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Can somebody please check if my proof is okay?

Prove that $||z_1|-|z_2|| \leq |z_1-z_2|$.

Attempt: This proof seems similar to the triangle inequality proof.

$$\begin{split} |z_1-z_2|^2 &= (z_1-z_2) \overline{(z_1-z_2)} \\ &= (z_1-z_2)(\bar z_1 - \bar z_2) \\ &= z_1\bar z_1 - z_1 \bar z_2 - \bar z_1 z_2 + z_2 \bar z_2\\ &=|z_1|^2 - 2Re(z_1 \bar z_2) + |z_2|^2\\ & \geq |z_1|^2 -2|z_1 \bar z_2| + |z_2|^2 \\ &= |z_1|^2 - 2|z_1||z_2| + |z_2|^2 \\ &= (|z_1| - |z_2|)^2. \end{split}$$

Taking the square root of both sides gives $|z_1|-|z_2| \leq |z_1-z_2|$

Can I just take the norm of both sides so that $||z_1|-|z_2|| \leq ||z_1-z_2|| = |z_1-z_2|$?

Thank you.

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In general you cannot simply take absolute sign to an inequality ($-2 \le1$ does not imply $|-2|\le 1$).

However, in this case, there are some ways to get around:

  • Use $(|z_1| - |z_2|)^2 = (\big||z_1| - |z_2|\big|)^2$ before taking square root.

  • Interchange the role of $z_1, z_2$ in $|z_1| - |z_2| \le |z_1-z_2|$ to give $$|z_2| - |z_1| \le |z_2-z_1| = |z_1 - z_2|.$$ Then $||z_1| - |z_2|| = |z_1| - |z_2|$ or $|z_2|-|z_1|$ are both $\le |z_1-z_2|$.

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We have $$|z_1| = |z_1-z_2+z_2| \leqslant |z_1-z_2|+|z_2|, $$ so $|z_1|-|z_2|\leqslant |z_1-z_2|$. Similarly, $$|z_2| = |z_2-z_1+z_1| \leqslant |z_1-z_2| + |z_1|, $$ so that $-(|z_1|-|z_2|)\leqslant |z_1-z_2|$. From this it follows that $$\big||z_1|-|z_2|\big|\leqslant |z_1-z_2|. $$

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