Complex Current Divider

Since both paths are in phase, I looked for a value of x that would give both paths the same complex impedance. When you add up the impedances of the path on the left and the path on the right you find that both paths have a real part of "1". You just need to equate the expressions for the imaginary parts, which gives you a value of x = $\frac{2}{3}$

Let the left and right impedances be $Z_{1,\:2}$ , respectively. They both consist of a series connection: $\begin{aligned} Z_1&=(1+j)+x(3j)=1+j(1+3x) &&&\Rightarrow &&&&\sphericalangle (Z_1)&=\arctg2(1+3x,\:1) \\[.5em] Z_2&=(1+2j)+(1-x)(3j)=1+j(5-3x) &&&\Rightarrow &&&&\sphericalangle (Z_2)&=\arctg2(5-3x,\:1) \end{aligned}$ Let $I_{1,\:2}$ be the currents through $Z_{1,\:2}$ from bottom to top, respectively. Then we get via current divider: $\begin{aligned} I_1&=\frac{Z_2}{Z_1}\cdot I_2=\left|\frac{Z_2 I_2}{Z_1}\right|e^{ j\left(\sphericalangle(Z_2) - \sphericalangle(Z_1) + \sphericalangle(I_2)\right) }&&&\Rightarrow &&&& \sphericalangle(I_1)&=\sphericalangle(Z_2) - \sphericalangle(Z_1) + \sphericalangle(I_2) \end{aligned}$ If both current phases $\sphericalangle(I_{1,\:2})$ are equal (mod $2\pi$ ), we can solve for $x$ : $\begin{aligned} k&\in\mathbb{Z}:&&&\sphericalangle(I_1)&=\sphericalangle(I_2)+2\pi k &&&\Rightarrow &&&&\sphericalangle(Z_2)&=\sphericalangle(Z_1)+2\pi k &&&\Rightarrow &&&&1+3x&=5-3x &&&\Rightarrow &&&&x&=\frac{2}{3}\approx\boxed{0.667} \end{aligned}$