How to decompose a complex number into a sum of two unitary modulus complex numbers?

Question

Is it possible to decompose any complex number $z = x + iy\in \mathbb{C}$ with $0\leq|z|\leq2$ into a sum of two unitary modulus exponentials ? i.e. $ z = e^{i\phi_1} + e^{i\phi_2}$ ?

I tried to decompose the problem $x + iy = \cos(\phi_1) + \cos(\phi_2) + i(\sin(\phi_1) + \sin(\phi_2)) $ into a set of two real equations but is seems that they are not linear :

\begin{eqnarray} \cos(\phi_1) + \cos(\phi_2) & = &x \\ \sin(\phi_1) + \sin(\phi_2) & = & y \end{eqnarray}

If it is possible, are there any known algorithm ? I tried the usual trigonometric transformations without success. And formulating the problem in terms of modulus and phase rather than real and imaginary parts made it seem more complex.

Thanks in advance.

Answer 1

This document contains a closed-form solution for a slightly more general version of this problem: find $\theta_1,\theta_2$ such that $$\alpha_1\exp(i\theta_1) + \alpha_2\exp(i(\theta_2+\theta_1))=x+i y$$ with $\alpha_1,\alpha_2>0$ given, and $\alpha_1=\alpha_2=1$ in the context of this question.

The solution is $$\theta_2=\arccos\left(\frac{x^2+y^2-\alpha_1^2-\alpha_2^2}{2\alpha_1\alpha_2}\right)$$ $$\theta_1=\arctan(x/y) - \arctan\left(\frac{\alpha_2\sin(\theta_2)}{\alpha_1+\alpha_2\cos(\theta_2)}\right)$$

Answer 2

No. it is not, because$$\left|e^{i\phi_1}+e^{i\phi_2}\right|\leqslant1+1=2.$$

Answer 3

You can go like this:

first consider $z$ is a real number, let $\cos \theta = z/2$,so we have $z = e^{i\theta} + e^{-i\theta}$.

In general,let $z = re^{i\theta}$,and $r = e^{i\alpha} + e^{-i\alpha}$(As above). Then $z = e^{i(\theta + \alpha)} + e^{i(\theta- \alpha)}$

Answer 4

Geometrically, it is perfectly possible: in the Argand-Cauchy plane, consider the image $M$ of $z$: $0\le OM=r\le 2$, and the intersections $I$ and $J$ of the circle of radius $OM$, centred at $M$. Elementary geometry shows that intersection points exist because$M$ is at distance $ \le 1$ from the unit circle, the quadrilateral $OIMJ$ is a rhombus and $$\overrightarrow{OM}=\overrightarrow{OI}+\overrightarrow{OJ}.$$ If $z_I, z_J$ are the affixes of these points, this vector relation translates as $\;z=z_I+z_J$.

Answer 5

The perpendicular bisector of the segment connecting $z$ with $0$ intersects the unit circle in two points $e^{i\theta_1}$, $e^{i\theta_2}$ satisfying $e^{i\theta_1}+e^{i\theta_2}=z$.

Answer 6

Let's derive the solution ourselves. Given $z = x + iy$ we search for two imaginary numbers $\rho_1, \rho_2$ such that $|\rho_1| = |\rho_2| = 1$ and $\rho_1 + \rho_2 = z$.

Start out with enforcing that the norms are equal: $(\rho_1 + \rho_2) \overline{(\rho_1 + \rho_2)} = \rho_1\overline{\rho_1} + \rho_2\overline{\rho_2} + \rho_1\overline{\rho_2} + \overline{\rho_1}\rho_2 = 2 + 2Re[\rho_1\overline{\rho_2}] = z\overline{z}$

which we can solve for $Re[\rho_1\overline{\rho_2}]$. Of course this is just the cosine of the difference in the angles, $cos(\phi_1 - \phi_2)$. Anyway, plugging in the fact that $\rho_1\overline{\rho_2}$ has also norm $1$, we can then solve for $Im[\rho_1\overline{\rho_2}]$, yielding the full value of $\rho_1\overline{\rho_2}$.

Multiplying the original equation by $\overline{\rho_1}$, we get $1 + \overline{\rho_1}\rho_2= \overline{\rho_1}z$, yielding the $\rho_1 = \overline{z}^{-1}(\rho_1\overline{\rho_2} + 1)$.