Consider the signal $ x[n]=a^{n}u[n] $
$ X(z)=\sum_{n=-\infty}^{\infty}a^{n}u[n]z^{-n}=\sum_{n=0}^{\infty}(az^{-1})^{n} $
for convergence,
$ \left| az^{-1}\right|<1 $, or equivalently $ \left| z\right|>\left| a\right| $
$ X(z)=\sum_{n=0}^{\infty}(az^{-1})^{n}=\frac{1}{1-az^{-1}}=\frac{z}{z-a} $, $ \left| z\right|>\left| a\right| $
Now let $ x[n]=-a^{n}u[-n-1] $
$ X(z)=-\sum_{n=-\infty}^{\infty}a^{n}u[-n-1]z^{-n}=-\sum_{n=-\infty}^{-1}a^{n}z^{-n} $
$ =-\sum_{n=1}^{\infty}a^{-n}z^{n}=1-\sum_{n=0}^{\infty}(a^{-1}z)^{n} $
$ \left| a^{-1}z\right|<1 $, or equivalently $ \left| z\right|<\left| a\right| $
