How do you simplify $i^{35}$ $i^35$ ?

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Answer 1 · 2015-11-16T12:34:32

$i^{35} = - i$ $i^35 = -i$

$i^{2} = - 1$ $i^2=-1$
$i^{4} = {(i^{2})}^{2} = {(- 1)}^{2} = + 1$ $i^4 = (i^2)^2 = (-1)^2 = +1$

$i^{35} = i^{32} \cdot i^{2} \cdot i$ $i^35 = i^32*i^2*i$

$= {(i^{4})}^{8} \cdot i^{2} \cdot i$ $= (i^4)^8*i^2*i$

$= 1^{8} \cdot (- 1) \cdot i$ $=1^8*(-1)*i$

$= - i$ $=-i$

How do you simplify i35i^35?