How do you simplify $i^{29}$ $i^29$ ?

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Answer 1 · 2016-03-17T09:50:53

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$i^{2} = - 1$ $i^2=-1$ .
$i^{29} = i X i^{28} = i ({(i^{2})}^{14}) = i {(- 1)}^{14} = i$ $i^29=i X i^28=i((i^2)^14)=i(-1)^14=i$

How do you simplify i29i^29?