Question

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

Answer 1

`z = 2-3i`
Now, `z^2-4z+13 = (2-3i)^2-4(2-3i)+13`
`=4+9i^2-12i-8+12i+13`
`=4-9+5 = 0...[As i^2 = -1]`
`:. z^2-4z+13 = 0`
Now, `4z^3-3z^2+169 = 4z^3-16z^2+13z^2+52z-52z+169`
`=4z^3-16z^2+52z+13z^2-52z+169`
`=4z(z^2-4z+13) + 13(z^2-4z+13) `
`=4z(0)+13(0) = 0`
`:. 4z^3-3z^2+169 = 0.`