Question
Answer
$$8i^{119}+3i^{102}-10i^{81}-3i^{29}=-21i-3$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}8i^{119}+3i^{102}-10i^{81}-3i^{29}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-8i-3-10i-3i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-21i-3\end{aligned} $$ | |
| ① | $$ 8i^{119} = 8 \cdot i^{4 \cdot 29 + 3} =
8 \cdot \left( i^4 \right)^{ 29 } \cdot i^3 =
8 \cdot 1^{ 29 } \cdot (-i) =
8 \cdot -i = -8i $$ |
| ② | $$ 3i^{102} = 3 \cdot i^{4 \cdot 25 + 2} =
3 \cdot \left( i^4 \right)^{ 25 } \cdot i^2 =
3 \cdot 1^{ 25 } \cdot (-1) =
3 \cdot -1 = -3 $$ |
| ③ | $$ -10i^{81} = -10 \cdot i^{4 \cdot 20 + 1} =
-10 \cdot \left( i^4 \right)^{ 20 } \cdot i^1 =
-10 \cdot 1^{ 20 } \cdot i =
-10 \cdot i $$ |
| ④ | $$ -3i^{29} = -3 \cdot i^{4 \cdot 7 + 1} =
-3 \cdot \left( i^4 \right)^{ 7 } \cdot i^1 =
-3 \cdot 1^{ 7 } \cdot i =
-3 \cdot i $$ |
| ⑤ | Combine like terms: $$ \color{blue}{-8i} \color{red}{-10i} \color{red}{-3i} -3 = \color{red}{-21i} -3 $$ |
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