# Euler’s formula

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What could eiθ mean? Assuming it means anything, and that the result will be a complex number, we could write

${e^{i\theta }} = u(\theta ) + iv(\theta ) = u + iv$

Where u and v must be functions of θ. Now let us, of some obscure reason, take the derivative of the above with respect to θ If we treat i as any constant, we get

$i{e^{i\theta }} = u' + iv'$

If we take the derivative again, we get

${i^2}{e^{i\theta }} = u'' + iv''$

or

$- {e^{i\theta }} = u'' + iv''$

and thus

${e^{i\theta }} = - u'' - iv''$

Comparing this to the original function we get that

$u = - u'',\quad v = - v''$

So, what functions would, if we take the derivative twice, give the function back but with a negative sign?

The obvious candidates are the sine and cosine functions. How to find which is which? We know that e0=1, so if θ=0 then u must be cosine, and v must be sine, since cos(0)=1 and Asin(0)=0. We thus have

${e^{i\theta }} = \cos \theta + iA\sin \theta$

But what value would A have? If we square both sides we get

${({e^{i\theta }})^2} = {e^{i2\theta }} = \cos (2\theta ) + iA\sin (2\theta )$

and

${(\cos \theta + iA\sin \theta )^2} = {\cos ^2}\theta - A^2{\sin ^2}\theta + 2iA\cos \theta \sin \theta$

Looking at the imaginary part we must thus have that

$\cos (2\theta ) = {\cos ^2}\theta - A^2{\sin ^2}\theta$

But the double angle formula gives us that A=±1.  (If we instead look at the imaginary part we have A in both sides, so we cannot use that to determine the value of A. ) We select 1, and that will finally give us

${e^{i\theta }} = \cos \theta + i\sin \theta$

This is a very important formula, and arguably one of the most beautiful in mathematics, and it will be used extensively  in these pages.

How about is we would have chosen A= –1? That would actually work pretty fine.  What is the difference between i and –i? In a sense nothing but the sign. All the maths would essentially remain the same.

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Next page : Euler's identityLast modified: Feb 20, 2021 @ 20:36