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) $e^{e^{i\theta}}$
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complex-analysis
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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
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Just write $e^{\cos \theta + i \sin \theta} = e^{\cos \theta} e^{i \sin \theta}$ and use the formula for $e^{ix}$ with $x = \sin \theta$.
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0Thanks for your help! May I ask, is there a reason why sometimes it's written in that form? – 2011-12-02
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6The real question is why anybody would want to consider $e^{e^{i\theta}}$ (other than just to give you practice in using Euler's formula). The answer might be that they want to see what the conformal map $z \to e^z$ does to the unit circle. To plot the image as a parametric curve, you might want to take the real and imaginary parts, and those are $e^{\cos \theta} \cos(\sin \theta)$ and $e^{\cos \theta} \sin(\sin \theta)$. – 2011-12-02
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0It could be an interesting exercise to see if it can help solve certain integrals involving iterated trig functions. – 2011-12-02
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2You mean like $J_0(x) = \frac{1}{2\pi} \int_0^{2 \pi} \cos(x \cos(t))\ dt$ (where $J_0$ is the Bessel function of the first kind and order $0$) and its relatives? – 2011-12-02
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1@Aleks, like [this](http://math.stackexchange.com/a/61672)? – 2011-12-03
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0That's a very nice example:) – 2011-12-03