i dont know if anything discovered this yet?

Question

let angle in radian be %
e^j%=cos%+Jsin%
Given %=pi radian
e^jpi=-1
taking natural Log
jpi=ln(-1)
j=ln(-1)/pi
if sqaure root of -1=j
square root of -1pi=ln(-1)
squaring by both sides
(jpi)^2=ln^2(-1)
j^2pi^2=ln^2(-1)
-pi^2=ln^2(-1)
1=ln^2(-1)/-pi^2
If this is so we can use this to represent trgonometriv function such as cos and sin so
-ln^2(-1)/pi^2=sin90
-ln^2(-1)=pi^2
ln^2(-1)=-pi^2
ln^2(-1)-9.869(if we sun into the earlier eqution we get one too)
cospi=ln^2(-1)/-pi^2
and likewise for son pi we add a pi^2
using euler equation
e^j2pi=cos2pi+jsin2pi
e^2pij=1
if we ever attempt to solve for j
2pijlne-ln(10
j=o( so imagnary number can exsist at o for certin values of pi
given the earlier deduction i made if
ln^2(-1)=-9.869
it shws thatln^2(-1) is a sum of two productsw hich we can use a spread sheet ne notable thing one value in the product must always stay as negative and the other positive

soory for my mistake
2pijlne=ln(1) not ln10 as shown

i did some deduction myself one studying imaginary numbers

Answer 1

I think that your deduction falls down near the end. You correctly get to e^(2*pi*j) = 1 but this doesn't give 2*pi*j = 0. This is because with complex numbers ln is not a single valued function. In fact ln1 = 0 + n*2pi*j.

Answer 2

thats it?! i knew that when i was 10!