How can one prove $||z_1|-|z_2||\le|z_1+z_2|$?

Question

How can it be shown that for the complex numbers $z_1$ and $z_2$: $$||z_1|-|z_2||\le|z_1+z_2|$$ My text provides a hint that $z_1=z_1+z_2+(-z_2)$, and $z_2=z_1+(-z_1)+z_2$. $${}$$

Answer 1

Recall the triangle inequality: $$|z_1+z_2|\le |z_1|+|z_2|$$ so using it we have $$|z_1|=|(z_1+z_2)-z_2|\le|z_1+z_2|+|z_2|$$ hence we find $$|z_1|-|z_2|\le |z_1+z_2|\tag1$$ and interchange the role of $z_1$ and $z_2$ we find $$|z_2|-|z_1|\le |z_1+z_2|\tag2$$ and finally $(1)$ and $(2)$ gives the desired result.

Answer 2

$$|z_1|=|z_1-z_2 +z_2|\leq |z_1-z_2| +|z_2|$$ so $$|z_1|-|z_2|\leq |z_1-z_2| $$ now by symmetry $$|z_2|-|z_1|\leq |z_1-z_2| $$

This gives $$||z_2|-|z_1||\leq |z_1-z_2| $$ as this holds for all $z_1$ and $z_2$ we can change the sign of $z_2$ to get also
$$||z_2|-|z_1||\leq |z_1+z_2| $$

Answer 3

In general $\left|a+b\right|\leq\left|a\right|+\left|b\right|$.

Applying that on $z_{1}=z_{1}+z_{2}+\left(-z_{2}\right)$ gives:

$$\left|z_{1}\right|=\left|z_{1}+z_{2}+\left(-z_{2}\right)\right|\leq\left|z_{1}+z_{2}\right|+\left|-z_{2}\right|=\left|z_{1}+z_{2}\right|+\left|z_{2}\right|$$

so that $$\left|z_{1}\right|-\left|z_{2}\right|\leq\left|z_{1}+z_{2}\right|$$

Switching $z_{1}$ and $z_{2}$ gives: $$\left|z_{2}\right|-\left|z_{1}\right|\leq\left|z_{1}+z_{2}\right|$$

This together allows the conclusion that $$\left|\left|z_{1}\right|-\left|z_{2}\right|\right|\leq\left|z_{1}+z_{2}\right|$$