Let $g(z) = 1/(1+e^{-z})$ be the logistic function. Show that $1-g(z)=g(-z)$

Question

I am having trouble with this problem. I am able to work it out to the point where I have either an extra $1$ or with $e^z$ and $e^{-z}$ and also the extra $1$

Let $g(z)= 1/ (1+e^{-z})$ Show that $1-g(z)=g(-z)$

from this I say $1 - g(z) $ = $g(-z)$

$1 - (1/(1+e^{-z})) = 1/(1+e^z)$ ### this becomes a positive $z$

$1 - (1+e^{-z})^{-1} = (1+e^z)^-1$ ### we know that $1/x = x^{-1}$ property

$1 - (1+e^z) = (1+e^{-z})$ ### distribute negative

$1 - 1+e^z = 1+e^{-z}$ ### $0 + e^z = 1 + e^{-z}$ and now I'm stuck

else

$1 - 1+e^z = 1+e^{-z}$ ###

$0 + e^z = 1 + 1/e^z$

and I'm still stuck

Answer 1

$1-g(z)=1-\frac 1 {1+e^{-z}}=\frac {e^{-z}} {1+e^{-z}}$. Multiply numerator and denominator by $e^{z}$ to get $1-g(z)=\frac 1 {1+e^{z}}=g(-z)$.