solve-for-x-C-sin-x-z-z-a-bi-re-i-

Question Number 68207 by mr W last updated on 07/Sep/19

s o l v e f o r x \in C

\sin x = z (z = a + b i = {r e}^{i θ})

Commented by mathmax by abdo last updated on 07/Sep/19

s i n x = z \Leftrightarrow \frac{e^{i x} - e^{- i x}}{2 i} = z \Leftrightarrow e^{i x} - e^{- i x} = 2 i r e^{i θ}

l e t t = e^{i x} \Rightarrow i x = l n (t) + i 2 k π w i t h k \in Z

(e) \Rightarrow t - t^{- 1} = 2 i r e^{i θ} \Rightarrow t^{2} - 1 = 2 i r e^{i θ} t \Rightarrow

t^{2} - 2 {i r e}^{i θ} t - 1 = 0 \to Δ^{^{'}} = {({i r e}^{i θ})}^{2} + 1 = - r^{2} e^{2 i θ} + 1

t_{1} = i r e^{i θ} + \sqrt{1 - r^{2} e^{2 i θ}} a n d t_{2} = i r e^{i θ} - \sqrt{1 - r^{2} e^{2 i θ}}

i x = l n (t_{1}) + i 2 k π \Rightarrow - x = {i l n t}_{1} - 2 k π \Rightarrow x = - {i l n t}_{1} + 2 k π

= - i l n (r e^{i θ} - i \sqrt{1 - r^{2} e^{2 i θ}}) + 2 k π

i x = {l n t}_{2} + i 2 k π \Rightarrow - x = {i l n t}_{2} - 2 k π \Rightarrow x = - {i l n t}_{2} + 2 k π

= - i l n (i r e^{i θ} - \sqrt{1 - r^{2} e^{2 i θ}}) + 2 k π

Answered by Smail last updated on 07/Sep/19

\frac{e^{i x} - e^{- i x}}{2} e^{- i π / 2} = {r e}^{i θ}

e^{2 i x} - 1 - 2 {r e}^{i x} \times e^{i (θ + π / 2)} = 0

{(e^{i x} - {r e}^{i (θ + π / 2)})}^{2} = 1 + r^{2} e^{2 i (θ + π / 2)}

e^{i x} = \underline{+} {(1 + r^{2} e^{2 i (θ + π / 2)})}^{1 / 2} + {r e}^{i (θ + π / 2)}

x = - i l n ({r e}^{i (θ + π / 2)} \underline{+} \sqrt{1 + r^{2} e^{2 i (θ + π / 2)}})

x = - {i s i n h}^{- 1} ({r e}^{i (θ + π / 2)})

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