What is the inverse of f(z)=(z^2-z)/(z+1) ?

The inverse of f(z)=(z^2-z)/(z+1) is (1+z+sqrt(z^2+6z+1))/2 ,(1+z-sqrt(z^2+6z+1))/2

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inverse of f(z)=(z^2-z)/(z+1)

inverse

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

\frac{1 + z + z ^{2} + 6 z + 1}{2}, \frac{1 + z - z ^{2} + 6 z + 1}{2}

Solution steps

y = \frac{z ^{2} - z}{z + 1}

Replace

z

with

y

z = \frac{y ^{2} - y}{y + 1}

Solve for

y, z = \frac{y ^{2} - y}{y + 1}

\frac{1 + z + z ^{2} + 6 z + 1}{2}, \frac{1 + z - z ^{2} + 6 z + 1}{2}

What is the inverse of f(z)=(z^2-z)/(z+1) ?
The inverse of f(z)=(z^2-z)/(z+1) is (1+z+sqrt(z^2+6z+1))/2 ,(1+z-sqrt(z^2+6z+1))/2