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I am stuck on the following problem:

Let $w_1,w_2$ are distinct complex numbers such that $|w_1|=|w_2|=1$ and $w_1+w_2=1$.Then the triangle in the complex plane with $w_1,w_2,-1$ as vertices

  1. must be isosceles ,but not necessarily equilateral

  2. must be equilateral

I have to decide which of the aforementioned options is correct.

I tried with $\,\,w_1=e^{i\theta},w_2=e^{i \phi},w_3=e^{i \pi}$ but messed it up . Can someone explain how to tackle it?

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    $\begingroup$ Draw a picture. $\endgroup$ – Ma Ming Jun 27 '13 at 11:26
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$$w_k=e^{i\phi_k}\;,\;\;k=1,2\;,\;\;\phi_k\in\Bbb R\implies 1=w_1+w_2=\cos\phi_1+i\sin\phi_1+\cos\phi_2+i\sin\phi_2\implies$$

$$\sin\phi_1=\sin(-\phi_2)\iff \begin{cases}\phi_1=-\phi_2\\{}\\\phi_1=\pi+\phi_2\end{cases}$$

In the first case we have

$$2\cos\phi_1=1\implies \phi_1=\pm\frac\pi3$$

In the second one we get

$$1=\cos\phi_1+\cos(\phi_1-\pi)=0\implies\;\;\text{contradiction}$$

Well, finish the argument...

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Using $w_1=\cos A+i\sin A,w_2=\cos B+i\sin B$

$\sin A+\sin B=0\implies \sin A=-\sin B$ $\implies \cos A=\pm\sqrt{1-\sin^2A}=\pm\sqrt{1-\sin^2B}=\pm\cos B$

As $\cos A+\cos B=1,\cos A\ne-\cos B\implies \cos A=\cos B=\frac12$

So, $\sin A=\pm\sqrt{1-\cos^2A}=\pm\frac{\sqrt3}2 \implies \sin B=-\sin A=\mp \frac{\sqrt3}2$

So, the vertcies are $(-1,0),(\frac12, \frac{\sqrt3}2), (\frac12, -\frac{\sqrt3}2)$

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  • $\begingroup$ Also $(-\frac12, \frac{\sqrt3}2), (-\frac12, -\frac{\sqrt3}2)$ are valid vertices. There are two possible triangles. $\endgroup$ – user51196 Jun 27 '13 at 11:56
  • $\begingroup$ @noether, As $w_1+w_2=1,$ it can not be $-\frac12,$ right? $\endgroup$ – lab bhattacharjee Jun 27 '13 at 12:57
  • $\begingroup$ you are right, I have read $|w_1 + w_2| = 1$, but now I have realized that is without $|\cdot|$. I will remove my answer then :P $\endgroup$ – user51196 Jun 27 '13 at 13:13

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