Better way to solve cos⁡(γ′2)=g0 and eiβ′sin⁡(γ′2)=g1 for γ′,β′?Where g0 is real, and g1 is some complex number. I triedγ′=2cos−1(g0), cos⁡(β′)sin⁡(γ′2)=Re(g1) (or) sin⁡(β′)sin⁡(γ′2)=Im(g1)Then for each γ′, there should be two corresponding β′ s. However, it turned out that my solution doesn't quite work

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Better way to solve cos⁡(γ′2)=g0 and eiβ′sin⁡(γ′2)=g1 for γ′,β′?Where g0 is real, and g1 is some complex number. I triedγ′=2cos−1(g0),   cos⁡(β′)sin⁡(γ′2)=Re(g1) (or) sin⁡(β′)sin⁡(γ′2)=Im(g1)Then for each γ′, there should be two corresponding β′ s. However, it turned out that my solution doesn&#039;t quite work

Lynne Trussell · Accepted Answer

ei,β=±g11−g02  and  β=−ilog⁡g11−g02+kπ

Matthew Rodriguez · Accepted Answer

Solve one for sin⁡γ′2 , solve the other for cos⁡γ′2, and then use sin2x+cos2x=1

Vasquez · Accepted Answer

Clearly γ′±2arccos⁡g0(mod4π), which is consistent with sin⁡γ′2=±|g1| if g02+|g1|2=1, in which case eiβ′=sgng1 (except for the case g0=±1,g1=0, where β′ is arbitrary). This constrains β′ to either the even or odd multiples of π

Better way to solve \cos(\frac{\gamma '}{2})=g_0 and P

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