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### Euler Equation and DeMoivre's Theorem

```
Date: 05/18/99 at 19:01:03
From: Anthony
Subject: e, pi, and i

Is there a proof of e^(i*Pi) +1 = 0?
```

```
Date: 05/19/99 at 09:44:10
From: Doctor Anthony
Subject: Re: e, pi, and i

The derivation is not too difficult if you are familiar with the
basics of complex numbers and exponential functions.

Then  dz/dx = -sin(x) + i*cos(x)

= i(cos(x) + i*sin(x))         (since i^2 = -1)

= i*z

So   dz/z = i*dx

Now integrate both sides

ln(z) = i*x + c    From (1); when x=0, z=1   so c=0

ln(z) = i*x

z = e^(i*x)    but z = cos(x) + i*sin(x), So

cos(x) + i*sin(x) = e^(i*x)       ......(2)

Put x = pi in (2) and we get:

-1 + 0 = e^(i*pi)

and so     e^(i*pi) + 1 = 0

This is the Euler equation.

Now returning to (2) we have

[cos(x)+i*sin(x)]^n = [e^(i*x)]^n

=  e^(i*nx)

=  cos(nx) + i*sin(nx)

and this is the statement of DeMoivre's theorem.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus
High School Exponents
High School Imaginary/Complex Numbers

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