Question

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

Answer

$$f=1/(z*(z^g*a^2*m^2+1)*(z^g*a^2*m^2+2))$$

Solution


Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[fz=\frac{1}{({z}^{g}{a}^{2}{m}^{2}+1)({z}^{g}{a}^{2}{m}^{2}+2)}\]
Divide both sides by \(z\).
\[f=\frac{\frac{1}{({z}^{g}{a}^{2}{m}^{2}+1)({z}^{g}{a}^{2}{m}^{2}+2)}}{z}\]
Simplify  \(\frac{\frac{1}{({z}^{g}{a}^{2}{m}^{2}+1)({z}^{g}{a}^{2}{m}^{2}+2)}}{z}\)  to  \(\frac{1}{z({z}^{g}{a}^{2}{m}^{2}+1)({z}^{g}{a}^{2}{m}^{2}+2)}\).
\[f=\frac{1}{z({z}^{g}{a}^{2}{m}^{2}+1)({z}^{g}{a}^{2}{m}^{2}+2)}\]