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How can $|{\cos(n x)+i \sin(n x)}|$ = 1 (for $n$ and $x$ positive)? I'm not very good with complex numbers yet, so I would appreciate if someone could please explain how this absolute value comes to be.

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    $z=x+iy$ so $|z|=\sqrt{x^2+y^2},\or\ |z|^2=(x+iy)(x-iy)$ now for your case $|cos(nx) +i sin(nx)|=\sqrt{cos^2(nx)+sin^2(nx)}$ – Khosrotash Jul 29 '15 at 05:57

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First we need Euler's Formula:

$$e^{i\theta} = cos(\theta) + i\sin(\theta)$$

Using this formula:

$$|\cos(nx) + i\sin(nx)| = |e^{in\theta}| = 1$$

Though the above may not make it obvious for a beginner, so lets define the modulus properly:

$$z = x + iy \implies |z|^2 = x^2 + y^2$$

Using this definition:

$$|\cos(nx) + i\sin(nx)| = \sqrt{\cos^2(nx) + \sin^2(nx)} = 1$$

where the last evaluation comes from the fact that $\cos^2(x) + \sin^2(x) = 1$.