3 years ago

I^89 simplified. Complex number.

1 answer:

3 0

This problem can be simplified by remembering that since $i= \sqrt{-1}$ , we have $i^{4}=(-1)^{2}=1$ . Then we have $i^{89}=i^{88+1}=i^{88}i^{1}=(i^{4})^{22}i=(1)^{22}i=i$ .

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