Question #4f142

3 Answers
Dec 16, 2017

Not true. This is Euler's formula, and it states

e^(ix) = cosx + isinxeix=cosx+isinx

Explanation:

Check this page for a proof of the relation:

https://andromeda.rutgers.edu/~loftin/nafal11/euler-formula.pdf

The proof is a straightforward application of the Taylor expansion of e^(ix)eix

Dec 16, 2017

The right equation is e^(ix)=cos(x)+isin(x)eix=cos(x)+isin(x). It is called Euler's formula.

Explanation:

There are many proofs. You can either expand each function to Taylor series or use the limit definition of e^xex and trigonometric form of complex numbers.

Expanding:
e^x=1+x+x^2/(2!)+x^3/(3!)+...
plug ix into exponent
e^(ix)=1+ix+(ix)^2/(2!)+(ix)^3/(3!)+...
calculate powers of i
e^(ix)=1+ix-x^2/(2!)-(ix^3)/(3!)+...
separate real and imaginary part
e^(ix)=(1-x^2/(2!)+x^4/(4!)+-...)+i(x-x^3/(3!)+x^5/(5!)+-...)
these are expansions of cos(x) and sin(x)

Limit definition:
e^x=lim_(n->oo)(1+x/n)^n
plug ix into exponent
e^(ix)=lim_(n->oo)(1+(ix)/n)^n
the number (1+ix/n) has absolute value
sqrt(1+x^2/n^2)
and argument (angle)
tan(x/n)~~x/n
That's because x/n is very small.

Now when you multiply complex numbers, you multiply absolute values and add angles (De Moivre's theorem and its proof) , so the number (1+ix/n)^n has absolute value
(1+x^2/n^2)^(n/2)~~1+(nx^2)/(2n^2)=1+x^2/(2n)
and argument (angle)
n*tan(x/n)~~n*x/n=x

When n->oo the absolute value approaches 1 and angle approaches x.
Therefore the number lies on the unit circle on the complex plane, and its coordinates are cos(x) and sin(x).

Not convinced enough? You can check
wiki for Euler's formula
wiki for De Moivre's formula
wolframalpha knows stuff - try typing in cos(x), sin(x), e^x, e^(ix) etc.
or just google anything... internet is full of wisdom

Rhys ยท
Dec 22, 2017

A few alternate proofs...

Explanation:

1st order differential equation proof:

Let z = cos theta + isin theta

=>(dz)/(d theta) = -sintheta + icostheta = i^2sintheta + icostheta = i(cos theta + isin theta)

We see that:

=> (dz)/(d theta) = iz

=> (dz)/(d theta) - iz = 0

This is just a homogeneous 1st order linear differential eqaution

Auxiliary equation: k - i = 0

=> k = i

The solution to 1st order: y_(c) = Ae^(kx)

=> z = Ae^(itheta)

For theta=0 we have:

Ae^(0i)=cos 0 + isin 0

Ae^(0)=1 + 0i

A=1

color(red)(z = e^(itheta) = cos theta + isin theta)

Separable differential equation proof:

z = costheta + i sintheta

(dz)/(d theta) = -sintheta + icostheta = i(costheta + isintheta)=iz

=> dz = iz d theta

=> dz/z = i d theta

=> int dz /z = int i d theta

=> ln z = itheta

color(red)( z = e^(itheta))