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I'm stuck on a proof. There's a step that says:

$$ \left| \Im\left(\frac{1-e^{i(N+1)x}}{1-e^{ix}}\right)\right| \leq \left| (\frac{1-e^{i(N+1)x}}{1-e^{ix}}) \right|,\quad \text{with } N \in \mathbb{N} $$

I've tried to see this "intuitively" but I'm not able to see this as clear as my professor sees it. I'd appreciate any help with this.

Thanks.

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    $\begingroup$ The absolute value of a complex number is always greater than the absolute value of its parts. $|a+bi|=\sqrt{a^2+b^2}\geq |a|,|b|$. $\endgroup$ – Michael Burr Oct 3 '15 at 23:36
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Hint. The inequality basically states

$$\left|\Im(z)\right| \leq \left|z\right| = \sqrt{\Re(z)^2+\Im(z)^2} \qquad z \in \Bbb C$$

For an intuitive understanding just think of the complex plane.

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For $a,b\in\mathbb R$, we have $|a+bi|^2 = a^2 + b^2 \ge b^2$, so $|a+bi| \ge |b|$.

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Basically for all $a,b \in \mathbb R$ we have $$|a+bi|=\sqrt{a^2+b^2} \geq \sqrt{b^2} = |b|$$

I feel this is written a little clearer than the other answers.

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