Explanation on how we use polar coordinates/differential equations to prove Euler's formula, leading onto Euler's identity. Thanks

Aman NAGUNov 12, 2019Wow, thank you so much. The math behind this elegant formula is amazing. Thanks once again!

JamsDec 14, 2019Let z be the complex number on the unit circle in the complex plane. Thus, z= cos(φ)+isin(φ). (1) iz = icos(φ)-sin(φ)(2) dz/dφ = -sin(φ)+icos(φ)Notice how (1) and (2) are equal to eachother.(3) dz/dφ = iz(3) is in the form f'(x) = a*f(x). This equation implies that the derivative of f(x), f'(x), is equal to its original function f(x) when a=1 ( f(x)=f'(x) ). Therefore, f(x)=e^x, when a=1. Moreover, when f(x)*a =f'(x), the chain rule is applied. Thus, f(x)=e^(ax).Going back to (3) one sees that a=i. Consequently, z=e^(iφ)Finally, e^(iφ)=cos(φ)+isin(φ)QED

Will this do as a proof?

Wow, thank you so much. The math behind this elegant formula is amazing. Thanks once again!

@Aman NAGU

ℕℙ

Let z be the complex number on the unit circle in the complex plane.

Thus,

z= cos(φ)+isin(φ).(1) iz = icos(φ)-sin(φ)

(2) dz/dφ = -sin(φ)+icos(φ)

Notice how (1) and (2) are equal to eachother.

(3) dz/dφ = iz

(3) is in the form f'(x) = a*f(x).

This equation implies that the derivative of f(x), f'(x), is equal to its original function f(x) when a=1

( f(x)=f'(x) ). Therefore, f(x)=e^x, when a=1.

Moreover, when f(x)*a =f'(x), the chain rule is applied. Thus, f(x)=e^(ax).

Going back to (3) one sees that a=i. Consequently, z=e^(iφ)

Finally,

e^(iφ)=cos(φ)+isin(φ)QED