How would you get $e^{\cos\theta+i\sin\theta}=e^{\cos\theta}(\cos(\sin\theta)+i(\sin(\sin\theta)))$
Euler's formula (alternate form) $e^{e^{i\theta}}$
2
$\begingroup$
complex-analysis
-
0You're reading a pair of parentheses in WolframAlpha's answer that aren't there. WA yields $e^{\cos\theta}\cos(\sin\theta) + i\sin(\sin\theta)e^{\cos\theta} = e^{\cos\theta}\cos(\sin\theta) + e^{\cos\theta}i\sin(\sin\theta) = e^{\cos\theta}(\cos(\sin\theta) + i\sin(\sin\theta))$, which is what your notes gave you. – 2011-12-02
1 Answers
6
Just write $e^{\cos \theta + i \sin \theta} = e^{\cos \theta} e^{i \sin \theta}$ and use the formula for $e^{ix}$ with $x = \sin \theta$.
-
0That's a very nice example:) – 2011-12-03