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Why can't we write complex numbers as $2^{i \theta} $ or $-40^{i \theta} $? Why does it have to be $e$?

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Not all complex numbers have that form, only those on the unit circle... –  The Chaz 2.0 Oct 22 '13 at 14:27
    
We can. But we are used to parametrize the unit circle with $\theta \in [0, 2\pi]$, not $\theta \in [0, 2\pi/\log 2]$. –  martini Oct 22 '13 at 14:29
    

4 Answers 4

up vote 3 down vote accepted

We know that complex numbers can be expressed as $$\cos { \theta } +i\sin { \theta } ,$$ and we want to express this form as $$a^{i\theta}=\cos { \theta } +i\sin { \theta } .$$ Take the second derivatives fo the both equation: $$\frac { { d }^{ 2 } }{ d{ \theta }^{ 2 } } \left( { a }^{ i\theta } \right) =\frac { { d }^{ 2 } }{ d{ \theta }^{ 2 } } (\cos { \theta } +i\sin { \theta } )\\ -{ a }^{ i\theta }{ \left( \ln { a } \right) }^{ 2 }=-(\cos { \theta } +i\sin { \theta } )=-{ a }^{ i\theta }\\ a=e$$

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$e$ is a special number. See here.

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we don't write it like that because $e$ is special, $e$ is special because of that :D –  Vicfred Oct 23 '13 at 4:07

Because of Euler's formula $e^{ix} = \cos{x}+i\sin{x}$.

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We can, actually : $2^i$ means $\cos\ln2+i\sin\ln2$, for instance... :-) The only difference is that, in the case of e, its natural logarithm ‘disappears’, thus becoming ‘invisible’ in the expression’s final form.

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