# Finding vertices when circumcentre is given for equilateral triangle by complex number method

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?

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.

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