Question
Answer
$$siniln\dfrac{1-i}{1+i}=-i^3ln^2s$$
Explanation
| $$ \begin{aligned}siniln\frac{1-i}{1+i}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}siniln(-i) \xlongequal{ } \\[1 em] & \xlongequal{ }sin-i^2ln \xlongequal{ } \\[1 em] & \xlongequal{ }-i^3ln^2s\end{aligned} $$ | |
| ① | Divide $ \, 1-i \, $ by $ \, 1+i \, $ to get $\,\, -i $. ( view steps ) |
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