Question

how can proving this ?

e^(pi*i) = -1

Answer 1

so this mean that

e^(pi*i) = i^2 make ln of both sides

ln e^(pi*i) = ln i^2

(pi*i)ln e = 2ln i

pi*i = 2ln i divide both sides by 2

(pi*i)/2 = ln i can being this really true that logarithm natural of ,,i" is equal (i*pi)/2 ?

Answer 2

@TheSmartOne
@Kainui
@mathmate
@ganeshie8
@welshfella

Answer 3

Use the expansion of trigonometric and exponential functions.
\(e^{x}=1+x/1!+x^2/2!+...\)
\(e^{i\pi}=1+(i\pi)/1!+(i\pi)^2/2!+...\)
\(=1+i^2(\pi)^2/2!+...+(i\pi)/1!+(i\pi)^3/3!+...\)
\(=1-(\pi)^2/2!+...+i(\pi)/1!-i(\pi)^3/3!+...\)
\(=cos(\pi)+(i)sin(\pi)\)
\(=-1+0\)
\(=-1\)

Answer 4

ty. @mathmate this is really GREAT - thank you very much !!!

Answer 5

You're welcome! :)

Answer 6

Infinite sums is the only way I have seen it derived.

Answer 7

hi

Answer 8

hello

Answer 9

so you accept it now how is here proven ?

Answer 10

It is correct that \(e^{i\pi} = -1\), but it is not true that \(\ln(-1) = i\pi\), because it is not unique. I could also have said that \(\ln(-1) = 3i\pi = 5i\pi = \cdots\)

So because of domain problems, we only think about logarithms in real numbers. Otherwise, it leads to very contradictory results.

Answer 11

but in place of (-1) i used i^2

Answer 12

So exponential functions in complex numbers are OK because they are unique, but logarithmic functions cannot be used at the elementary level.

Answer 13

Here is another way to think why logarithms should not be used for complex numbers:

We know that \(i = 1i = i^4i = i^5\). Now because \(i = i^5\) we can say that \(\ln(i) = \ln(i^5)\). So \(\ln(i) = 5 \ln(i)\) and finally \(\ln(i) = 0\). But \(e^0 \ne i\) so this is false.

Answer 14

can i reposting these all on OS newly please - bc. this your opinion about this is very very interesting

Answer 15

Parth this is possible or not reposting it ?

Answer 16

Yes, please do.

Answer 17

ok. but how ?

Answer 18

I can do it for you.

Answer 19

can you tel me how ?bc. this is

ohhh. ok please do it than now

Answer 20

thank you very much