write-z-1-3-i-in-e-i-

Question Number 208398 by mokys last updated on 14/Jun/24

w r i t e z = \frac{1}{\sqrt{3} + i} i n e^{i θ}

Answered by A5T last updated on 14/Jun/24

z = \frac{\sqrt{3} - i}{{(\sqrt{3})}^{2} - {(i)}^{2}} = \frac{\sqrt{3} - i}{4}, ∣ z ∣= \sqrt{{(\frac{\sqrt{3}}{4})}^{2} + {(\frac{- 1}{4})}^{2}} = \frac{1}{4}

t a n θ = \frac{\frac{- 1}{4}}{\frac{\sqrt{3}}{4}} = \frac{- 1}{\sqrt{3}} \Rightarrow θ = \frac{- π}{6}

\Rightarrow z = \frac{e^{\frac{- i π}{6}}}{2}

Answered by mathzup last updated on 15/Jun/24

z = \frac{\sqrt{3} - i}{4} = \frac{\sqrt{3}}{4} - \frac{1}{4} i (= x + i y)

∣ z ∣= \sqrt{{(\frac{\sqrt{3}}{4})}^{2} + {(- \frac{1}{4})}^{2}} = \sqrt{\frac{3}{16} + \frac{1}{16}}

= \frac{1}{2} w e h a v e z =∣ z ∣ e^{i a r c t a n (\frac{y}{x})}

= \frac{1}{2} e^{i a r c t a n (- \frac{1}{. \sqrt{3}})} = \frac{1}{2} e^{- \frac{i π}{6}}

Answered by lepuissantcedricjunior last updated on 15/Jun/24

z = \frac{1}{\sqrt{3} + i} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2} + i \frac{1}{2}} = \frac{\frac{1}{2}}{e^{i \frac{π}{6}}} = \frac{1}{2} e^{- \frac{π}{6}}

=> e^{i θ} = \frac{1}{2} e^{- i \frac{π}{6}}

=> θ = - \frac{π}{6}