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

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

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inverse of-1/z 1/(z-1)

inverse

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

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

Solution steps

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

Replace

z

with

y

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

Solve for

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

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

f (x) = - x - 10

\frac{9}{5} c + 32

f (x) = \frac{4 x ^{2} - 4}{x + 1}

(- 6, 12), (2, - 20)

(3, 9), (0, 5)

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