Question

What is the Euler's Identity in a nutshell?

I know it has to do with a relationship with trigonometric functions, but I still don't really understand it.

Answer 1

There are many, many, things named after Euler. However, when we talk about Euler's Identity, we think mostly of

`e^(ipi) +1 = 0`

Where

`{(pi " is the circle constant"),(e" is the base of the natural logarithm"),(i^2 = -1) :}`

Question: Is this related to trigonometric functions?

Well, since we have complex numbers and are talking about trigonometric functions, let's find a formula connecting them.

Let `z` be a complex number, `z=a+bi`, as visualised below.

If we take `(r,theta)` to be the Polar coordinates of `z`, then we can see that

`{(b=rsintheta),(a=rcostheta) :}`

Hence, `z=rcostheta+irsintheta=r(costheta+isintheta)`.

Now, let `r=1` and `theta=pi`.

`z=1(cospi+isinpi)=-1`

So, we have found out that if we use `pi` for the angle coordinate, we reach a complex number with imaginary part zero.

Now, we can spot a somewhat surprising equality here:

`cospi+isinpi = e^(ipi)`

We still didn't prove that `e^(ipi)` is equal to `-1` yet. We just agreed to it.

Maybe proving a general case will be helpful:

`cosx+isinx=e^(ix)`

This is pretty unfamiliar, right? These don't really resemble each other. What's some possible connection between them?

Well, for a fact, we know that the Taylor series for `sinx`, `cosx` and `e^x` at 0 (also called Maclaurin series) are similar, aren't they?

Here they are:

`e^x = sum_(n=0)^oo (x^n)/(n!)=x^0/(0!) + x^1/(1!) + x^2/(2!)+...`

`sinx = sum_(n=0)^oo (-1)^n (x^(2n+1))/((2n+1)!)=x^1/(1!)-x^3/(3!)+x^5/(5!)-...`

`cosx= sum_(n=0)^oo (-1)^n (x^(2n))/((2n)!)=x^0/(0!)-x^2/(2!) + x^4/(4!)-...`

Generally speaking, Taylor series at some point `a` approximates a function of `x` by representing near the point `x=a`.

For info on how to get these series, you can watch this video:

I highly recommend this channel; it offers intuitive proof which can be easily visualised.

Anyway, back on track, we can use this formula to find `e^(ix)`:

`e^(ix)=sum_(n=0)^oo ((ix)^n)/(n!)=1+ix-x^2/(2!)-ix^3/(3!)+x^4/(4!)+...`

Knowing that the powers of `i` repeat in this way:

`i^(4n+k) = i^(k)`, `n in ZZ`

We can sum up the real part and imaginary part:

`e^(ix) = (color(blue)(1-x^2/(2!)+x^4/(4!) - ...)) + i(color(red)(x-x^3/(3!)+x^5/(5!)-...))`

This resembles our Maclaurin series for the trigonometric functions!
Therefore, we have

`e^(ix)=color(blue)cosx+icolor(red)sinx`

As such, if we allow `x=pi`, we reach Euler's Identity:

`e^(ipi) +1 = 0`