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# How do you prove Euler's formula?

Updated: 12/15/2022

Wiki User

11y ago

Euler's formula states that eix = cos(x) + isin(x) where i is the imaginary number and x is any real number.

First, we get the power series of eix using the formula:

ez = Î£âˆžn=o zn/n! where z = ix. That gives us:

1 + ix + (ix)2/2! + (ix)3/3! + (ix)4/4! + (ix)5/5! + (ix)6/6! + (ix)7/7! + (ix)8/8! + ...

which from the properties of i equals:

1 + ix - x2/2! - ix3/3! + x4/4! + ix5/5! - x6/6! - ix7/7! + x8/8! + ...

which equals:

(1 - x2/2! + x4/4! - x6/6! + x8/8! - ...) + i(x - x3/3! + x5/5! - x7/7! + ...).

These two expressions are equivalent to the Taylor series of cos(x) and sin(x). So, plugging those functions into the expression gives cos(x) + isin(x).

Q.E.D.

Of course, were you to make x = Ð¿ in Euler's formula, you'd get Euler's identity:

eiÏ€ = -1

It depends on your definitionThe question presumes that someone has already given a definition of the exponential function for complex numbers. But has this definition been given? If so, what is this official definition?

The answer above assumes that the exponential function is defined, for all complex numbers, by its power series. (Or, at least, that someone else has already proved that the power series definition is equivalent to whatever we're taking to be the official definition).

So the answer depends critically on what the definition of the exponential function for complex numbers is.

Suppose you know everything about the real numbers, and you're trying to build up a theory of complex numbers. Most people probably view it this way: Mathematical objects such as complex numbers are out there somewhere, and we have to find them and work out their properties. But mathematicians look at it slightly differently. If we can construct a thing which has all the properties we feel the field of complex numbers should have, then for all practical purposes the field of complex numbers exists. If we can define a function on that field which has all the properties we think exponentiation on the complex numbers should have, then that's as good as proving that complex numbers have exponentials. Mathematicians are comfortable with weird things like complex numbers, not because they have proved that they exist as such, but because they have proved that their existence is consistent with everything else. (Unless everything else is inconsistent, which would be a real pain but is very unlikely.)

So how do we construct this exponential function? One approach would be simply to define it by exp(x+iy) = ex(cos(y)+i.sin(y)). Then the answer to this question would be trivial: exp(iy) = exp(0+iy) = e0(cos(y)+i.sin(y)) = cos(y)+.sin(y). But you'd still have some work to do to prove things like exp(z+w) = exp(z).exp(w). Alternatively, you could define the exponential function by its power series. (There's a theorem that lets you calculate the radius of convergence, and that tells us the radius is infinite, i.e. the power series works everywhere.) Or maybe you could try something like proving that the equation dw/dz = w has a unique solution up to a multiplicative constant, and defining the exponential function to be the solution which satisfies w=1 at z=0.

Let z = cos(x) + i*sin(x)

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

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

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

= i*(z)

therefore dz/dx = iz

(1/z)dz = i dx

INT((1/z)dz = INT(i dx)

ln(z) = ix+c

z = e(ix+c)

Substituting x=0, we get cos(0) + i*sin(0) = e(0i+c)

1+0 = e(0+c)

e(0+c) = 1

therefore c=0

z = eix

eix = cos(x) + i*sin(x)

Substituting x=pi, we get e(i*pi) = -1 + 0

e(i*pi) = -1

e(i*pi) + 1 = 0

qED

Wiki User

11y ago