If z_1+z_2+z_3=0 and |z_1|=|z_2|=|z_3|=1 , show that…

If $z_1+z_2+z_3=0$ and $|z_1|=|z_2|=|z_3|=1$, show that $z_1,z_2,z_3$ are vertices of an equilateral triangle inscribed in the unit circle.

Step 1 · Use the zero-sum condition

$z_3=-(z_1+z_2)$, so $|z_1+z_2|=|z_3|=1$.

Step 2 · Expand $|z_1+z_2|^2$

$$ |z_1+z_2|^2=|z_1|^2+|z_2|^2+2\,\text{Re}(z_1\bar z_2)=2+2\,\text{Re}(z_1\bar z_2) $$

Setting this equal to $1$:

$$ \text{Re}(z_1\bar z_2)=-\frac12 $$

Step 3 · The same holds for every pair, by symmetry

The zero-sum condition is symmetric in $z_1,z_2,z_3$, so identical reasoning (pairing any two and using the third) gives:

$$ \text{Re}(z_1\bar z_2)=\text{Re}(z_2\bar z_3)=\text{Re}(z_3\bar z_1)=-\frac12 $$

Step 4 · Compute all three side lengths

$$ |z_1-z_2|^2=|z_1|^2+|z_2|^2-2\,\text{Re}(z_1\bar z_2)=2-2\left(-\frac12\right)=3 $$

and identically $|z_2-z_3|^2=|z_3-z_1|^2=3$. So:

$$ \boxed{|z_1-z_2|=|z_2-z_3|=|z_3-z_1|=\sqrt3} $$

Step 5 · Conclusion

All three sides are equal — the triangle is equilateral, inscribed in the unit circle, with side length $\sqrt3$. Equivalently, since $\text{Re}(z_1\bar z_2)=\cos(\arg z_1-\arg z_2)=-\frac12$, consecutive vertices are separated by exactly $120^\circ$.

Verification

$z_1=1,\ z_2=e^{2\pi i/3},\ z_3=e^{4\pi i/3}$: sum $=1+\left(-\frac12+i\frac{\sqrt3}2\right)+\left(-\frac12-i\frac{\sqrt3}2\right)=0$ ✓. Side: $|z_1-z_2|=\left|\frac32-i\frac{\sqrt3}2\right|=\sqrt{\frac94+\frac34}=\sqrt3$ ✓.