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Math Help - please help, imaginary numbers

  1. #1
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    Smile please help, imaginary numbers

    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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  2. #2
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    i^n  = i^{\bmod (n,4)} \quad  \Rightarrow \quad i^{75}  = i^{\bmod (75,4)}  = i^3 <br />
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Plato View Post
    i^n  = i^{\bmod (n,4)} \quad  \Rightarrow \quad i^{75}  = i^{\bmod (75,4)}  = i^3 <br />
    very nice!

    Quote Originally Posted by mirdita59 View Post
    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:

    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)
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    (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.
    i^{75}  = i^{72 + 3}  = \left( {i^{72} } \right)i^3  = \left( {i^4 } \right)^{18} i^3  = i^3 <br />
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