Question
Answer
$$i^{75}+i^{80}+i^{85}+i^{90}=0$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}i^{75}+i^{80}+i^{85}+i^{90}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-i+1+i-1 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}0\end{aligned} $$ | |
| ① | $$ i^{75} = i^{4 \cdot 18 + 3} =
\left( i^4 \right)^{ 18 } \cdot i^3 =
1^{ 18 } \cdot (-i) =
-i = -i $$ |
| ② | $$ i^{80} = i^{4 \cdot 20 + 0} =
\left( i^4 \right)^{ 20 } \cdot i^0 =
1^{ 20 } \cdot 1 =
1 $$ |
| ③ | $$ i^{85} = i^{4 \cdot 21 + 1} =
\left( i^4 \right)^{ 21 } \cdot i^1 =
1^{ 21 } \cdot i =
i $$ |
| ④ | $$ i^{90} = i^{4 \cdot 22 + 2} =
\left( i^4 \right)^{ 22 } \cdot i^2 =
1^{ 22 } \cdot (-1) =
-1 = -1 $$ |
| ⑤ | Combine like terms: $$ \, \color{blue}{ -\cancel{i}} \,+ \, \color{blue}{ \cancel{i}} \,+ \, \color{green}{ \cancel{1}} \, \, \color{green}{ -\cancel{1}} \, = \color{green}{0} $$ |
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