How do you simplify $i^{333}$ $i^333$ ?

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Answer 1 · 2016-01-13T23:04:24

$i^{333} = i$ $i^333=i$

Using the exponent rules:

$a^{n + m} = a^{n} \cdot a^{m}$ $a^(n+m)=a^n*a^m$

$a^{n \cdot m} = {(a^{n})}^{m}$ $a^(n*m)=(a^n)^m$

$i^{333} = i \cdot i^{332} = i \cdot {(i^{2})}^{\frac{332}{2}} = i \cdot {(i^{2})}^{166}$ $i^333=i*i^332=i*(i^2)^(332/2)=i*(i^2)^166$

as $i^{2} = - 1$ $i^2=-1$

$:.i*(i^2)^166=i*(-1)^166$

as $(-1)^(2p)=1$

$:.i*(-1)^166=i*1=i$

How do you simplify i^333i333i^333?