Nature of a triangle with vertices $z_1, z_2$ and $-1$ such that $|z_1|=|z_2|=1=z_1+z_2$

Question

If $z_1$ and $z_2$ are distinct complex number such that $|z_1|=|z_2|=1$ and $z_1+z_2=1$, then the triangle in the complex plane with $z_1,z_2$ and $-1$ as vertices must be:

1. equilateral.
2. right angled.
3. be isosceles, but not necessarily equilateral.
4. obtuse angled.

How to solve this?

Answer 1

Hint: Since $|z_1| = |z_2| = 1$, $z_1 = \cos \theta_1 + i \sin \theta_1$ and $z_2 = \cos \theta_2 + i \sin \theta_2$.

Since $z_1 + z_2 = 1 + 0i$, $\cos \theta_1 + \cos \theta_2 = 1$ and $\sin \theta_1 = -\sin \theta_2$. This means that either $\theta_2 = -\theta_1$ or $\theta_2 = \pi + \theta_1$. What happens in each case?

Answer 2

Let $z_1=a+ib$. Since $$ z_1+z_2=1=|z_1|^2=|z_2|^2, $$ then $z_2=1-a-ib$ and $$ a^2+b^2=1=(a-1)^2+b^2, $$ i.e. $$ a=\frac12,\ b=\pm\frac{\sqrt{3}}{2}. $$ Hence $$ z_1=\frac{1+i\sqrt{3}}{2},\ z_2=\frac{1-i\sqrt{3}}{2}. $$ Now we have $$ |z_1+1|=|z_2+1|=|z_1-z_2|=\sqrt{3}, $$ i.e. only 1. holds true.

Answer 3

$$z_1=1-z_2 \Rightarrow 1=|z_1|^2=|1-z_2|^2=(1-z_2)\overline{(1-z_2)}=1-z_2-\bar{z_2}+z_2\bar{z_2}=2-z_2-\bar{z_2}$$

Thus

$$z_2+\bar{z_2}=1 \Rightarrow \mbox{Re}(z_2)=\frac{1}{2}$$ Exactly the same way you get $\mbox{Re}(z_1)=\frac{1}{2}$.

Combining this with $|z_1|=|z_2|=1$ and $z_1 \neq z_2$ you can find the two complex numbers.