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My confusion lies with this : http://www.wolframalpha.com/input/?i=modulus+%28cos%282+pi+r_1%29%2Bcos%282+pi+r_2%29%2Bi+%28sin%282+pi+r_1%29%2Bsin%282+pi+r_2%29%29%29+squared

I was looking at alternate representations, and I was confused how to go from $|Exp(2\pi i R_1) + Exp(2\pi i R_2)|^2$ to a representation using Real and Imaginary parts, as on the link.

To clear confusion: I was looking at $(-Im(\sin(2 \pi r_1))-Im(\sin(2 \pi r_2))+Re(\cos(2 \pi r_1))+Re(\cos(2 \pi r_2)))^2+(Im(\cos(2 \pi r_1))+Im(\cos(2 pi r_2))+Re(\sin(2 \pi r_1))+Re(\sin(2 \pi r_2)))^2$

3 Answers3

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Using Prosthaphaeresis Formulas

$$\cos2\pi r_1+\cos2\pi r_2=2\cos(r_1+r_2)\pi\cdot\cos(r_1-r_2)\pi$$

$$\sin2\pi r_1+\sin2\pi r_2=2\sin(r_1+r_2)\pi\cdot\cos(r_1-r_2)\pi$$

So using $|x||y|=|xy|$ (where $x,y$ are complex numbers),

$$\left|\cos2\pi r_1+\cos2\pi r_2+i(\sin2\pi r_1+\sin2\pi r_2)\right|$$ $$=2|\cos(r_1-r_2)\pi|\cdot|\cos(r_1+r_2)\pi+i\sin(r_1+r_2)\pi|$$

Now what is $|\cos\phi+i\sin\phi|=?$

lab bhattacharjee
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  • I'm sorry, I'll add to the question but I was actually confused with the $(-Im(\sin(2 \pi r_1))-Im(\sin(2 \pi r_2))+Re(\cos(2 \pi r_1))+Re(\cos(2 \pi r_2)))^2+(Im(\cos(2 \pi r_1))+Im(\cos(2 pi r_2))+Re(\sin(2 \pi r_1))+Re(\sin(2 \pi r_2)))^2$ –  Jan 14 '14 at 17:21
  • @Anonymous, I think, it tries to mean Im$(\cos2\pi r_1+i\sin2\pi r_1)=\sin2\pi r_1$ etc. – lab bhattacharjee Jan 14 '14 at 17:40
  • I don't think it means that, as it says @Im(\sin(2\pi r)$ and specifies that Im is the imaginary part of it's arguments... –  Jan 14 '14 at 17:48
  • @Anonymous, But have you noticed how they have defined Im$(z)$? Now are $r_1,r_2$ real? – lab bhattacharjee Jan 14 '14 at 17:59
  • They defined Im(z) as being the imaginary part of z, right? And yes, in my case r1 and r2 are real, and I know that $Im(\cos(2\pi r + i \sin 2\pi r) = \sin(2\pi r)$ but $Im(sin(2\pi r)$ is something else, in the case they are assuming r to be complex, right? –  Jan 14 '14 at 18:10
  • @Anonymous, yes they are . Now $\sin(ix)=i\sinh(x)$ – lab bhattacharjee Jan 14 '14 at 18:13
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Another way :$$F=\left|\cos A+\cos B+i(\sin A+\sin B)\right|$$

$$=\sqrt{(\cos A+\cos B)^2+(\sin A+\sin B)^2}$$

$$=\sqrt{2+2\cos(A-B)}$$

Using Double-Angle Formulas $\displaystyle \cos2A=2\cos^2A-1$,

$$F=\sqrt{2\cdot 2\cos^2\frac{A-B}2}=2\left|\cos\frac{A-B}2\right|$$

lab bhattacharjee
  • 274,582
  • I was actually interested in the $(-Im(\sin(2 \pi r_1))-Im(\sin(2 \pi r_2))+Re(\cos(2 \pi r_1))+Re(\cos(2 \pi r_2)))^2+(Im(\cos(2 \pi r_1))+Im(\cos(2 pi r_2))+Re(\sin(2 \pi r_1))+Re(\sin(2 \pi r_2)))^2$... –  Jan 14 '14 at 17:22
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Perhaps I am missing something, but the answer seems like it can be presented in a much simpler fashion than the other two answers.

Euler's formula states that $$e^{ix} = \cos{x}+i\sin{x},$$ explaining the separation of a complex exponential into real and imaginary components. Thus, for $$\left| e^{2\pi i r_1} + e^{2\pi i r_2} \right| ^2$$ we expand the two complex exponentials using Euler's formula, resulting in $$\left| \cos{(2\pi r_1)} + i\sin{(2\pi r_1)} + \cos{(2\pi r_2)} + i\sin{(2\pi r_2)}\right| ^2.$$ Note that if we rearrange the terms and factor an $i$, we are left with $$\left| \cos{(2\pi r_1)} +\cos{(2\pi r_2)} + i(\sin{(2\pi r_1)} + \sin{(2\pi r_2)})\right| ^2$$ which was your original input on WolframAlpha. In this case, the real part is $$\cos{(2\pi r_1)} +\cos{(2\pi r_2)}$$ and the imaginary part is $$\sin{(2\pi r_1)} + \sin{(2\pi r_2)}.$$

As for a proof of Euler's formula, if you scroll down to where it says "Using power series" in the Wikipedia link you will find the proof which is usually encountered first in the calculus sequence.

Aaron Taylor
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