solve-inside-C-z-4-1-i-1-i-3-

Question Number 61652 by maxmathsup by imad last updated on 05/Jun/19

s o l v e i n s i d e C z^{4} = \frac{1 - i}{1 + i \sqrt{3}}

Commented by maxmathsup by imad last updated on 06/Jun/19

w e h a v e ∣ 1 - i ∣= \sqrt{2} \Rightarrow 1 - i = \sqrt{2} e^{- \frac{i π}{4}}

∣ 1 + i \sqrt{3} ∣ = 2 \Rightarrow 1 + i \sqrt{3} = 2 {\frac{1}{2} + \frac{i \sqrt{3}}{2}} = 2 e^{\frac{i π}{3}} \Rightarrow \frac{1 - i}{1 + i \sqrt{3}} = \frac{\sqrt{2}}{2} e^{- \frac{i π}{4}} e^{- \frac{i π}{3}}

= \frac{\sqrt{2}}{2} e^{- i (\frac{π}{4} + \frac{π}{3})} = \frac{\sqrt{2}}{2} e^{- \frac{i 7 π}{12}} l e t z = r e^{i θ}

(e) \Rightarrow r^{4} e^{i 4 θ} = 2^{- \frac{1}{2}} e^{- i (\frac{7 π}{12})} \Rightarrow r = 2^{- \frac{1}{8}} a n d 4 θ = - \frac{7 π}{12} + 2 k π \Rightarrow

θ_{k} = - \frac{7 π}{48} + \frac{k π}{2} a n d 0 ⩽ k ⩽ 3 s o t h e s o l u t i o n s a r e

Z_{k} = 2^{- \frac{1}{8}} e^{i (\frac{k π}{2} - \frac{7 π}{48})} w i t h 0 ⩽ k ⩽ 3 .

Answered by MJS last updated on 06/Jun/19

\frac{1 - i}{1 + i \sqrt{3}} = \frac{1 - \sqrt{3}}{4} - i \frac{1 + \sqrt{3}}{4} = \frac{\sqrt{2}}{2} e^{- i \frac{7 π}{12}}

z^{4} = \frac{\sqrt{2}}{2} e^{- i \frac{7 π}{12}}

z = \frac{1}{\sqrt[8]{2}} e^{i π \times {- \frac{31}{48}, - \frac{7}{48}, \frac{17}{48}, \frac{41}{48}}}

Commented by maxmathsup by imad last updated on 06/Jun/19

t h a n k s s i r m j s