Complex Equation Solving

Using Euler's formula $\large e^{ix} = \cos(x) + i \sin(x)$ , the given equation can be rewritten as

$A\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right) + B\left(\dfrac{\sqrt{3}}{3} + \dfrac{1}{2}\right) = \sqrt{2}\left(\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}i\right) \Longrightarrow$

$A(1 + \sqrt{3}i) + B(\sqrt{3} + i) = 2 - 2i \Longrightarrow (A + \sqrt{3}B) + (\sqrt{3}A + B)i = 2 - 2i$ .

Equating the respective real and imaginary components on either side of this last equation gives us that

$A + \sqrt{3}B = 2$ and $\sqrt{3}A + B = -2$ .

Adding these two equations then gives us that $(A + B)(1 + \sqrt{3}) = 0 \Longrightarrow A + B = 0 \Longrightarrow A = -B$ .

As $A,B \ne 0$ we can conclude that $\dfrac{A}{B} = \boxed{-1}$ .

Solving for $A,B$ , we find that $A = -(1 + \sqrt{3})$ and $B = 1 + \sqrt{3}$ .