Why can't we write complex numbers as $2^{i \theta} $ or $-40^{i \theta} $? Why does it have to be $e$?
Asked
Active
Viewed 2,710 times
2
-
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
-
http://en.wikipedia.org/wiki/E_%28mathematical_constant%29#e_in_calculus and http://math.stackexchange.com/questions/797/whats-so-natural-about-the-base-of-natural-logarithms – lab bhattacharjee Oct 22 '13 at 15:33
4 Answers
4
$e$ is a special number. See here.
Ahaan S. Rungta
- 7,706
-
1we don't write it like that because $e$ is special, $e$ is special because of that :D – Vicfred Oct 23 '13 at 04:07
3
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: $$\frac { { d }^{ 2 } }{ d{ \theta }^{ 2 } } \left( { a }^{ i\theta } \right) =\frac { { d }^{ 2 } }{ d{ \theta }^{ 2 } } (\cos { \theta } +i\sin { \theta } )\\ -{ a }^{ i\theta }{ \left( \ln { a } \right) }^{ 2 }=-(\cos { \theta } +i\sin { \theta } )=-{ a }^{ i\theta }\\ a=e$$
newzad
- 4,855
3
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.
Lucian
- 48,334
- 2
- 83
- 154