Question
Answer
$$-3i^{76}-3i^{97}+7i^{48}-i^{25}=-4i+4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-3i^{76}-3i^{97}+7i^{48}-i^{25}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-3-3i+7-i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-4i+4\end{aligned} $$ | |
| ① | $$ -3i^{76} = -3 \cdot i^{4 \cdot 19 + 0} =
-3 \cdot \left( i^4 \right)^{ 19 } \cdot i^0 =
-3 \cdot 1^{ 19 } \cdot 1 =
-3 \cdot 1 $$ |
| ② | $$ -3i^{97} = -3 \cdot i^{4 \cdot 24 + 1} =
-3 \cdot \left( i^4 \right)^{ 24 } \cdot i^1 =
-3 \cdot 1^{ 24 } \cdot i =
-3 \cdot i $$ |
| ③ | $$ 7i^{48} = 7 \cdot i^{4 \cdot 12 + 0} =
7 \cdot \left( i^4 \right)^{ 12 } \cdot i^0 =
7 \cdot 1^{ 12 } \cdot 1 =
7 \cdot 1 $$ |
| ④ | $$ -i^{25} = - i^{4 \cdot 6 + 1} =
- \left( i^4 \right)^{ 6 } \cdot i^1 =
- 1^{ 6 } \cdot i =
- i $$ |
| ⑤ | Combine like terms: $$ \color{blue}{-3i} \color{blue}{-i} + \color{red}{7} \color{red}{-3} = \color{blue}{-4i} + \color{red}{4} $$ |
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