Question
Answer
$$\dfrac{1}{(z^2+4)^2}=\dfrac{1}{z^4+8z^2+16}$$
Explanation
| $$ \begin{aligned}\frac{1}{(z^2+4)^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{z^4+8z^2+16}\end{aligned} $$ | |
| ① | Find $ \left(z^2+4\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ z^2 } $ and $ B = \color{red}{ 4 }$. $$ \begin{aligned}\left(z^2+4\right)^2 = \color{blue}{\left( z^2 \right)^2} +2 \cdot z^2 \cdot 4 + \color{red}{4^2} = z^4+8z^2+16\end{aligned} $$ |
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