1.If z=32+i2z = \dfrac{\sqrt{3}}{2} + \dfrac{i}{2}z=23+2i (i=−1i = \sqrt{-1}i=−1), then (1+iz+z5+iz8)9(1 + iz + z^5 + iz^8)^9(1+iz+z5+iz8)9 is equal to:a.−1-1−1b.111c.000d.(−1+2i)9(-1 + 2i)^9(−1+2i)9Login to continueOnly logged in users canattempt or see the solution.
