let-z-2-3-i-2-3-calculate-z-n-and-arg-z-n-

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] .