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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