_3$---(2)

From first equation $z_2+z_3=2-7i$ And from the second equation $(1+i)^2+(2-7i)^2-2z_2z_3=(1+i)(2-7i)+z_2z_3$ which implies $z_2z_3=-18-7i$

So the equation whose roots are $z_2,z_3$ is $z^2-(2-7i)z+(-18-7i)=0$. The solutions are shown here : http://www.wolframalpha.com/input/?i=z%5E2-(2-7i)z%2B(-18-7i)%3D0

Is my method correct?

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You have$$z_3-z_0=3i$$ so $$z_2=z_0+3ie^{\frac{2\pi}{3}}$$ and $$z_1=z_0+3ie^{-\frac{2\pi}{3}}$$

Alternatively, a diagram and simple trigonometry will suffice.

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  • $\begingroup$ I like your method but is my one correct too? (I know it's lengthy) $\endgroup$
    – user220382
    Sep 5 '16 at 21:59
  • $\begingroup$ Where does the second equation come from? $\endgroup$ Sep 6 '16 at 3:53
  • $\begingroup$ It is the condition for a triangle to be equilateral $\endgroup$
    – user220382
    Sep 6 '16 at 6:55
  • $\begingroup$ @DavidQuinn I checked.Even my method is correct and matching with yours.I will accept yours for its conciseness :-). $\endgroup$
    – user220382
    Sep 6 '16 at 23:24

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