Hi!
I'll start right away. I have a question about Complex Numbers.
I have a complex number, or a bunch of complex numbers, like you can see below. Here I need to know the absolute value of the number B squared, but in the result there is still the imaginery unit in it. I said that all the variables are real numbers greater than zero, so it should be possible to give me a correct term for the absolute value of B.
Or what is my mistake here? How do I use complex numbers with variables in it? How do I make sure that the variables are part of the reals numbers an grater than zero?
Input
\[Alpha] \[Element] Reals; \[Lambda] \[Element] Reals; L \[Element] Reals; \[Alpha] > 0; \[Lambda] > 0; L > 0;
\[CapitalOmega] = (\[Alpha] - I*\[Lambda])/(\[Alpha] + I*\[Lambda])*Exp[\[Alpha]*L];
c = (2*I*\[Lambda])/(I*\[Lambda] + \[Alpha])*Exp[(\[Alpha] - I*\[Lambda])*L/2]/(1 + \[CapitalOmega]^2);
d = c*\[CapitalOmega];
B = c*Exp[(-\[Alpha] + I*\[Lambda])*L/2] + d*Exp[(\[Alpha] + I*\[Lambda])*L/2] - 1;
Abs[B]^2
Output
Abs[-1 + (
2 I E^(L \[Alpha] + 1/2 L (\[Alpha] - I \[Lambda]) +
1/2 L (\[Alpha] + I \[Lambda])) (\[Alpha] -
I \[Lambda]) \[Lambda])/((1 + (
E^(2 L \[Alpha]) (\[Alpha] - I \[Lambda])^2)/(\[Alpha] +
I \[Lambda])^2) (\[Alpha] + I \[Lambda])^2) + (
2 I E^(1/2 L (\[Alpha] - I \[Lambda]) +
1/2 L (-\[Alpha] + I \[Lambda])) \[Lambda])/((1 + (
E^(2 L \[Alpha]) (\[Alpha] - I \[Lambda])^2)/(\[Alpha] +
I \[Lambda])^2) (\[Alpha] + I \[Lambda]))]^2
In[7]:= Simplify[
Abs[-1 + (
C E^(L \[Alpha] +
1/2 L (\[Alpha] + I \[Lambda])) (\[Alpha] -
I \[Lambda]))/(\[Alpha] + I \[Lambda]) + (
2 I E^(1/2 L (\[Alpha] - I \[Lambda]) +
1/2 L (-\[Alpha] + I \[Lambda])) \[Lambda])/((1 + (
E^(2 L \[Alpha]) (\[Alpha] - I \[Lambda])^2)/(\[Alpha] +
I \[Lambda])^2) (\[Alpha] + I \[Lambda]))]^2]
Out[7]= Abs[-1 + (
C E^(1/2 L (3 \[Alpha] + I \[Lambda])) (\[Alpha] -
I \[Lambda]))/(\[Alpha] + I \[Lambda]) + (
2 I \[Lambda])/((1 + (
E^(2 L \[Alpha]) (\[Alpha] - I \[Lambda])^2)/(\[Alpha] +
I \[Lambda])^2) (\[Alpha] + I \[Lambda]))]^2
I hope you can help me. Thanks in advance! :)
Best regards, Thomas