We know that $z_1+z_2+z_3=0$ and $|z_1|=|z_2|=|z_3|=1$ where $z_1,z_2,z_3$ are complex numbers. How can we show that the images of $w_1=z_1^4*z_2^3 , w_2=z_2^4*z_3^3 , w_3=z_3^4*z_1^3$ form an equilateral triangle?

Note: the conditions have been edited and the statement now is I believe correct

  • $\begingroup$ is there an intended assumption that the $z_i$ all lie on the unit circle? $\endgroup$ – David Holden Aug 19 '14 at 0:31
  • $\begingroup$ Sorry, the conditions I gave were completely wrong. See edited version. $\endgroup$ – Matheo Aug 19 '14 at 0:34
  • $\begingroup$ You're right David Holden $\endgroup$ – Matheo Aug 19 '14 at 0:38

First, note that the first three $z$ must themselves form an equilateral triangle. To see this, note they all lie on the unit circle, but their sum is zero. Then $$\dfrac{z_2}{z_1}+\dfrac{z_3}{z_1}=e^{i\phi}+e^{i\theta}=-1$$ for some real $\phi,\theta$ since both quotients also have unit modulus. For this expression to be real, we need $\phi=-\theta$ so that the quotients are complex conjugates. But this implies $\cos{\phi}=-\dfrac12\implies \phi =\pm \dfrac{2\pi}{3}$. So $z_2$ and $z_3$ are at angles of $2\pi/3$ relative to $z_1$, and so form an equilateral triangle.

To see that this holds true for the three $w$, observe that all three are of unit modulus since they are products of numbers of unit modulus. Furthermore, their sum is zero:

$$w_1+w_2+w_3=z_1^4 z_2^3+z_2^4 z_3^3+z_3^4 z_1^3 =z_1^7\left(e^{2i\pi}+e^{2i\pi/3}+e^{-8i\pi/3}\right)=z_1^7\left(1+e^{2i\pi/3}+e^{-2i\pi/3}\right)=0$$

Consequently the $w$'s satisfy the same assumptions as the $z$'s, and we conclude that they form an equilateral triangle as well.

  • $\begingroup$ Nice! Lovely idea to get $z_1^7$ out of the parenthesis. $\endgroup$ – Matheo Aug 19 '14 at 1:21

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