If `z=2-3i` show that `z^2=4z+13=0` and hence find the value of `4z^3-3z^2+169.`

Answer

N/a

Explanation

`z=2-3i`
`rArr" "z-2=-3i`
`rArr" "(z-2)^(2) = 9i^(2)`
`rArr" " z^(2) - 4z+4=-9`
` rArr" " z^(2)-4z +4-9`
`rArr" " z^(2) -4z+ 13= 0" "...(1)"Hence Proved"`
Now , `4z^(3) - 3z^(2) +140`
`" "=4z(z^(2))-3z^(2) +140`
`" "=4z(4z-13)-3z^(2)+140`
`" "["From eq."(1)z^(2)=4z-13]`
`" "=16z^(2)-52z-3z^(2)+140`
`" "=13z^(2)-52z+140`
`" "=13(4z-13)-52z+140`
`" "["From eq. "(i)z^(2)=4z-13]`
`" "=52z-169-52z+140`
`" "=-29`.