# Problem proving Euler's formula?

## Homework Statement

Prove ei$\theta$ = cos(θ) + isin(θ)

## The Attempt at a Solution

Let g = cos(θ) + isin(θ)
$\frac{dg}{dθ}$ = -sin(θ) + icos(θ)
=> $\frac{dg}{dθ}$ = ig
=> $\frac{dθ}{dg}$ = $\frac{1}{ig}$
=> $\frac{dθ}{dg}$ = -i$\times$$\int$$\frac{1}{g}$
=> θ = -i$\times$ln|g| + c
=> iθ - ic = ln|g|
=> |g| = e$^{iθ - ic}$
g(0)=1, => c = 0
When g=1,θ=0
=> |g| = e$^{iθ}$

Here's where the problem arises. Wouldn't that mean that g = $\pm$e$^{iθ}$
Help is appreciated. Thanks.

Last edited:

pasmith
Homework Helper

## Homework Statement

Prove ei$\theta$ = cos(θ) + isin(θ)

## The Attempt at a Solution

Let g = cos(θ) + isin(θ)
$\frac{dg}{dθ}$ = -sin(θ) + icos(θ)
=> $\frac{dg}{dθ}$ = ig
=> $\frac{dθ}{dg}$ = $\frac{1}{ig}$
=> $\frac{dθ}{dg}$ = -i$\times$$\int$$\frac{1}{g}$
=> θ = -i$\times$ln|g| + c
=> iθ - ic = ln|g|
=> |g| = e$^{iθ - ic}$
g(0)=1, => c = 0
=> |g| = e$^{iθ}$

Here's where the problem arises. Wouldn't that mean that g = $\pm$e$^{iθ}$
Help is appreciated. Thanks.
Use the series definition of $e^x$. Set $x = i\theta$ and compare the real and imaginary parts of the result to the taylor series of cosine and sine.

Use the series definition of $e^x$. Set $x = i\theta$ and compare the real and imaginary parts of the result to the taylor series of cosine and sine.

I'm afraid I have to use straight calculus to solve.

HallsofIvy
Homework Helper
Then show that
1) $e^{i\theta}$ and $cos(\theta)+ isin(\theta)$
have the same second derivative

ii) The two functions have the same value at $\theta= 0$

iii) The derivatives of the two functions have the same value at $\theta= 0$.

Then show that
1) $e^{i\theta}$ and $cos(\theta)+ isin(\theta)$
have the same second derivative

ii) The two functions have the same value at $\theta= 0$

iii) The derivatives of the two functions have the same value at $\theta= 0$.
Finding the second derivative gives the same problem with the modulus. I also can't really use the value of θ as proof (for the first or second derivative), as I subbed it in to find the value of c from the integral.

Edit: would simply stating +-$e^{i\theta}$ = $cos(\theta)+ isin(\theta)$, subbing in θ = 0 and stating therefore $e^{i\theta}$ = $cos(\theta)+ isin(\theta)$ be sufficient?

Last edited:
Ray Vickson
Homework Helper
Dearly Missed
Finding the second derivative gives the same problem with the modulus. I also can't really use the value of θ as proof (for the first or second derivative), as I subbed it in to find the value of c from the integral.
You have been mislead by the "|.|" notation; although the symbol is the same, the meaning is very different in the real and complex cases. Your equation ##\int dg/g = \ln |g|## is false when ##g## is complex. In fact ##g = \cos(\theta) + i \sin(\theta)## satisfies ##|g| = 1 ## for all ##\theta##.

I cannot understand why you refuse to use the series expansion approach. Has somebody told you that you are not allowed to do that?

HallsofIvy
Homework Helper
I should have said "(1) Show that they satisfy the same second order differential equation" rather than have the same second derivative (they do but you have to use the statement you want to prove to show that).

The first derivative of $y= e^{i\theta}$ is $y'= ie^{i\theta}$ and the second derivative is $y''= -e^{i\theta}= -y$.

The first derivative of $y= cos(\theta)+ isin(\theta)$ is $y'= -sin(\theta)+ icos(\theta)$ and the second derivative is $y''= -cos(\theta)- isin(\theta)= -y$.

Since both functions satisfy $y''= -y$, it is sufficient to show that they have the same value and first derivative at $\theta= 0$.

You have been mislead by the "|.|" notation; although the symbol is the same, the meaning is very different in the real and complex cases. Your equation ##\int dg/g = \ln |g|## is false when ##g## is complex. In fact ##g = \cos(\theta) + i \sin(\theta)## satisfies ##|g| = 1 ## for all ##\theta##.

I cannot understand why you refuse to use the series expansion approach. Has somebody told you that you are not allowed to do that?
I apologize if I am mistaken but I am doing advanced high school calculus and we are supposed to prove it using calculus and by treating i as a constant. Integrating complex functions is not part of the curriculum in any way but these are the instructions I have been given.

Ray Vickson