Equation in complex numbers

Question

Find all complex numbers such that: $$|z_1|=|z_2|=|z_3|$$ $$z_1+z_2+z_3=1$$ $$z_1\cdot z_2 \cdot z_3=1$$ There is solution with vectors or Vietes formulas. Can we solve this problem with using only algebra?

Answer 1

Let $n=|z_1|=|z_2|=|z_3|$. Then $n^3=|z_1z_2z_3|=1$, so $n=1$ and the $z_k$ have unit modulus. Since $z_1=z_2=z_3$ implies $1=|z_1+z_2+z_3|=|3z_1|=3$, and the equations are unchanged under permutations of the $z$'s, we can assume WLOG that $z_3\ne1$. Then let $z_k'=\frac{z_k}{1-z_3}$ and $z_4'=\frac1{1-z_3}$, so that:

$$z_1+z_2+z_3=1\implies \frac{z_1+z_2}{1-z_3}=z_1'+z_2'=1$$

Note that $|z_1'|=|z_2'|=|z_3'|=|z_4'|$. Solving for $z_2'$ and multiplying by the conjugate, we get:

$$|z_2'|^2=|1-z_1'|^2=1-z_1'-\bar z_1'+|z_1'|^2\implies z_1'+\bar z_1'=1$$

So $z_2'=\bar z_1'$. Since it is also true that $-z_3'+z_4'=1$ and $|z_3'|=|z_4'|$, in an exactly equivalent manner we get $-z_3'=\bar z_4'$. Finally, to relate these pairs to each other:

$$(z_1'-\bar z_1')^2=(z_1'+\bar z_1')^2-4|z_1'|^2=(z_4'+\bar z_4')^2-4|z_4'|^2=(z_4'-\bar z_4')^2$$

where the central equality is due to $z_1'+\bar z_1'=1=z_4'+\bar z_4'$ and $|z_1'|=|z_4'|$. Thus either $$z_1'-\bar z_1'=z_4'-\bar z_4'\implies 2z_1'-1=2z_4'-1\implies z_1'=z_4'$$ or $$-(z_1'-\bar z_1')=z_4'-\bar z_4'\implies 2z_2'-1=2z_4'-1\implies z_2'=z_4'.$$

Using our remaining permutation freedom of $z_1,z_2$, we can assume WLOG that $z_1'=z_4'$. Then applying the definitions, we get $\frac1{1-z_3}=\frac{z_1}{1-z_3}\implies z_1=1$, and so $z_2=-z_3$.

We still have yet to use the third equation, $z_1z_2z_3=1$, and we do so now. Since $z_1=1$, this reduces to $z_2^2=-1$, so $z_2=i$ and $z_3=-i$ or vice versa.

Thus up to permutations, $\{z_1,z_2,z_3\}=\{1,i,-i\}$ is the unique solution.

Answer 2

since $z_1, z_2, z_3$ are on the unit circle and $z_1z_2z_3 = 1$ we can take $$z_1 = e^{it}, z_2 =e^{is} \text{ and } z_3 = e^{-i(s+t)}, 0 \le t \le s < 2\pi \text{ and } t + s \le 2\pi$$

the third condition says $$ e^{it} + e^{is} + e^{-i(s+t)} = 1 \tag 1$$

imaginary part of $(1)$ is

$\begin{align} 0 &=\sin t + \sin s - \sin(s+t) \\ &= 2\sin (t/2+s/2)(\cos(s/2-t/2) - \cos(s/2+t/2)\\ &= 4\sin (t/2+s/2)\sin s/2 \sin t/2\\ \end{align}$ the possible solutions are $$t = 0, s = 0, t + s = 0, t+s = 2\pi$$

the real part of $(1)$ is

$$ \cos t + \cos s + \cos(t+s) = 1 \tag 2$$

case $t = 0$

from $(2)$ we get $\cos s = 0,$ so $s = \pi/2, s = 3\pi/2$ that is $$z_1 = 1, z_2 = \pm i, z_3 = \mp i $$

case $s = 0$

we also have $t=0$ and $(2)$ cannot be satisfied.

case $t+s = 2\pi$

from $(2),$ we get $\cos t + \cos t + 1 = 1$ so $\cos t = 0$ which gives you $t = \pi/2, s = 3\pi/2$

$$z_1 = \pm i, z_2 = \mp i, z_3 = 1 $$

the complete solutions is the triples $$ 1, i, -i $$

Answer 3

In more general case: $|z|^2=z\bar{z}$

We have $$z_1z_2z_3=1\,\,\,(1)$$ so we have also $$\bar{z_1}\bar{z_2}\bar{z_3}=1\,\,\,(2)$$ Multiplying $(1)$ and $(2)$ we get: $|z_1|^2|z_2|^2|z_3|^2=1$ thus $|z_1|=|z_2|=|z_3|$. Thus: $z_k=e^{i\theta_k}$