# Better way to solve $\cos\left(\frac{\gamma'}{2}\right) = g_0$ and $e^{i\beta'} \sin\left(\frac{\gamma'}{2}\right) = g_1$ for $\gamma',\beta'$?

I'm trying to solve the two equations to solve for: $$\gamma',\beta'$$: \begin{align} \cos\left(\frac{\gamma'}{2}\right) = g_0\qquad e^{i\beta'}\sin\left(\frac{\gamma'}{2}\right) = g_1 \end{align} Where $$g_0$$ is real, and $$g_1$$ is some complex number. I tried \begin{align*} \gamma'=2\cos^{-1}(g_0),\qquad\cos(\beta')\sin\left(\frac{\gamma'}{2}\right)=Re(g_1),\ \text{(or)}\ \sin(\beta')\sin\left(\frac{\gamma'}{2}\right)=Im(g_1) \end{align*} Then for each $$\gamma'$$, there should be two corresponding $$\beta'$$s. However, it turned out that my solution doesn't quite work. I'm wondering is there another way I can solve this?

Thanks!

• If you make that "or" an "and" you get one solution for $\beta'$. What makes you say this doesn't work?
– anon
Commented Mar 18, 2021 at 22:07

$$e^{i\beta}=\pm\frac{g_1}{\sqrt{1-g_0^2}}$$ and $$\beta=-i\log\frac{g_1}{\sqrt{1-g_0^2}}+k\pi.$$
• Thanks for the answer! Should the sign of the first term of $\beta$ be $\pm$?
Solve one for $$\sin{\frac{\gamma '}{2}}$$, solve the other for $$\cos{\frac{\gamma '}{2}}$$, and then use $$\sin^2 x + \cos^2 x = 1$$.
Clearly $$\gamma^\prime\pm2\arccos g_0(\mod4\pi)$$, which is consistent with $$\sin\tfrac{\gamma^\prime}{2}=\pm|g_1|$$ iff $$g_0^2+|g_1|^2=1$$, in which case $$e^{i\beta^\prime}=\operatorname{sgn}g_1$$ (except for the case $$g_0=\pm1,\,g_1=0$$, where $$\beta^\prime$$ is arbitrary). This constrains $$\beta^\prime$$ to either the even or odd multiples of $$\pi$$.