Question
Answer
$$7i^{10}-6i^{49}+12i^{112}+4i^{111}=-10i+5$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}7i^{10}-6i^{49}+12i^{112}+4i^{111}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-7-6i+12-4i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-10i+5\end{aligned} $$ | |
| ① | $$ 7i^{10} = 7 \cdot i^{4 \cdot 2 + 2} =
7 \cdot \left( i^4 \right)^{ 2 } \cdot i^2 =
7 \cdot 1^{ 2 } \cdot (-1) =
7 \cdot -1 = -7 $$ |
| ② | $$ -6i^{49} = -6 \cdot i^{4 \cdot 12 + 1} =
-6 \cdot \left( i^4 \right)^{ 12 } \cdot i^1 =
-6 \cdot 1^{ 12 } \cdot i =
-6 \cdot i $$ |
| ③ | $$ 12i^{112} = 12 \cdot i^{4 \cdot 28 + 0} =
12 \cdot \left( i^4 \right)^{ 28 } \cdot i^0 =
12 \cdot 1^{ 28 } \cdot 1 =
12 \cdot 1 $$ |
| ④ | $$ 4i^{111} = 4 \cdot i^{4 \cdot 27 + 3} =
4 \cdot \left( i^4 \right)^{ 27 } \cdot i^3 =
4 \cdot 1^{ 27 } \cdot (-i) =
4 \cdot -i = -4i $$ |
| ⑤ | Combine like terms: $$ \color{blue}{-6i} \color{blue}{-4i} + \color{red}{12} \color{red}{-7} = \color{blue}{-10i} + \color{red}{5} $$ |
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