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Reading through my digital signal processing text, I came across this statement: $$X(e^{j\omega})=\sum_{n=0}^4e^{-j\omega n}=e^{-j2\omega}\frac{\sin(5\omega/2)}{\sin(\omega/2)}$$ I was able to reduce it to the following using Euler's formula: $$e^{-2j\omega}(1+2\cos\omega+2\cos2\omega)$$ I'm thinking something along the lines of a double-angle identity, but I'm hoping there's an easier way that I'm not realizing.

Parcly Taxel
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user50420
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  • what does it mean $e^{j\omega}$? – André Porto Oct 29 '18 at 01:29
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    $e^{j\omega} = \cos \omega + j \sin \omega$; $j$ is used for the imaginary unit $\sqrt {-1}$ when used for electrical calculations ($i$ stands for electrical current). – bjcolby15 Oct 29 '18 at 01:34
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    If you break the complex exponential into sine and cosine, then you can use the identities shown in this answer on the separate real and imaginary components, then recombine the complex exponential. Note, however, that one of the ways to prove those identities (as shown in this answer) is by using the geometric series formula with the complex exponential as a complex exponential. – Blue Oct 29 '18 at 01:45

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Summing the geometric series gives $$\sum_{n=0}^4e^{-nj\omega}=\frac{1-e^{-5j\omega}}{1-e^{-j\omega}}$$ Now we have $$1-e^{jx}=2\sin(x/2)e^{j(x-\pi)/2}{}^*$$ and thus $$\frac{1-e^{-5j\omega}}{1-e^{-j\omega}}=\frac{2\sin(-5\omega/2)e^{j(-5\omega-\pi)/2}}{2\sin(-\omega/2)e^{j(-\omega-\pi)/2}}=e^{-2j\omega}\frac{\sin5\omega/2}{\sin\omega/2}$$

*A geometric derivation of this identity is shown above. The magnitude of $1-e^{-jx}$ can be calculates using the cosine rule as $\sqrt{1^2+1^2-2(1)(1)\cos x}$, which simplifies to $\sqrt{2-2\cos x}=2\sin\frac x2$; the argument is an easy application of angles in a triangle and is $\frac{x-\pi}2$.

Parcly Taxel
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