# |COMPLEXE NUMBERS| Show that?... thank you!

## Show that: Is z ∈ ℤ, than is ${z}^{-} 1 = {z}^{-} / | z {|}^{2}$ The z in the numerator is supposed to be the complex conjugate of z. Thank you!

Jan 14, 2018

Yes

#### Explanation:

take $z = x + i y$
then ${z}^{-} 1 = \frac{1}{x + i y}$
$| z {|}^{2} = {x}^{2} + {y}^{2} = \left(x + i y\right) \left(x - i y\right) = z {z}^{_}$

hence we get
RHS

${z}^{_} / \left(z {z}^{_}\right) = \frac{1}{z} = {z}^{-} 1$ = LHS
hence proved
Hope u find it helpful :)