I want to confirm following equations solves to know

dϕ = ϕl - ϕ0;
exp1 = (p0 Exp[I ϕ0] + pl Exp[I ϕl])^2 == Icos;
exp2 = (p0 Exp[I ϕ0] + pl Exp[I ϕl] I)^2 == Isin;
exp3 = dϕ = ϕl - ϕ0;
Solve[exp1 && exp2 && exp3, ϕl - ϕ0]

however,the output is

 {dϕ -> -ϕ0 + ϕl

Actually, the answer is

dϕ =-arctan[(2Isin-pl^2-p0^2)/(2Icos-pl^2-p0^2)]

How can I get the correct result?


closed as off-topic by happy fish, Artes, gwr, Feyre, Daniel Lichtblau Mar 10 '17 at 15:19

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  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – happy fish, gwr, Feyre, Daniel Lichtblau
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  • 2
    $\begingroup$ what is Isin and Icos ? arctan? cos? Are these supposed to be trig functions? If so, try with UpperCase first letter? Also watch for the missing space between if so. Also, can't solve for \[Phi]l - \[Phi]0 it has to be single variable you are solving for, $\endgroup$ – Nasser Mar 10 '17 at 3:30
  • $\begingroup$ Isin and Icos are real number ,and arctan is trig function, $\endgroup$ – YOUNGHI Mar 11 '17 at 14:43
  • $\begingroup$ and I have difned dϕ = ϕl - ϕ0, which suposed to be solved. $\endgroup$ – YOUNGHI Mar 11 '17 at 14:45

You have redundant equations, and wrong syntax. Try this

exp1 = (p0 Exp[I ϕ0] + pl Exp[I ϕl])^2 == Icos;
exp2 = (p0 Exp[I ϕ0] + pl Exp[I ϕl] I)^2 == Isin;
Solve[exp1 && exp2, {ϕl, ϕ0}]
  • $\begingroup$ I have done it, but got undesirable result, what's wrong with me, please $\endgroup$ – YOUNGHI Mar 13 '17 at 1:26
  • $\begingroup$ @YOUNGHI Add Clear[ϕ0,ϕ1,Icos,Isin,pl,exp1,exp2,p0] before $\endgroup$ – yarchik Mar 13 '17 at 8:37

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