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 = 0ix=y z=x+xibut, 2 i ( x + i y ) = 2 i z = 0 ⇒ z = 0 , z = 2 π n※ ?

Question

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  =  0ix=y z=x+xibut,   2  i  (  x  +  i  y  )  =  2  i  z  =  0  ⇒  z  =  0  ,  z  =  2      π    n※ ?

esperoanow · Accepted Answer

The exponential is no longer injective in the complex plane. When you have       e          2      i      x      −      2      y        =  1  =      e    0   you can&#039;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

grenivkah3z · Accepted Answer

You have made a mistake in that  2  i  x  −  2  y  =  0    ⟹    x  =  −  i  y  ≠  −  y

Contradiction in solving sin ⁡ ( z ) = 0 ? sin ⁡<!-- ⁡

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