I'm trying to solve these two equations: $$ \cos\left(\frac{\gamma'}{2}\right) = g_0\qquad e^{i\beta'}\sin\left(\frac{\gamma'}{2}\right) = g_1 $$ Where $g_0$ is real and $g_1$ is complex. From the discussions on the math channel, we can find that $$ \gamma'=2\cos^{-1}(g_0),\qquad e^{i\beta'}=\pm\frac{g_1}{\sqrt{1-g_0^2}}\qquad\beta'=-i\log\frac{g_1}{\sqrt{1-g_0^2}}+k\pi,\ \text{where}\ k\in\mathbb{Z}$$ However, I'm still having trouble understanding the solution when $k$ is an odd number, say if it equals $1$, then I'm trying to verify the solution using Mathematica:
Given that
g_0 = 0.8974187611347583, g_1= -0.33540402107889095+0.28660723962849705*I
We can find $\gamma' = 0.913825694123124 , \beta_1'= 2.434484141132717 , \beta_2'=5.5760767947225105$ (differ by $1*\pi$).
However, when I tried to use Mathematica to verify the solutions, only $\beta_1'$ and $\gamma'$ works, by which I mean $e^{i\beta'}\sin\left(\frac{\gamma'}{2}\right)-g_1=0$.
When I plug $\beta_2', \gamma'$ into the equation, it returns 0.670808 - 0.573214 I. Why the second answer doesn't work? Thanks a lot for the help!!
PS: This is what I tried: Exp[I*5.5760767947225105]* Sin[0.913825694123124/2] - (-0.33540402107889095 + 0.28660723962849705*I) // Simplify
Solvereveals this; the general solution involves distinct arbitrary multiples of $4\pi$ and $2\pi$, as well as a coupled choice of sign. $\endgroup$