• Oct 28th 2007, 12:32 PM
mirdita59
can anyone simplify i^75? I am not sure I am doing it wright on my own. Help will be appreciated. Thank you!
• Oct 28th 2007, 12:46 PM
Plato
$i^n = i^{\bmod (n,4)} \quad \Rightarrow \quad i^{75} = i^{\bmod (75,4)} = i^3
$
• Oct 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
$

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)
• Oct 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
$