Question

Question #4f142

Answer 1

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

`e^(ix) = 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)`

Answer 2

The right equation is `e^(ix)=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^x` 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

Answer 3

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))`