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

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

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inverse of+z z/(z^2+1)

inverse

+ z \frac{z}{z ^{2} + 1}

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

Solution steps

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

Replace

z

with

y

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

Solve for

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

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

\frac{( x + 3 ) ^{2}}{x ^{2} - 2 x}

f (x) = \frac{7 x - 4}{x + 3}

f (x) = \frac{( 6 x - 16 )}{- 5 x}

g (x) = \frac{x + 1}{9}

g (x) = \frac{7}{x} + 5

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