What is the Euler's Identity in a nutshell?

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#