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QUESTION: We know that
e^i(pi)=-1
this gives,
e^2i(pi)=1

taking log to base e on both sides ,
we get,

2i(pi)=0

which means i=0. Can u explain me what went wrong in the above calculation ?(From what i know,taking log for e^2i(pi)is defined, since e^2i(pi)=1 gives [e^2i(pi)] is a real number and thus logarithm is defined) .

ANSWER: Questioner:Surya
Country:Karnataka, India
Category:Advanced Math
Private:No
Subject:

Complex numbers

Question:

We know that
e^i(pi)=-1
this gives,
e^2i(pi)=1

taking log to base e on both sides , we get,

2i(pi)=0

>>>>>>>>>>>>>>>>>> No you don't get that.  The complex logarithm is defined differently -- it is multiple-valued, sort of like the inverse trig functions.


which means i=0. Can u explain me what went wrong in the above calculation ?(From what i know,taking log for e^2i(pi)is defined, since e^2i(pi)=1 gives [e^2i(pi)] is a real number and thus logarithm is defined) .

There is a cute book on Mathematical Fallacies, which you might look up.  Perhaps this belongs there.


---------- FOLLOW-UP ----------

QUESTION: Thanks for the previous answer.

Qs. i = (-1)^1/2
   
applying euler's formula , e^i(pi)=-1
we get,
     i = (e^i(pi))^1/2 =[e^(pi/2)]i
       = (e^(pi/2))^{[e^(pi/2)]^i}
    
   so, i = [e^(pi/2)]^[e^(pi/2)]^[e^(pi/2)].............

As a result we find that i tends to infinity .
And we also have many theories on i

Why can't it be categorised as an "indeterminate form" .

Answer
---------- FOLLOW-UP ----------

QUESTION: Thanks for the previous answer.

Qs. i = (-1)^1/2
>>>>>>>>>>>>>>> You cannot write that, as explained earlier.  Fractional exponents are not well defined for complex numbers.
   
applying euler's formula , e^i(pi)=-1
we get,
    i = (e^i(pi))^1/2 =[e^(pi/2)]i
       = (e^(pi/2))^{[e^(pi/2)]^i}
    
   so, i = [e^(pi/2)]^[e^(pi/2)]^[e^(pi/2)].............

As a result we find that i tends to infinity .
And we also have many theories on i

Why can't it be categorised as an "indeterminate form" .

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