# Sum of angles in triangle is equal to $\pi$

I have to prove that sum of the angles in triangle is equal to $$\pi$$ using complex numbers. Can anyone give me some hints on how to do it? Thank you!

• – Martin R Oct 20 '19 at 13:09
• is there any more elementary proof than this? – user672596 Oct 20 '19 at 13:20

Hint: If the vertices of the triangle are represented by complex numbers $$\ z_1\$$, $$\ z_2\$$ and $$\ z_3\$$, and the angles at these vertices are $$\ \theta_1\$$, $$\ \theta_2\$$, and $$\ \theta_3\$$, respectively, then \begin{align} \frac{z_2-z_1}{\left\vert z_2-z_1\right\vert}&=e^{i\theta_1}\frac{z_3-z_1}{\left\vert z_3-z_1\right\vert}\ ,\\ \frac{z_3-z_2}{\left\vert z_3-z_2\right\vert}&=e^{i\theta_2}\frac{z_1-z_2}{\left\vert z_1-z_2\right\vert}\ , \text{and}\\ \frac{z_1-z_3}{\left\vert z_1-z_3\right\vert}&=e^{i\theta_3}\frac{z_2-z_3}{\left\vert z_2-z_3\right\vert} .\\ \end{align} What happens if you multiply these equations together?