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

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

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inverse of z^{-1}((2z)/((z-2)^2))

inverse

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

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

Solution steps

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

Replace

z

with

y

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

Solve for

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

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

y = \frac{x}{x + 6}

\frac{1}{2} - e^{1 - \frac{1}{2} x}

f (A) = 2 π r^{2} + 2 π r

f (x) = \frac{15}{2} lo g_{3} (x - 9) + 4

y = 2 lo g_{3} (x)

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