If e 2 π i = 1 then is 2 π i = 0?So with Euler identity I used 2 π ( τ ) instead and got e 2 π i = 1, and took natural logarithm 2 π i = 0? But is see online the answer is 6.28.... i.

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If       e          2      π      i        =  1 then is   2  π  i  =  0?So with Euler identity I used   2  π  (  τ  ) instead and got       e          2      π      i        =  1, and took natural logarithm   2  π  i  =  0? But is see online the answer is   6.28....  i.

Eynardfb0 · Accepted Answer

In   (  0  ,  ∞  ), every number has one and only one real logarithm. So, in   (  0  ,  ∞  ), it makes sense to assert that       e    x    =  y    ⟺    x  =  log  ⁡  yHowever, every complex numbers has infinitely many logarithms. In particular every number of the form   2  k  π  i (with   k  ∈      Z  ) is a logarithm of   1. So, your approach does not work here.Another way of seeing this is: in       R  , the exponential function is injective, but not in       C

inurbandojoa · Accepted Answer

The exponential function   f  (  x  )  =      e    x   is a group homomorphism from   (      C    ,  +  ) to   (      C    ∖  {  0  }  ,  ⋅  ). And yes, this group homomorphism is not injective, for   f  (  2  π      i    )  =  f  (  0  )  =  1. In fact,   Ker  ⁡  (  f  )  =  2  π      i        Z

If e^(2 pi i)=1 then is 2pi i=0?

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