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For the function $G(w)=\frac{1}{2}(\sqrt{2}-\sqrt{2}e^{iw})$ , show that;

$\qquad\mathrm{Re}\,G(w)=\sqrt2\sin^2(w/2)\quad$ and $\quad\operatorname{Im}\,G(w)=-1/\sqrt2\sin w$.

I really need help with understanding this. Anything you could do to help would be greatly appreciated.

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Do you know $e^{i\theta}=\cos\theta+i\sin\theta$ and $\sin$'s half-angle formula? (I'm not sure the amplitudes here are quite correct...) – anon Aug 24 '12 at 1:35
Im aware of eiθ=cosθ+isinθ but didn't know about the sine half angle formula. I just googled it and found this sin(B/2)=±√([1−cos B] / 2). Ive also just made sure that the above equations are the same as the question sheet i copied them from. Im still confused though. – john kash Aug 24 '12 at 1:51
Are you sure it isn't $$G(w)=\frac{1}{2}(\sqrt{2}-\sqrt{2}e^{iw})~?$$ – anon Aug 24 '12 at 2:01
I agree with anon – Aang Aug 24 '12 at 2:06
Oh my... im so sorry about that. I actually checked it and didnt even realise. – john kash Aug 24 '12 at 2:14
up vote 0 down vote accepted

Assumption: $w\in \Bbb R$

$G(w)=\frac{1}{\sqrt 2} (1-\cos w-i\sin w)=\frac{1}{\sqrt 2}(1-\cos w)+i(-\frac{1}{\sqrt 2}\sin w)=\sqrt 2 \sin^2(w/2)+i(-\frac{1}{\sqrt 2}\sin w)$

Thus, Re$(G(w))=\sqrt 2 \sin^2(w/2)$ and Im$(G(w))=-\frac{1}{\sqrt 2}\sin w$

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My question is wrong ill correct it now. sorry. – john kash Aug 24 '12 at 2:15

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