sin-z-200-find-z-

Question Number 19419 by NEC last updated on 10/Aug/17

\sin z = 200

f i n d z

Commented by NEC last updated on 10/Aug/17

y e s . i k n o w t h a t .

i f i t s a c o m p l e x n u m b e r i^{'} l l l i k e

t o s e e i t . T h a n k s

Answered by sma3l2996 last updated on 10/Aug/17

s i n z = \frac{e^{i z} - e^{- i z}}{2 i} = 200 \Leftrightarrow e^{i z} - e^{- i z} = 400 i; z = a + i b

e^{i a} e^{- b} - e^{- i a} e^{b} = 400 i

e^{- b} (c o s a + i s i n (a)) - e^{b} (c o s a - i s i n a) = 400 i

c o s (a) (e^{- b} - e^{b}) + i s i n (a) (e^{- b} + e^{b}) = 400 i

c o s (a) = 0

s i n (a) (e^{- b} + e^{b}) = 400

a = \frac{π}{2} + k π

{(- 1)}^{k} (e^{- b} + e^{b}) = 400

e^{2 b} + 1 = {(- 1)}^{k} 400 e^{b}

e^{2 b} + {(- 1)}^{k + 1} 400 e^{b} + 1 = 0

e^{2 b} + {(- 1)}^{k + 1} 2 \times 200 e^{b} + 200^{2} - 200^{2} + 1 = 0

{(e^{b} + {(- 1)}^{k + 1} 200)}^{2} = 4 \times 10^{4} - 1

e^{b} = \sqrt{39999} + {(- 1)}^{k} 200

b = l n (\sqrt{39999} + {(- 1)}^{k} 200)

Commented by NEC last updated on 11/Aug/17

w o w! t h a n k s

h o w i s a = \frac{π}{2} + k π ?

Commented by sma3l2996 last updated on 11/Aug/17

b e c a u s e c o s (a) = 0

Commented by NEC last updated on 11/Aug/17

t h a n k s \dots . i t s u n d e r s t o o d .

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