solve-sinz-2-zfromC-

Question Number 37296 by math khazana by abdo last updated on 11/Jun/18

s o l v e s i n z = 2 z f r o m C

Commented by prof Abdo imad last updated on 17/Jun/18

s i n z = 2 \Leftrightarrow \frac{e^{i z} - e^{- i z}}{2} = 2 \Leftrightarrow e^{i z} - e^{- i z} = 4 \Rightarrow

l e t t = e^{i z} \Rightarrow t - \frac{1}{t} = 4 \Rightarrow t^{2} - 1 = 4 t \Rightarrow

t^{2} - 4 t - 1 = 0

Δ^{^{'}} = 4 + 1 = 5 \Rightarrow t_{1} = 2 + \sqrt{5} a n d t_{2} = 2 - \sqrt{5}

e^{i z} = 2 + \sqrt{5} \Rightarrow i z = l n (2 + \sqrt{5}) \Rightarrow z = - i l n (2 + \sqrt{5})

e^{i z} = 2 - \sqrt{5} \Rightarrow i z = l n (2 - \sqrt{5})

= l n (- 1) + l n (- 2 + \sqrt{5}) = i π + l n (- 2 + \sqrt{5}) \Rightarrow

z = π - i l n (- 2 + \sqrt{5}) s o t h e s o l u t i o n s f o r

t h i s e q u a t i o n a r e

- i l n (2 + \sqrt{5}) a n d π - i l n (- 2 + \sqrt{5}) .