prove that e^(i*(pi)) + 1 = 0?

thats easy,

BY Eulers formula, you kow that e^(i*pi)= cos(pi)+ i*sin(pi)

and we know that cos(pi) = -1

and sin(pi) = 0

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

= -1

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

Hence Proved

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

Euler Formula: e^(pi*i) = -1

http://mathforum.org/library/drmath/view/51921.htm...

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! - ...

Cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + x^8/8! - ...

e^x = 1 + x/1! + x^2/2! + x^3/3! + x^4/4! + x^5/5! + ...

(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

etc.

= 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

equation:

= 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)!

= 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

cos(pi) = -1 and sin(pi) = 0

(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:

e^(i*pi) = cos(pi) + i*sin(pi)

= -1 + 0i

= -1

Euler's identity (your equation) is derived from Euler's formula (e^ix = cos x + i*sin x)

so e^(i*pi) = cos(pi) + i sin (pi)

=> e^(i*pi) = -1 (because cos(pi) = -1 and sin(pi) = 0)

To prove Euler's formula see the proof on Wikipedia

z = cos(x) + i*sin(x) ......(1)

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)

First:

e^(ix) = cos(x) + i*sin(x)

e^(i*pi) = cos(pi) + i*sin(pi)

cos(pi) = -1 and sin(pi) = 0.

e^(i*pi) = -1

so:

-1 + 1 = 0

e^(i*pi)=-1

ln e^(i*pi)= ln -1

i*pi= ln -1

i*pi/ ln -1=0

0=0

ln -1 is undefined

it just does