For any two complex number `z_1a n d\ z_2` prove that: `|z_1-z_2|geq|z_1|-|z_2|`

Explanation

Let `z_1=r_1(\cos A+i \sin A)` and `z_2=r_2(\cos B+i \sin B)`


Further, `\|z_1|=r_1` and `\|z_2\|=r_2`

`\|z_1-z_2\|`
`=\sqrt (r_1^2+r_2^2-2 r_1 r_2 \cos (A-B))`
`\geq \sqrt (r_1^2+r_2^2-2 r_1 r_2)`

as
`\cos (A-B) \leq 1`
`=r_1-r_2`
So, `\|z_1-z_2\| \geq\|z_1\|-\|z_2\|`, the equality occurs when `\cos (A-B)=1` i.e., when `A=B`