Question
Answer
$$-10i^{55}+6i^{32}-4i^{27}-3i^{60}=14i+3$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-10i^{55}+6i^{32}-4i^{27}-3i^{60}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}10i+6+4i-3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}14i+3\end{aligned} $$ | |
| ① | $$ -10i^{55} = -10 \cdot i^{4 \cdot 13 + 3} =
-10 \cdot \left( i^4 \right)^{ 13 } \cdot i^3 =
-10 \cdot 1^{ 13 } \cdot (-i) =
-10 \cdot -i = 10i $$ |
| ② | $$ 6i^{32} = 6 \cdot i^{4 \cdot 8 + 0} =
6 \cdot \left( i^4 \right)^{ 8 } \cdot i^0 =
6 \cdot 1^{ 8 } \cdot 1 =
6 \cdot 1 $$ |
| ③ | $$ -4i^{27} = -4 \cdot i^{4 \cdot 6 + 3} =
-4 \cdot \left( i^4 \right)^{ 6 } \cdot i^3 =
-4 \cdot 1^{ 6 } \cdot (-i) =
-4 \cdot -i = 4i $$ |
| ④ | $$ -3i^{60} = -3 \cdot i^{4 \cdot 15 + 0} =
-3 \cdot \left( i^4 \right)^{ 15 } \cdot i^0 =
-3 \cdot 1^{ 15 } \cdot 1 =
-3 \cdot 1 $$ |
| ⑤ | Combine like terms: $$ \color{blue}{10i} + \color{blue}{4i} + \color{red}{6} \color{red}{-3} = \color{blue}{14i} + \color{red}{3} $$ |
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