The principal argument of z=1+cos (((6π)/5))+isin (((6π)/5)) is = ?


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Question Number 109147 by ajfour last updated on 21/Aug/20

$T h e p r i n c i p a l a r g u m e n t o f$ $z = 1 + \cos (\frac{6 π}{5}) + i \sin (\frac{6 π}{5}) i s = ?$
	Answered by Dwaipayan Shikari last updated on 21/Aug/20

	$t a n θ = \frac{s i n \frac{6 π}{5}}{1 + c o s \frac{6 π}{5}} = t a n \frac{3 π}{5}$ $θ = k π + \frac{3 π}{5}$ $a r g (z) = k π + \frac{3 π}{5} (k \in Z)$ $a r g (z) = - π + \frac{3 π}{5} = - \frac{2 π}{5}$
	Commented by ajfour last updated on 21/Aug/20

	$i t h i n k a n s w e r s h o u l d b e θ = - \frac{2 π}{5} .$
	Commented by Dwaipayan Shikari last updated on 21/Aug/20

	$Y e s . I h a v e e d i t e d m y m i s t a k e$
	Answered by Dwaipayan Shikari last updated on 21/Aug/20

	$z = 1 - (\frac{\sqrt{5} - 1}{4}) - i (\frac{\sqrt{10 - 2 \sqrt{5}}}{4})$ $z = \frac{- \sqrt{5} + 5}{4} - i (\frac{\sqrt{10 - 2 \sqrt{5}}}{4})$ $z = \frac{5 - \sqrt{5}}{4} - i (\frac{\sqrt{10 - 2 \sqrt{5}}}{4})$ $a r g (z) = {t a n}^{- 1} (- \frac{\sqrt{10 - 2 \sqrt{5}}}{5 - \sqrt{5}}) = - \frac{2 π}{5}$
	Answered by pticantor last updated on 21/Aug/20

	$z = 1 + e^{i θ} = e^{i \frac{θ}{2}} (e^{- i \frac{θ}{2}} + e^{i \frac{θ}{2}})$ $= 2 c o s (\frac{θ}{2}) e^{i \frac{θ}{2}}$ $t a k e θ = \frac{6 π}{5} a n d a r g (z) = \frac{θ}{2} = \frac{3 π}{5}$
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