Solve-in-C-1-iz-1-iz-n-e-i-n-n-R-n-N-

Question Number 127081 by slahadjb last updated on 26/Dec/20

S o l v e i n C

{(\frac{1 + i z}{1 - i z})}^{n} = e_{}^{i θ_{n}}

θ_{n} \in R

n \in N

Answered by Dwaipayan Shikari last updated on 26/Dec/20

{(\frac{1 + i z}{1 - i z})}^{n} = {(\frac{\sqrt{1 + z^{2}}}{\sqrt{1 + z^{2}}} . \frac{e^{({t a n}^{- 1} z) i}}{e^{- ({t a n}^{- 1} z) i}})}^{n} = e^{i θ_{n}}

\Rightarrow e^{2 n ({t a n}^{- 1} z) i} = e^{i θ_{n}} \Rightarrow θ_{n} = 2 {n t a n}^{- 1} z = 2 n (2 π + {t a n}^{- 1} z)

Commented by Olaf last updated on 26/Dec/20

I suppose z \in C, ∣ 1 \pm i z ∣ \neq \sqrt{1 + z^{2}}

Commented by Escritor last updated on 26/Dec/20

E x c e l l e n t

Answered by mathmax by abdo last updated on 26/Dec/20

{(\frac{1 + iz}{1 - iz})}^{n} = e^{i θ_{n}} = e^{i (θ_{n} + 2 k π)} \Rightarrow \frac{1 + {iz}_{k}}{1 - {iz}_{k}} = e^{i (\frac{θ_{n} + 2 k π}{n})} \Rightarrow

1 + {iz}_{k} = e^{i (\frac{θ_{n} + 2 k π}{n})} - {ie}^{i (\frac{θ_{n} + 2 k π}{n})} z_{k} \Rightarrow i (1 + e^{i (\frac{θ_{n} + 2 k π}{n})}) z_{k} = e^{i (\frac{θ_{n} + 2 k π}{n})} - 1 \Rightarrow

z_{k} = - \frac{1 - e^{i (\frac{θ_{n} + 2 k π}{n})}}{1 + e^{i (\frac{θ_{n} + 2 k π}{n})}} = - \frac{1 - \cos (\frac{θ_{n} + 2 k π}{n}) - isin (\frac{θ_{n} + 2 k π}{n})}{1 + \cos (\frac{θ_{n} + 2 k π}{n}) + isin (\frac{θ_{n} + 2 k π}{n})}

= - \frac{{2 \sin}^{2} (\frac{θ_{n} + 2 k π}{2 n}) - 2 isin (\frac{θ_{n} + 2 k π}{2 n}) \cos (\frac{θ_{n} + 2 k π}{2 n})}{{2 \cos}^{2} (\frac{θ_{n} + 2 k π}{2 n}) + 2 isin (\frac{θ_{n} + 2 k π}{2 n}) \cos (\frac{θ_{n} + 2 k π}{2 n})}

- \frac{- isin (\frac{θ_{n} + 2 k π}{2 n}) e^{i (\frac{θ_{n} + 2 k π}{2 n})}}{\cos (\frac{θ_{n} + 2 k π}{2 n}) e^{i (\frac{θ_{n} + 2 k π}{2 n})}} = itan (\frac{θ_{n} + 2 k π}{2 n}) with k \in [[0, n - 1]]

Commented by slahadjb last updated on 28/Dec/20

Z^{n} = e^{i (θ + 2 k Π)} ⇏ Z = e^{i \frac{(θ + 2 k Π)}{n}}

e x a m l e Z = i n = 2

Z^{2} = i^{2} = e^{i (Π + 2 k Π)}

e^{i \frac{(Π + 2 k Π)}{2}} = i o r - i