1.

If z = 2 + i, prove that `z^(3) + 3z^(2) - 9z + 8 = (1 + 14i)`.

Answer» We have
`z = 2 + i rArr z - 2 = i rArr (z-2)^(2) = i^(2) rArr z^(2) - 4z + 5 = 0." "...(i)`
`therefore" "z^(3) + 3z^(2) - 9z + 8`
`= z(z^(2)-4z+5) + 7(z^(2) - 4z + 5) + 14z - 27`
`=(z xx 0) + (7 xx 0) + 14z - 27 = (14z - 27)" "["using (i)"]`
`= 14 (2 + i) - 27 = (1 + 14i)`.
Hence, `z^(3) + 3z^(2) - 9z + 8 = (1 + 14i)`.


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