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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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    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

1 Answers 1

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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