A complex number z is defined by z = (1/2)(cos θ + isin θ),such that z^n = (1/2^n ) (cos nθ + isin nθ) Using De Moivre′s theorem,or otherwise, show that (i) Σ_(r=0) ^∞ (1/4^r ) sin 2rθ is a convergent geometic progression. (ii) Σ_(r=0) ^∞ (1/4^r ) sin 2r = ((14 sin 2θ)/(17−16cos 2θ))


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Question Number 87497 by Rio Michael last updated on 04/Apr/20

$A complex number z is defined by z = \frac{1}{2} (\cos θ + i \sin θ), such that$ $z^{n} = \frac{1}{2^{n}} (\cos n θ + i \sin n θ)$ $Using De {Moivre}^{'} s theorem, or otherwise, show that$ $(i) \underset{r = 0}{\sum^{\infty}} \frac{1}{4^{r}} \sin 2 r θ is a convergent geometic progression .$ $(ii) \underset{r = 0}{\sum^{\infty}} \frac{1}{4^{r}} \sin 2 r = \frac{14 \sin 2 θ}{17 - 16 \cos 2 θ}$
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