What is the inverse of z 1/(1-z^{-1)} ?

The inverse of z 1/(1-z^{-1)} is (z+sqrt(z^2-4z))/2 ,(z-sqrt(z^2-4z))/2

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inverse of z 1/(1-z^{-1)}

inverse

z \frac{1}{1 - z ^{- 1}}

\frac{z + z ^{2} - 4 z}{2}, \frac{z - z ^{2} - 4 z}{2}

Solution steps

y = z \frac{1}{1 - z ^{- 1}}

Replace

z

with

y

z = y \frac{1}{1 - y ^{- 1}}

Solve for

y, z = y \frac{1}{1 - y ^{- 1}}

\frac{z + z ^{2} - 4 z}{2}, \frac{z - z ^{2} - 4 z}{2}

f (x) = 11 - \frac{x}{4} h (x) = - 4 (x - 11)

f (x) = 59 ln (x)

f (x) = 8 x

y = (x + 5)^{2}

\frac{x + 9}{x ^{2} - 81}

What is the inverse of z 1/(1-z^{-1)} ?
The inverse of z 1/(1-z^{-1)} is (z+sqrt(z^2-4z))/2 ,(z-sqrt(z^2-4z))/2