can anyone simplify i^75? I am not sure I am doing it wright on my own. Help will be appreciated. Thank you!
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$\displaystyle i^n = i^{\bmod (n,4)} \quad \Rightarrow \quad i^{75} = i^{\bmod (75,4)} = i^3 $
Originally Posted by Plato $\displaystyle i^n = i^{\bmod (n,4)} \quad \Rightarrow \quad i^{75} = i^{\bmod (75,4)} = i^3 $ very nice! Originally Posted by mirdita59 can anyone simplify i^75? I am not sure I am doing it wright on my own. Help will be appreciated. Thank you! another approach: $\displaystyle i^{75} = i \cdot i^{74} = i \cdot \left( i^2 \right)^{37} = -i$ (sorry for giving away the answer Plato, but splitting it up this way kind of made the answer obvious anyway)
Originally Posted by Jhevon (sorry for giving away the answer, but splitting it up this way kind of made the answer obvious anyway) Well of course and here is the other way. $\displaystyle i^{75} = i^{72 + 3} = \left( {i^{72} } \right)i^3 = \left( {i^4 } \right)^{18} i^3 = i^3 $
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