Answer
$$(3-2zi)^2=4i^2z^2-12iz+9$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3-2zi)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}9-12iz+4i^2z^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}4i^2z^2-12iz+9\end{aligned} $$ | |
| ① | Find $ \left(3-2iz\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 3 } $ and $ B = \color{red}{ 2iz }$. $$ \begin{aligned}\left(3-2iz\right)^2 = \color{blue}{3^2} -2 \cdot 3 \cdot 2iz + \color{red}{\left( 2iz \right)^2} = 9-12iz+4i^2z^2\end{aligned} $$ |
| ② | Combine like terms: $$ 4i^2z^2-12iz+9 = 4i^2z^2-12iz+9 $$ |
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