Modulus Larger than Real Part

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Theorem

Let $z \in \C$ be a complex number.


Then the modulus of $z$ is larger than the real part $\operatorname{Re} \left({z}\right)$ of $z$:

$\quad \left\vert{z}\right\vert \ge \left\vert{\operatorname{Re} \left({z}\right) }\right\vert$


Proof

By the definition of a complex number, we have:

$z = \operatorname{Re} \left({z}\right) + i \operatorname{Im} \left({z}\right)$

Then:

\(\displaystyle \left\vert{z}\right\vert\) \(=\) \(\displaystyle \sqrt {\left({\operatorname{Re} \left({z}\right) }\right)^2 + \left({\operatorname{Im} \left({z}\right) }\right)^2}\) Definition of Modulus
\(\displaystyle \) \(\ge\) \(\displaystyle \sqrt {\left({\operatorname{Re} \left({z}\right) }\right)^2 }\) Square of Real Number is Non-Negative, as $\operatorname{Im}\left({z}\right)$ is real
\(\displaystyle \) \(=\) \(\displaystyle \left\vert{\operatorname{Re} \left({z}\right) }\right\vert\) Square of Real Number is Non-Negative, as $\operatorname{Re}\left({z}\right)$ is real

$\blacksquare$


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