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Miscellaneous Exercise 1 (Subjective) · Q174

Q.Evaluate (1−i+i2)−15(1-i+i^2)^{-15}

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1−i+i2=1−i−1=−i1-i+i^2=1-i-1=-i. So (1−i+i2)−15=(−i)−15=dfrac1(−i)15(1-i+i^2)^{-15}=(-i)^{-15}=\\dfrac{1}{(-i)^{15}}. (−i)15=(−1)15i15=−i15(-i)^{15}=(-1)^{15}i^{15}=-i^{15}. Since 15=4(3)+315=4(3)+3, i15=i3=−ii^{15}=i^3=-i, so $(- …

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