Question

(complexe numbers) Show that? Thank you!

Is z ∈ C \ {0},
than is `|z|/z` a square root of `z^-/z`

Answer 1

yes

Explanation:

multiply and divide `RHS` by `z`
so we get
`zz^-/(z)^2` = `|z|^2/(z)^2` = `(|z|/z)^2`
hence
`|z|/z` will be a square root of the `z^-/z` which is `(LHS)^2`
hope u find it helpful :)