If $a,b,c,d$ are complex number with the same modulus such that $a+b+c=d$, then $d$ is equal to one of the other three.

Question

I've come up with a solution, but it's not so insightful, so I was wondering if anyone has a cleaner not-so-brute-force solution:

Equivalently (since the case where all of them ar zero is trivial), if $\alpha,\beta,\gamma$ are complex numbers in the unitary circle such that $\alpha + \beta + \gamma = 1$ then one of them is equal to one. Then one could write $\alpha = a_\alpha + b_\alpha$ and so on, and solve the system of equations for the real part, the imaginary part and the modulus.

Is there a neater form of solving it? Like using the triangle inequality or something like so?

Answer 1

Replacing $c$ by $-c$, we can rewrite the equation as $a + b = c + d$. Assume that all four of these numbers lie on the unit circle, and (by rotation) assume that $a+b$ is a nonnegative real number. If this is positive, then you can check that it determines the unordered pair $\{a, b\}$ uniquely, and thus either $d = a$ or $d = b$. On the other hand, if $a+b=c+d=0$, then $d = -c$.

Answer 2

Here's an algebraic solution that is somewhat simpler than brute force $sin$ and $cos$ comparisons, but also isn't very insightful. We are given that $$e^{\alpha i} +e^{\beta i} +e^{\gamma i} = 1 $$ Applying complex conjugation to both sides we get $$e^{-\alpha i} +e^{-\beta i} +e^{-\gamma i} = 1 $$

Expanding $(1-e^{\alpha i})(1-e^{\beta i})(1-e^{\gamma i})$ gives $$ 1 - e^{\alpha i} -e^{\beta i} -e^{\gamma i} + e^{(\alpha +\beta)i} +e^{(\beta + \gamma)i} +e^{(\gamma + \alpha)i}-e^{(\alpha + \beta + \gamma)i} =$$ $$ 1 - e^{\alpha i} -e^{\beta i} -e^{\gamma i} + (e^{-\alpha i} +e^{-\beta i} +e^{-\gamma i}- 1)\cdot e^{(\alpha + \beta + \gamma)i}=$$ $$ 0 + 0\cdot e^{(\alpha + \beta + \gamma)i}= $$ $$ 0$$

Therefore, some factor of the form $(1-e^{\theta i})$ is zero.

Answer 3

Reformulate the question to:

If $A, B, C, D$ are points on the unit circle such that $ A + B + C + D = 0$, and $ A + B \neq 0, A + D \neq 0$, then $ A = - C$

Consider the 4 distinct points: $ 0, A, A+B,$ and $ A+B+C = - D $.
This is a quadrilateral whose sides are all length 1, so it is a rhombus.
The complex number representing going from $0$ to $A$ is just $A$, and the complex number representing going from $-D$ to $A+B$ is $-C$.
These 2 are opposite sides of a rhombus, so they are equal. Hence $ A = - C $.