What is the inverse of (z^2)/((z-1)(z-3)) ?

The inverse of (z^2)/((z-1)(z-3)) is (2z+sqrt(z(z+3)))/(z-1),(2z-sqrt(z(z+3)))/(z-1)

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inverse of (z^2)/((z-1)(z-3))

inverse

\frac{z ^{2}}{( z - 1 ) ( z - 3 )}

\frac{2 z + z ( z + 3 )}{z - 1}, \frac{2 z - z ( z + 3 )}{z - 1}

Solution steps

y = \frac{z ^{2}}{( z - 1 ) ( z - 3 )}

Replace

z

with

y

z = \frac{y ^{2}}{( y - 1 ) ( y - 3 )}

Solve for

y, z = \frac{y ^{2}}{( y - 1 ) ( y - 3 )}

\frac{2 z + z ( z + 3 )}{z - 1}, \frac{2 z - z ( z + 3 )}{z - 1}

2^{x^{2} - 1}

11 cos (2 x) + 5

\frac{2 s + 3}{s ( 2 s + 1 )}

f (x) = \frac{1}{7} x^{2} + 3

f (x) = x^{2 - 10 + 15}

What is the inverse of (z^2)/((z-1)(z-3)) ?
The inverse of (z^2)/((z-1)(z-3)) is (2z+sqrt(z(z+3)))/(z-1),(2z-sqrt(z(z+3)))/(z-1)