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Date: 11/12/97 at 19:53:21
From: Jill
Subject: e^(i*pi)+1=0 Where did it come from?

Is there a way to solve the equation: e^(i*pi)+1=0 without using or 
using very little calculus? I am in a math history class and we must 
go about solving this problem in order to define it and show where 
it came from. Can you help?

Date: 01/23/98 at 17:03:02
From: Doctor Sonya
Subject: Re: e^(i*pi)+1=0 Where did it come from?

Hi Jill.  I'm sorry it took us so long to answer your question.  I 
hope you managed to turn your paper in anyway. Just in case you're 
still interested, here's a little bit of information for you.  
Unfortunately, most of the information about e^(i*Pi) not only uses 
calculus, but analysis and complex analysis as well.

The Math Forum has some information on e^(i*pi) at:


I'm going to give you a reason the equation is true. Keep in mind that 
this is not a proof, for to prove it rigorously you need a pretty deep 
understanding of complex numbers and how they work. If you want to 
know more, you can always write us back. I hope this will at least 
convince you that the equation is true.

First I'll have to introduce a calculus fact you may or may not know.  
Most of the functions you deal with in calculus are polynomials, but 
there are several very important ones that are not: sin, cos, e, and 
ln are the most well known. It turns out that you can write these 
functions as polynomials if you want to.

   Sin(x) = x/1! - x^3/3! + x^5/5! - x^7/7! + x^9.9! - ...

(The notation 5! is said, "five factorial" and means 5*4*3*2*1.  
Likewise, 3! = 3*2*1, 6! = 6*5*4*3*2*1, etc.)

These all continue onto infinity, and the more terms I include, the 
closer the polynomial is to the non-polynomial function. For instance, 
if I want to calculate sin(238), I just plug 238 into the polynomial 
in place of x, and calculate out as many terms as I want to. These are 
called Taylor polynomials. 

Now, I haven't proven to you that these polynomials actually equal the 
things they're supposed to equal, but you'll have to take my word for 
it. If you want to know more, please write us back.

The other thing you need to know is that i^2 = -1. 

I'm going to calculate e^(1*pi) using the Taylor polynomial for x; 
all I have to do is plug in 1*pi for x.
 e^(i*pi) = 1 + (i*pi)/1! + (i*pi)^2/2! + (1*pi)^3/3! + (i*pi)^4/4! + 

Now I'm going to multiply all the i's out. Remember that: 

  i^2 = -1
  i^3 = -i
  i^4 = 1
  i^5 = i
  i^6 = -1
      = 1 + i(pi/1!) - pi^2/2! - i(pi^3/3!) + pi^4/4! + pi^5/5! - ...

Now I put the terms in a different order and pull the i's out of the 
  = 1 - pi^2/2! + pi^4/4! - ... + i[pi/1! - pi^3/3! + pi^5/5! - ...]

And lo and behold, we now have the Taylor polynomials for sin(pi) and 

  = cos[pi] + i*sin[pi].

Now, if you use your calculator (or your head) to find the values of 
cos(pi) and sin(pi), you'll see that


(If you use a calculator, make sure it's in the radians mode. If you 
want to use degrees, pi radians is 180 degrees.) 

Plugging these back into the equation we wrote way up at the top, 
you'll get that:

           = -1 + 0i 
           = -1

Again, if you want a more thorough answer, please write us back.

-Doctors Pete and Doctor Sonya,  The Math Forum
Associated Topics:
High School Calculus

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