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\begin{align}\lvert \cos\theta(\cos\theta -i\sin\theta)\rvert&\leq \lvert \cos\theta -i\sin\theta\rvert\\&=\sqrt{\cos^2\theta+\sin^2\theta}=1 \end{align} I know $0\leq\lvert \cos\theta \rvert\leq 1$, but why can we just remove $\lvert \cos \theta \rvert$ on the right hand side of the inequality?

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    $\begingroup$ because cos is bounded by 1. $\endgroup$ – user392395 Apr 29 '17 at 18:33
  • $\begingroup$ $|z_1z_2|=|z_1| |z_2|$ $\endgroup$ – Nosrati Apr 29 '17 at 18:38
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Note that since $\vert\cos(x)\vert\le1$ and $\vert x_1x_2\vert=\vert x_1\vert\cdot\vert x_2\vert$ (you should verify this yourself), then $$\left\vert\cos(x)(\cos(x)-i\sin(x))\right\vert=\left\vert\cos(x)\right\vert\cdot\left\vert(\cos(x)-i\sin(x))\right\vert\le1\cdot\left\vert(\cos(x)-i\sin(x))\right\vert=\sqrt{\cos^2(x)+\sin^2(x)}=1$$

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The modulus of a complex number $a + ib$ $(a,b \in \mathbb R)$ is $\sqrt{a^2 + b^2}$, and a complex number multiplied by a real number $c$ is $ac + i(bc)$.

$cos\theta$ and $sin\theta$ being real numbers,

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

Multiplying the complex number with $cos\theta$ and taking its modulus, $$|cos\theta(cos\theta - i sin\theta)| = |(cos\theta)^2 - i (cos\theta sin\theta)|$$ $$= \sqrt{(cos\theta)^4 + (cos\theta sin\theta)^2} = |cos\theta| \sqrt{(cos\theta)^2 + (sin\theta)^2}$$ $$= |cos\theta| \le 1$$ If your doubt is in the above step, for a number $a \in \mathbb R$ $$\sqrt{a^2} = |a|$$

So, $$|cos\theta(cos\theta - i sin\theta)| \le |cos\theta - i sin\theta|$$

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It is generally true that $|ab|=|a||b|$, so that in this case: $|cos(\theta)(cos(\theta)-isin(\theta))|=|cos(\theta)|(cos(\theta)-isin(\theta))|\leq|(cos(\theta)-isin(\theta))| $.

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    $\begingroup$ |ab|≤|a||b| You must mean $|ab|\color{red}{=}|a|\,|b|\,$. $\endgroup$ – dxiv Apr 29 '17 at 18:47

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