please help, imaginary numbers

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October 28th 2007, 12:32 PM
mirdita59

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!
October 28th 2007, 12:46 PM
Plato

$i^n = i^{\bmod (n,4)} \quad \Rightarrow \quad i^{75} = i^{\bmod (75,4)} = i^3 <br />$
October 28th 2007, 12:51 PM
Jhevon

Quote:

Originally Posted by Plato

$i^n = i^{\bmod (n,4)} \quad \Rightarrow \quad i^{75} = i^{\bmod (75,4)} = i^3 <br />$

very nice!

Quote:

Originally Posted by mirdita59

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)
October 28th 2007, 01:15 PM
Plato

Quote:

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.
$i^{75} = i^{72 + 3} = \left( {i^{72} } \right)i^3 = \left( {i^4 } \right)^{18} i^3 = i^3 <br />$

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