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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...)2012-08-24
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    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.2012-08-24
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    Are you sure it isn't $$G(w)=\frac{1}{2}(\sqrt{2}-\sqrt{2}e^{iw})~?$$2012-08-24
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    I agree with anon2012-08-24
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    Oh my... im so sorry about that. I actually checked it and didnt even realise.2012-08-24
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    @johnkash: please use $\LaTeX$ (which was there) instead of images of the square root sign. This version doesn't show what is under the sign as well. You also lost the opening dollar sign in the title. Also, $\sqrt 2 /2 e^{iw}$ is dangerous as some will think the $e^{iw}$ is in the numerator and some the denominator. Please add parentheses, make it $\sqrt 2 e^{iw}/2$ or (preferably) use \frac like this: \frac{\sqrt 2 e^{iw}}2 for $\frac{\sqrt 2 e^{iw}}2$2012-08-24
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    Your latest edit did not address my comments. I think it is still not clear whether the $e^{(iw)}$ is in the denominator or in the numerator of that term. Deleting the brackets in favor of parentheses doesn't address the order of operations.2012-08-24

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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.2012-08-24