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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1$z=x+iy$ 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 $\cos^2(x) + \sin^2(x) = 1$.
Nathan Marianovsky
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