QuestionReportQuestion: \[- i\]...−i- i−iA−1- 1−1Bi=−1i = \sqrt{- 1}i=−1C1+i2+i3−i6+i81 + i^{2} + i^{3} - i^{6} + i^{8}1+i2+i3−i6+i8D2−i2 - i2−iAnswer−1- 1−1ExplanationSolution⇒(i)2n=(i)4\Rightarrow (i)^{2n} = (i)^{4}⇒(i)2n=(i)4 ⇒2n=4\Rightarrow 2n = 4⇒2n=4 ⇒n=2\Rightarrow n = 2⇒n=2or (1+i)x−2i3+i+(2−3i)y+i3−i=i\frac{(1 + i)x - 2i}{3 + i} + \frac{(2 - 3i)y + i}{3 - i} = i3+i(1+i)x−2i+3−i(2−3i)y+i=i .
