2

I am given:

$$\left(2-e^{i \theta}\right)\left(2-e^{-i \theta}\right)$$

Which if you expand you get:

$$4 - 2(\cos \theta - i\sin \theta) -2 (\cos \theta + i\sin \theta ) + e^0 = 4 - 4\cos \theta + 1 = 5 - 4\cos \theta.$$

Is there a shortcut I could have taken by working directly in exponential form? I.e. not using Euler's formula?

qianqian07
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Makogan
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  • +1 to you posting : I see no better approach. There is the principle that $~e^{i\theta} + e^{-i\theta} = 2\cos(\theta),~$ but that isn't really much of a shortcut. Personally, your work is exactly what I would hope to see, from a problem solver reasonably new to the theory. – user2661923 Mar 18 '23 at 06:28

1 Answers1

4

Here is another way to approach it for the sake of curiosity:

\begin{align*} (2 - e^{i\theta})(2 - e^{-i\theta}) & = (2 - e^{i\theta})\overline{(2 - e^{i\theta})}\\\\ & = |2 - e^{i\theta}|^{2}\\\\ & = (2 - \cos(\theta))^{2} + \sin^{2}(\theta)\\\\ & = 5 - 4\cos(\theta) \end{align*}

Átila Correia
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