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This question already has an answer here:

If $z_1,z_2,z_3 \in \mathbb{C}$ and $|z_1|=|z_2|=|z_3|$ and $z_1+z_2+z_3=0$. Prove that $z_1,z_2,z_3$ are points of a isosceles triangle that is on a unit circle with the center in the coordinate beginning.

The answer is given in the following manner:$$z_1=|z_1|e^{i\varphi_1}\\ z_2=|z_2|e^{i\varphi_2} \\ z_3=|z_3|e^{i\varphi_3}\\ |z_1|=|z_2|=|z_3|=|z|$$

It goes on to state that this needs to be true, which I understand why: $$|z_1-z_2|=|z_2-z_3|=|z_3-z_1|=\sqrt{3}|z|$$ and then finding $$|z_1-z_2|=....$$

$$2|z||\sin\frac{\varphi_1-\varphi_2}{2}|=2|z||\sin\frac{\frac{2 \pi }{3}}{2}|????$$

How can we assume this??

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marked as duplicate by mrf, user147263, Yiorgos S. Smyrlis, Yagna Patel, Tim Raczkowski Oct 5 '15 at 23:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Do you mean equilateral triangle? $\endgroup$ – user170231 Oct 5 '15 at 15:56
  • $\begingroup$ See math.stackexchange.com/questions/1397066/… $\endgroup$ – lab bhattacharjee Oct 5 '15 at 15:57
  • $\begingroup$ Obviously, the points lie on a circle centered at the origin. And the second condition means that the centroid is also at the origin. Now show that if circumcentre and centroid coincide then triangle is equilateral $\endgroup$ – G-man Oct 5 '15 at 16:02
  • $\begingroup$ Also, if you know about the euler line, you can see that the orthocentre too conincides with the centroid. $\endgroup$ – G-man Oct 5 '15 at 17:34
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From the point of view of complex addition and absolute value the complex numbers behave like vectors. If the lengths of the vectors are the same and if the resulting vector is $0$ then the vectors (the first one is originated at $0$) form a triangle. A triangle with sides of equal length cannot be anything else but an equilateral triangle.

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  • $\begingroup$ Have to give +1 for the simplicity of the explanation. $\endgroup$ – Shailesh Oct 5 '15 at 16:04
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Suppose $i,j,k$ is a permutation of $1,2,3$. Then we have

\begin{eqnarray} \bar{z}_i (z_i+z_j+z_k) &=& |z_i|^2+ \bar{z}_i z_j+\bar{z}_i z_k= 0 \\ \overline{\bar{z}_j (z_i+z_j+z_k)} &=& |z_j|^2+ \bar{z}_i z_j+z_j \bar{z}_k= 0 \end{eqnarray} Subtracting and using the fact that $|z_i|=|z_j|$ gives $\bar{z}_i z_k = z_j \bar{z}_k$.

Since $|z_m-z_n|^2 = |z_m|^2+|z_n|^2- 2 \operatorname{re} (z_m \bar{z}_n)$, and all the quantities on the right hand side are invariant as long as $m\neq n$, we obtain the desired result.

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