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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" .

---------- 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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