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If a complex number z satisfies |z| = 1 and arg(z-1) = (2pi)/(3), then (omega is complex imaginarynumber) |
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Answer» `z^(2) + z` is PURELY imaginary number `|z|=1` Thismeans that z lies on a circle whose centre is at ORIGIN and radius is 1. Also ,`arg(z-1)= (2pi)/(3)` Thisimpliesthatz lies ona ray emanating (1,0) making an angle of `(2pi)/(3)` with positive real axis. `RARR /_PQX =(2pi)/(3)` ` rArr /_PQO = (pi)/(3)` and OP=OQThus, triangle OPQ is equilaterl.  `rArr OP = OQ = PQ` `rArr |z|=|z-1| =1` Also `arg(z) = 60^(@)` `therefore z = COS60^(@) + i sin 60^(@) = (1+ isqrt(3))/(2) = -omega^(2)` `z^(2)= omega^(4) = omega` `rArr z^(2) + z = omga - omega^(2) = ((-1+isqrt(3))/(2)) -((-1-isqrt(3))/(2)) = isqrt(3)` Therefore , `z^(2) +z` is purely imaginary number . |
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