Question
Answer
$$(e^{ipit}-e^{-ipit})^2=(e^{i^2pt}-e^{-i^2pt})^2$$
Explanation
| $$ \begin{aligned}(e^{ipit}-e^{-ipit})^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(e^{i^2pt}-e^{-i^2pt})^2\end{aligned} $$ | |
| ① | $$ i p i t = i^{1 + 1} p t = i^2 p t $$$$ i p i t = i^{1 + 1} p t = i^2 p t $$ |
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