# funny complex conjugate

Hi can anyone tell me what $\left|-iAe^{2ix}+Be^{-iy} \right|^{2}$ is equal to, I thought that it would be $$A^{2}(\sin^{2}x+\cos^{2}x)+B^{2}(\sin^{2}y+\cos^{2}y)+2AB$$

but I am reading that it is equal to $$A^{2}(\sin^{2}x+\cos^{2}x)+B^{2}(\sin^{2}y+\cos^{2}y) +2AB(\frac{1}{2}ie^{2ix}e^{-iy}e^{+arg(A)}e^{-arg(B)}-\frac{1}{2}ie^{-2ix}e^{iy}e^{-arg(A)}e^{arg(B)})$$ I am not sure why this second answer is correct.

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treat complex numbers as vectors.. – ABC Mar 24 '13 at 17:27
Compute $\lvert z \rvert^2 = z \cdot \overline{z}$, don't transform into trigonometric functions until you can't simplify exponentials anymore; only then apply trigonometric identities. – vonbrand Mar 24 '13 at 18:18
BTW, both are too complex, $\sin^2 x + \cos^2 x = 1$ always. And the $A^2$, $B^2$ are suspicious if $A$ and $B$ are assumed to be complex (as the last answer hints).. – vonbrand Mar 24 '13 at 18:19

The cross terms of $(z + w)\overline{(z+ w)}$ are given by $$-i Ae^{2ix} \bar{B}e^{iy} + i \bar{A}e^{-2ix} B e^{-iy}$$ which definitely isn't $2 AB$. Seeing as they are including $\arg A$ terms, we will assume $A$ and $B$ are complex and $A$ is actually $|A| e^{i \arg A}$ and $B = |B| e^{i \arg B}$. Then the whole thing should be:

$$|A|^2 + |B|^2 - i|A||B|(e^{i(\arg A-\arg B + 2x + y)} - e^{i(-\arg A + \arg B -2x - y)} )$$

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Sorry this was meant to be a comment. I'll leave it as an answer since it at least explains why it can't be the answer you have. – muzzlator Mar 24 '13 at 17:35
well Im not sure but I guess that A and B must be complex. I read on wiki that $|x|=xe^{i\arg x}$ but it is something that I am not familiar with. – user63407 Mar 24 '13 at 17:41
In that case you should write $|A|^2$ and $|B|^2$ instead of just $A^2$ and $B^2$. – muzzlator Mar 24 '13 at 17:43
can you explain why the cross term is like so, I thought that $|x+y|^{2}=(|x|+|y|)^{2}=|x|^{2}+|y|^{2}+2|x||y|$ – user63407 Mar 24 '13 at 17:49
You see the problem with that formula right? $0 = |0|^2 = |1 + (-1)|^2 = |1|^2 + |-1|^2 + 2|1||-1| = 4$ – muzzlator Mar 24 '13 at 17:51

Assuming $A,B$ to be real and using Euler's Identity/formula $$-iAe^{2ix}+Be^{-iy}=-iA(\cos 2x+i\sin2x)+B(\cos y-i\sin y)$$ $$=(A\sin2x+B\cos y)-i(A\cos2x+\sin y)$$

$$|-iAe^{2ix}+Be^{-iy}|^2=(A\sin2x+B\cos y)^2+(A\cos2x+\sin y)^2$$

$$=A^2(\sin^22x+\cos^22x)+B^2(\cos^2y+\sin^2y)+2AB(\sin2x\cos y+\cos2x\sin y)$$

$$=A^2+B^2+2AB\sin(2x+y)$$

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