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, for 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''


 - {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 )


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

Looking at the real part we must thus have that

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

The double angle formula gives us that A=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.

(If we instead look at the imaginary part we have A on both sides, so we cannot use that to determine the value of A. )

How about if we would have chosen A= –1? That would 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: Dec 28, 2023 @ 14:03