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I want to solve this transcendental complex equation, but I don't know the step by step to get E in form equation and value of E

f[E] = [exp^(i*p*b)/(16*A^2*B*H) {((A + H)^2)[(A + B)^2*exp^(2*i*(k (a - b) - q*a)) - (A - B)^2*exp^(-2*i*(k (a - b) - q*a))] + ((A - H)^2)[(A - B)^2*exp^(-2*i*(k (a - b) + q*a)) - (A + B)^2*exp^(2*i*(k (a - b) + q*a))] + 2 (A^2 - B^2) (A^2 - H^2)[exp^(2*i*q*a) - exp^(-2*i*q*a)]}] = 0

I need to get E function from equation above. And I have changed equation above into another form (trigonometry), which is as follows

tan(qa)=(A*H - i*B*H* tan⁡(k(a-b)))/(i*A*B - A^2*tan(⁡k(a-b)))

and, where

i = imajiner (sqrt(-1))
n = 2;
h = 1;
a = 0.01;
b = 0.02;
m = 1;
h = 1;
c = 137.036;
S1 = 0;
S2 = 0;
V1 = 0;
V2 = 50.000;
k = Sqrt[E^2 - m^2*c^4]/(h*c);
p = Sqrt[(E + m*c^2 - V2 + S2) (E - m*c^2 - V2 - S2)]/(h*c);
q = Sqrt[(E + m*c^2 - V1 + S1) (E - m*c^2 - V1 - S1)]/(h*c);
A = Sqrt[(E - m*c^2)/(E + m*c^2)];
B = Sqrt[(E - m*c^2 - V2 - S2)/(E + m*c^2 - V2 + S2)];
H = Sqrt[(E - m*c^2 - V1 - S1)/(E + m*c^2 - V1 + S1)];

I want to get value of E by using the variables that have been known above.

Help me to get E function and value of E, please. Thank you very much.

Aff_
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  • @ZeroTheHero don't swat it ,I have made much worse mistakes! – Wolphram jonny Jan 04 '21 at 04:48
  • Do you know how to numerically find roots of equations? For example, suppose you have a function $f(x)$, do you know how to find the values of $x$ where $f(x)=0$ numerically? – Philip Jan 04 '21 at 04:19
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    Newton-Raphson method https://en.wikipedia.org/wiki/Newton%27s_method – ZeroTheHero Jan 04 '21 at 04:22
  • @Zero isnt newton only for real valued functions? – Wolphram jonny Jan 04 '21 at 04:32
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    This isnt a question about physics – Triatticus Jan 04 '21 at 04:33
  • I don't know @Philip. But if the question is how to find roots of the simple equation analytically, may be I know. – Aff_ Jan 04 '21 at 04:38
  • I think, this is still a question about physics, it is quantum physics about eigenvalue. Because I get this question when I read a journal about physics. @Triatticus – Aff_ Jan 04 '21 at 04:43
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    @Wolphramjonny yes you are right I did not realize the function was complex (given the poor typesetting). – ZeroTheHero Jan 04 '21 at 04:45
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    @Afifatul_H it isn't about physics, it's about root finding of a complex valued function, there isn't any physics necessary to do that besides possibly discarding roots that are superfluous or that aren't actual solutions based on physical reasoning. – Triatticus Jan 04 '21 at 05:03
  • You can do this with Mathematica’s NSolve[] function. There seem to be lots of solutions. – G. Smith Jan 04 '21 at 05:13
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    Mathjax is the site standard for mathematical expressions. Your post is a lot harder to understand as the equation is hard to read and Mathjax would solve that problem (as well as being the standard on this site). – StephenG - Help Ukraine Jan 04 '21 at 05:17
  • Alright then. Thank you all, for responding and answering my questions. I am very sorry that the questions I ask should not be asked in this community – Aff_ Jan 04 '21 at 05:59

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