let z=(√(2−(√3))) −i(√(2+(√3))) calculate ∣z^n ∣ and arg(z^n )


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Question Number 41519 by maxmathsup by imad last updated on 08/Aug/18

$l e t z = \sqrt{2 - \sqrt{3}} - i \sqrt{2 + \sqrt{3}}$ $c a l c u l a t e ∣ z^{n} ∣ a n d a r g (z^{n})$
	Answered by alex041103 last updated on 09/Aug/18

	$W e k n o w t h a t ∣ z^{n} ∣=∣ z ∣^{n} .$ $\Rightarrow∣ z^{n} ∣=∣ z ∣^{n} = {(R e {(z)}^{2} + I m {(z)}^{2})}^{n / 2} =$ $= {(2 - \sqrt{3} + 2 + \sqrt{3})}^{n / 2} = 2^{n}$ $A l s o a r g (z^{n}) \equiv n a r g (z) (m o d 2 π) .$ $\Rightarrow a r g (z^{n}) \equiv n a r g (z) (m o d 2 π)$ $\equiv n (2 π - a r c t a n (\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2 - \sqrt{3}}}))$ $\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2 - \sqrt{3}}} = \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2 - \sqrt{3}}} \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2 + \sqrt{3}}} = \frac{2 + \sqrt{3}}{\sqrt{4 - 3}} = 2 + \sqrt{3}$ $\Rightarrow∣ z^{n} ∣= 2^{n} a n d a r g (z^{n}) = n (2 π - a r c t a n (2 + \sqrt{3})) (m o d 2 π)$
	Answered by maxmathsup by imad last updated on 09/Aug/18

	$w e h a v e {c o s}^{2} (\frac{π}{12}) = \frac{1 + c o s (\frac{π}{6})}{2} = \frac{1 + \frac{\sqrt{3}}{2}}{2} = \frac{2 + \sqrt{3}}{4} \Rightarrow c o s (\frac{π}{12}) = \frac{\sqrt{2 + \sqrt{3}}}{2} a l s o$ $w e h a v e {s i n}^{2} (\frac{π}{12}) = \frac{1 - c o s (\frac{π}{6})}{2} . \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{2 - \sqrt{3}}{4} \Rightarrow s i n (\frac{π}{12}) = \frac{\sqrt{2 - \sqrt{3}}}{2} \Rightarrow$ $z = 2 s i n (\frac{π}{12}) - 2 i c o s (\frac{π}{12}) = 2 c o s (\frac{π}{2} - \frac{π}{12}) - 2 i s i n (\frac{π}{2} - \frac{π}{12})$ $= 2 c o s (\frac{5 π}{12}) - 2 i s i n (\frac{5 π}{12}) = 2 e^{- i \frac{5 π}{12}} \Rightarrow z^{n} = 2^{n} e^{- n \frac{i 5 π}{12}} \Rightarrow$ $∣ z^{n} ∣ = 2^{n} a n d a r g (z^{n}) \equiv - \frac{5 n π}{12} [2 π] .$
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