Question

For what values of `m` does `z^3+(3+i)z^2-3z-(m+i) = 0` have at least one real root?

Answer 1

See below.

Explanation:

A possible solution is for `z = 0` in which case we need `m = -i` because then

`z^3 + (3+i)z^2 -3z=0 rArr z(z^2 + (3+i)z -3) = 0` hence

`z = 0 in RR`

NOTE

If `z = x + i 0` then

`x^3 + (3 + i) x^2 - 3 x - (m + i) = x^3+3x^2-3x-m+i(x^2-1)`

Now making `x = pm 1 rArr {(1+3-3-m=0),(-1+3+3-m=0):}`

hence `m = 1` or `m = 5` or `m = i`

Answer 2

`m=1" "` or `" "m = 5`

Explanation:

Given:

`z^3+(3+i)z^2-3z-(m+i) = 0`

Find values of `m` such that this has at least one real solution.

Assuming we are looking for suitable real values of `m`, we can equate imaginary parts to find:

`z^2-1 = 0`

So:

`z = +-1`

If `z=1` then:

`0 = z^3+(3+i)z^2-3z-(m+i)`

`color(white)(0) = 1+3+i-3-m-i`

`color(white)(0) = 1-m" "rarr" "m = 1`

If `z=-1` then:

`0 = z^3+(3+i)z^2-3z-(m+i)`

`color(white)(0) = -1+3+i+3-m-i`

`color(white)(0) = 5-m" "rarr" "m = 5`