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Reading a long solution I saw this a step that converts $1 + e^{2i}$ to $e^i \cos(1)$. How is this done?

How do I generalize this?

iveqy
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  • Can I know why was the upvote given?? – Jasser Oct 05 '14 at 13:36
  • @user291957 nothing since I don't know how to even begin trying things on this type of problem. How would you suggest a trial and error method would be done in this case? – iveqy Oct 05 '14 at 13:41

3 Answers3

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There's a factor of $2$ missing.

Generally,

$$1 + e^{2i\varphi} = e^{i\varphi}(e^{-i\varphi} + e^{i\varphi}) = e^{i\varphi}\cdot 2\cos \varphi$$

since

$$\cos \varphi = \frac{e^{i\varphi} + e^{-i\varphi}}{2}.$$

Daniel Fischer
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  • So I was trying this with $1-e^{-2i}$ to see if I got it right. I get: $-(1+e^{-2i}) = -e^{-i} (e^{-i} + e^i) = -e^{-i} \cos (1)$. However this is false according to wolfram alpha. Where am I wrong? https://www.wolframalpha.com/input/?i=1-e%5E%28-2i%29+%3D+-e%5E%28-i%29cos%281%29 – iveqy Oct 05 '14 at 14:01
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    $1 - e^{-2i} = (e^i - e^{-i})e^{-i} = 2ie^{-i}\sin 1$. Your first step was incorrect, $-(1+e^{-2i}) = -1 - e^{-2i}$. From then on, it's right, apart from the missing factor $2$. – Daniel Fischer Oct 05 '14 at 14:05
  • Thanks, but now I'm confused. If you look at method 2 in the accepted answer here http://math.stackexchange.com/questions/956968/calc-the-sum-of-sum-k-0-infty-frac-1kk-sin2k/956986?noredirect=1#comment1968038_956986 the result is $e^{-i} \cos (1)$. – iveqy Oct 05 '14 at 14:08
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    On the one hand, the factors $2$ cancel since one gets one in the numerator and one in the denominator. On the other, the answer contains a mistake. It's $\ln (1+e^{2i}) - \ln (1+e^{-2i})$, not $\ln (1+e^{2i}) - \ln (1-e^{-2i})$. – Daniel Fischer Oct 05 '14 at 14:12
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Using Euler's formula we get:

$$1 + e^{2i} = 1 + e^{i} \cdot e^{i} = 1 + e^i \cdot i \sin(1) + e^i \cdot \cos(1)$$

With this approach you can see, that the difference between the two expression given above is $1 + e^i \cdot i \sin(1)$.

Community
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user153012
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\begin{align}1+e^{2i}&=e^i(e^{-i}+e^i)\\&=e^i((\cos1-i\sin1)+(\cos1+i\sin1))&\\&=2e^i\cos1\end{align}

Akiva Weinberger
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