Contradiction in solving sin &#x2061;<!-- ⁡ --> ( z ) = 0 ? sin &#x2061;<!-- ⁡

hornejada1c 2022-07-08 Answered
Contradiction in solving sin ( z ) = 0?
sin ( z ) = 0
e i ( x + i y ) = e i ( x + i y )
e i x + y = e i x y
e i x y + i x y = 1
e 2 i x 2 y = e 0
2 i x 2 y = 0
ix=y
z=x+xi
but, 2 i ( x + i y ) = 2 i z = 0
z = 0 , z = 2 π n
※ ?
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Answers (2)

esperoanow
Answered 2022-07-09 Author has 11 answers
The exponential is no longer injective in the complex plane. When you have e 2 i x 2 y = 1 = e 0 you can't conclude 2 i x 2 y = 0. What happens is that 1 = e 2 π i n , so you have to solve 2 i x 2 y = 2 π i n this gives the missing solutions z = π n

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grenivkah3z
Answered 2022-07-10 Author has 6 answers
You have made a mistake in that
2 i x 2 y = 0
x = i y y

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