CV15

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Evaluate i^{{243}}\,

Powers of i\, are periodic with period 4, so replace 243 with its remainder on division by 4.

243=4(60)+3\,

So now

i^{{243}}=\left(i^{{4}}\right)^{{60}}i^{3}=(1)^{{60}}i^{3}=i^{3}=-i\,


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