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

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

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inverse of z/(z^2+1)

inverse

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

\frac{1 + 1 - 4 z ^{2}}{2 z}, \frac{1 - 1 - 4 z ^{2}}{2 z}

Solution steps

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

Replace

z

with

y

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

Solve for

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

\frac{1 + 1 - 4 z ^{2}}{2 z}, \frac{1 - 1 - 4 z ^{2}}{2 z}

\frac{3}{5}

\frac{( 3 - x ) ( x + 7 )}{2 x ( - 3 x - 36 )}

[24]

\frac{2 x + 3}{x - 2}

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

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