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Verify that $\sqrt{2}|z| \geq|\operatorname{Re} z|+|\operatorname{Im} z|$.
Hint: Reduce this inequality to $(|x|-|y|)^{2} \geq 0$.
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Solution. Note that
$$0 \leq(|\operatorname{Re} z|+|\operatorname{Im} z|)^{2}=|\operatorname{Re} z|^{2}-2|\operatorname{Re} z||\operatorname{Im} z|+|\operatorname{Im} z|^{2}$$
Thus
$$2|\operatorname{Re} z||\operatorname{Im} z| \leq|\operatorname{Re} z|^{2}+|\operatorname{Im} z|^{2}$$
and then
$$|\operatorname{Re} z|^{2}+2|\operatorname{Re} z||\operatorname{Im} z|+|\operatorname{Im} z|^{2} \leq 2\left(|\operatorname{Re} z|^{2}+|\operatorname{Im} z|^{2}\right)$$
That is
$$(|\operatorname{Re} z|+|\operatorname{Im} z|)^{2} \leq 2\left(|\operatorname{Re} z|^{2}+|\operatorname{Im} z|^{2}\right)=2|z|^{2}$$
and therefore,
$$|\operatorname{Re} z|+|\operatorname{Im} z| \leq \sqrt{2}|z|$$
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