Question: Give the order of each of the zeros of the given function: $(z^{2}+z-2)^{3}$
Answer:
$
\begin{align*}
&f(z) = (z^{2}+z-2)^{3} = (z-1)^{3}(z+2)^{3}\\
&\text{Let } f(z) = (z^{2}+z-2)^{3}=0 \text{, we have two zeros: }\\
&z_1=1\text{, }z_2=-2\\
&f'(z)=3(2z+1)(z^{2}+z-2)^{2}\\
&f''(z)=6(z^{2}+z-2)(5z^{2}+5z-1)\\
&f'''(z)=120z^{3}+180z^{2}-72z-66\\
\\
&\text{Case 1: }z_1=1\text{, }\\
&f(z_1)=0\\
&f'(z_1)=0\\
&f''(z_1)=0\\
&f'''(z_1)=162\neq0\\
&\text{order of zero at 1 is 3.}\\
\\
&\text{Case 2: }z_2=-2\text{, }\\
&f(z_2)=0\\
&f'(z_2)=0\\
&f''(z_2)=0\\
&f'''(z_2)=-162\neq0\\
&\text{order of zero at -2 is 3.}\\
\end{align*}
$
