I need to write the following expression $$H(\omega)=e^{-i \omega} + e^{-3i \omega}$$ in the form $$r(\omega)e^{i\phi(\omega)}$$ where $$r(\omega), \phi(\omega)\space$$ are real fuctions of omega. So, determine $$|H(\omega)|\space and \space arg(H(\omega)).$$ How should I look at this problem? I can't seem to work out how to get one length and argument from a sum of complex exponentials. (that is, if I computed H(w) correctly.)
2026-03-29 07:30:32.1774769432
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Express sum of complex exponentials in terms of length and argument
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Jan Eerland's answer works in the most general case, but here's a neat "trick" I like to use
$$ e^{-i\omega} + e^{-3i\omega} = e^{-2i\omega} (e^{-i\omega}+e^{i\omega}) = 2\cos(\omega)e^{-2i\omega} $$
Thus $r = 2\cos (\omega)$ and $\phi = -2\omega$
This works for any expression of the form $$ e^{ai\omega} + e^{bi\omega} = 2\cos\left({a-b\over2}\omega\right)e^{{a+b\over2}i\omega} $$
$$ e^{ai\omega} - e^{bi\omega} = 2i\sin\left({a-b\over2}\omega\right)e^{{a+b\over2}i\omega} $$
Well, assuming $\text{z}_\text{n}\in\mathbb{C}$ for all $\text{n}$:
$$\text{z}_1+\text{z}_2=\left|\text{z}_1\right|\cdot\exp\left(\left(\arg\left(\text{z}_1\right)+2\pi\cdot\text{k}_1\right)\cdot i\right)+\left|\text{z}_2\right|\cdot\exp\left(\left(\arg\left(\text{z}_2\right)+2\pi\cdot\text{k}_2\right)\cdot i\right)\tag1$$
Where $\text{k}_\text{n}\in\mathbb{Z}$, $\left|\text{z}_\text{n}\right|=\sqrt{\Re^2\left(\arg\left(\text{z}_\text{n}\right)\right)+\Im^2\left(\arg\left(\text{z}_\text{n}\right)\right)}$ and $0\le\arg\left(\text{z}_\text{n}\right)<2\pi$.
Using Euler's formula:
$$e^{\varphi i}=\cos\left(\varphi\right)+\sin\left(\varphi\right)\cdot i\tag2$$
So: