Why does e^iπ=-1 work?

1.93K views
0

I know that it’s true but why? I’ve never grasped this topic.

In: Mathematics

First, complex numbers are strange

Second, e^(i Pi)=-1 is bests understood by thinking of representing your complex numbers on a graph. The X axis is real numbers, and the Y axis is imaginary numbers. e^(i Pi) is better generalized as R e^(i theta) where R is the distance of your point from (0,0) (aka the magnitude of the complex number) and Theta is the angle(in radians) that you rotate it through on that graph. A circle has 2 Pi radians in it so if you rotate your point around the origin by Pi radians then it is 1 unit (our R) directly to the left of the origin so its -1 on the x axis

If you were to set theta to Pi/2 then the point would be 1 unit up on Y axis and 0 units over on the X axis so it’d be i. If theta is 3Pi/2 then you get -i, same as if its 7 Pi/2 because the circle resets every 2 Pi radians.

You can represent any complex number this way. 17-6i becomes 18.0278 e^(i 1.892 Pi). [Wolfram alpha even shows you the complex plane to help visualize the answer](https://www.wolframalpha.com/input/?i=17+-+6i)

e^(i π) = -1 is a particular result from a more useful, more general relationship:

e^(i t) = cos(t) + i sin(t).

This equation can be derived from the Taylor series of each function (Calculus incoming! You have been warned):

e^x = 1/0! + x/1! + x^(2)/2! + x^(3)/3! + x^(4)/4! + x^(5)/5! + …

= Sum(x^(n)/n!, for n=0 to infinity)

sin(x) = x/1! – x^(3)/3! + x^(5)/5! – …

= Sum((-1)^(n)x^(2n+1)/(2n+1)!, for n=0 to infinity)

sin(x) = 1/0! – x^(2)/2! + x^(4)/4! – …

= Sum((-1)^(n)x^(2n)/(2n)!, for n=0 to infinity)

Now, if we look at what e^(x) does with imaginary numbers, we get

e^(i x) = 1/0! + (ix)/1! + (ix)^(2)/2! + (ix)^(3)/3! + (ix)^(4)/4! + (ix)^(5)/5! + …

= 1/0! + ix/1! – x^(2)/2! – ix^(3)/3! + x^(4)/4! + ix^(5)/5! + …

If we separate out the real parts from the imaginary parts (We’ll skip the proof about absolute convergence. Suffice to say we can rearrange terms as we like), we get

e^(i x) = (1/0! – x^(2)/2! + x^(4)/4! + …) + i(x/1! – x^(3)/3! + x^(5)/5! + …)

= Sum((-1)^(n)x^(2n+1)/(2n+1)!) + i Sum((-1)^(n)x^(2n)/(2n)!)

= cos(x) + i sin(x)

If you haven’t taken calculus or gotten to infinite series yet, sorry if that went over your head. I’m not aware of any way to prove the identity without it. The key takeaway, though, is that

e^(i t) = cos(t) + i sin(t)

and when you plug π into that, you get

e^(i π) = cos(π) + i sin(π)

= -1 + i 0

= -1.

Likewise if you plug in 3π, 5π, 7π, etc.

If you ever had to study some calculus you might vaguely remember the following identities*:
> Sin[x]=(1/2)*(I*Exp[I*x]-I*Exp[-I*x])

> Cos[x]=(1/2)*(Exp[I*x]+Exp[-I*x])

Well, if we do some re-arrangements we can show that

> Exp[I*x]=Cos[x]+I*Sin[x]

and we can see that we’re almost there, we just have to plug in x=Pi.

>Exp[I*Pi]=Cos[Pi]+I*Sin[Pi]

which gives us our answer once we remember Sin[Pi]=0 and Cos[Pi]=-1.

That’s all well and good, but even though we derived the formula (from those Sin[x] and Cos[x] identities up at least) it doesn’t tell us much about what is going on. Why does Exp[I*Pi] go to -1 when we’re taught that Exp[x] always just makes an “exponential growth” curve that starts at 0 when x=-Infinity, goes through 1 when x=0, and then gets super-crazy-big the more x grows?

Well, what happens if we plug in Pi/2 instead of Pi?

> Exp[I*Pi/2]=Cos[Pi/2]+I*Sin[Pi/2]

> Exp[I*Pi/2]=0+I*(1)

We see that Exp[I*Pi/2]=I…? What’s going on?!?! If we plug in more numbers, like 3Pi/2 and Pi/4 and a whole bunch in between, and then graph them out on the complex plane we might start to see what is happening. As x increases, Exp[I*x] just keeps pushing our result further and further around the unit circle. At x=Pi (180 degrees) we’re halfway around the circle at point -1+(0)*I, whereas x=0*Pi (0 degrees) is at 0+(0)*I, and x=Pi/2 (90 degrees) is at 0+(1)*I etc.

So what happens if we do something like put the complex number x+I*y into the exponential function? Well, then we can use regular exponent expansion rules to show:

> Exp[x+I*y]=Exp[x]*Exp[I*y]

Which means that the exponential of a complex number is equal to the exponential of the real part (which does the exponential growth stuff we’re used to seeing) multiplied by the exponential of the imaginary part (which does the rotatey stuff that we noticed when messing around with Exp[I*Pi]).

So why does Exp[I*Pi]=-1 work the way it does? Because, unlike taking the exponential of a purely real number (e.g. x+I*0) which shows monatonic “exponential growth”, taking the exponential of a purely imaginary number (e.g. 0+I*x) rotates a point around the origin of the complex plane (and for x=Pi being 180-degrees specifically we end up back on the real number axis on the other side of the origin)!

* Sidenote: I’m using the notation of Sin[x], Exp[x], I^2 =(-1), Pi=3.14159… so that anyone can copy/paste it into [WolframAlpha.com](https://www.wolframalpha.com/input/?i=Sin%5Bx%5D) if they want to see the graphs of these things.

Exponential of a complex number “interprets” the imaginary part of that complex number as an angle — because this is the only natural way of defining the complex exponential function, so that it works (i.e. has a value, and respects the usual rules of exponentiation) for any complex numbers.

That angle is in radians, which is the natural angle in math (more on that below). 2π radians is a full circle, so π radians is half a circle. Therefore, e^iπ is rotated 180 degrees in the complex plane with respect to e^0 , which is 1. The point opposite to 1 in the complex plane is -1, QED.

Now why is it the case that the imaginary part of the exponent is an angle (in radians)? Well, it turns out that this is the (only) natural way to generalize “usual” (real) exponentiation to all complex numbers. Another poster derived the (not exactly ELI5-compliant) proof from the exponential definition as a sum of an infinite series; let me take another angle (no pun intended) from derivatives. A key property of what makes e the “natural” number for exponentiation, is that the derivative of e^x is itself. There is a similar property for the sine and cosine functions (but only if the angle is measured in radians); except you have to take the derivative four times to loop back to the original function — Just like you have to multiply i by itself four times to go back to 1.

Exponential and sine functions can therefore be seen as two aspects (in fact, they are two projections on orthogonal axes in the complex plane) of one and the same complex-exponential function, which is natural in the sense that it is the second-simplest function that is its own complex-derivative. (The simplest one is of course the zero function, but that’s boring.) This truth is expressed by the Euler formula, e^ix = cos x + i sin x, which gives you a way to compute the exponential of any given imaginary number — and knowing that e^(a + b) = e^a * e^b, which holds for any complex numbers a and b, you can compute separately the exponentials of the real part and imaginary part of any complex number z = x + i*y (the latter using the Euler formula), multiply them, and obtain the value of e^z = e^x * (cos y + i sin y).

(This is not the end of the depths of how sine and exponential functions are intertwined images of each other in the distorting mirror that is making a quarter of a turn in the complex plane. If you would like to know more, look up the Fourier transform and the Laplace transform.)

In conclusion, applying the Euler formula to π gives you e^iπ = cos π + i sin π = -1 + i * 0 = -1. Tada 🙂

[https://www.youtube.com/watch?v=v0YEaeIClKY](https://www.youtube.com/watch?v=v0YEaeIClKY) 3 blue 1 brown is amazing