Question

Question #27939

Answer 1

As Sudip Sinha has pointed out `-1+sqrt3i` is NOT a zero. (I neglected to check that.) The other zeros are `1-sqrt3 i` and `1`.

Explanation:

Because all of the coefficients are real numbers, any imaginary zeros must occur in conjugate pairs.
Therefore, `1-sqrt3 i` is a zero.

If `c` is a zero then `z-c` is a factor, so we could multiply

`(z-(1+sqrt3 i))(z-(1-sqrt3 i))` to get `z^2-2z+4`
and then divide `P(z)` by that quadratic.

But it's quicker to consider the possible rational zero for `P` first. Or add the coefficients to see that `1` is also a zero.

Answer 2

`1` and `1 - sqrt3 i`

Explanation:

There is an error in your question. The root should be `1 + sqrt3 i`. You can verify this by putting the value in the expression. If it is a root the expression should evaluate to zero.

The expression has all real coefficients, so by the Complex Conjugate Roots Theorem (https://en.wikipedia.org/wiki/Complex_conjugate_root_theorem), we have that the other complex root is `1 - sqrt3 i`,

Clearly, the third root (say `a`) has to be real, since it cannot have a complex conjugate; otherwise there will be 4 roots, which is not possible for a 3rd degree equation.

Note
`( z - ( 1 - sqrt3 i ) ) ( z - ( 1 + sqrt3 i ) )`
`= ( (z - 1) + sqrt3 i ) ( (z - 1) - sqrt3 i )`
`= ( (z - 1)^2 - (sqrt3 i)^2 )` ( Since `(z + a) (z - a) = z^2 - a^2`.)
`= z^2 - 2z + 1 - 3(-1)`
`= z^2 - 2z + 4`

We will try to get this factor in the expression.
We may write:
`P(z) = z^3 - 3z^2 + 6z - 4`
`= z ( z^2 - 2z + 4 ) - 1 ( z^2 - 2z + 4 )`
`= (z - 1) ( z^2 - 2z + 4 )`
`= (z - 1) ( z - ( 1 - sqrt3 i ) ) ( z - ( 1 + sqrt3 i ) )`

Answer 3

As an intro, I think that the root should be `color(blue)(1+sqrt3)` and not `color(red)(-1+sqrt3)`

On that basis my answer is :

`z in {1," "1+sqrt3," "1-sqrt3}`

Explanation:

By using the idea of complex conjugates and some other cool tricks.

`P(z)` is a polynomial of degree `3`. This implies that it should only have `3` roots.

One interesting fact about complex roots is that they never occur alone.They always occur in conjugate pairs.

So if `1+isqrt3` is one root, then its conjugate : `1-isqrt3` most certainly is a root too!

And since there is just one more root left, we can call that root `z=a`.
It is not a complex number because, complex roots always occur in pairs.
And since this is the last of the `3` roots, there cannot be any other pair after the first one!

In the end the factors of `P(z)` were easily found to be `[z-(1+isqrt3)] " , " [z-(1-isqrt3)] " and " (z-a)`

NB : Note that the difference between a root and a factor is that :
- A root could be `z=1+i`
But the corresponding factor would be `z-(1+i)`

The second trick is that, by factoring `P(z)` we should get something like this :

`P(z)=[z-(1+isqrt3)][z-(1-isqrt3)] (z-a)`

Next, expand the braces,

`P(z)=[z^2-z(1+isqrt3+1-isqrt3)+(1+isqrt3)(1-isqrt3)] (z-a)`

`=[z^2-z(2)+(1+3)] (z-a)`

`=[z^2-2z+4] (z-a)`

`=z^3+z^2(-a-2)+z(2a+4)-4a`

Next, we equate this to the original polynomial `P(z)=z^3-3z^2+6z-4`

`=>z^3+z^2(-a+2)+z(-2a+4)-4a=z^3-3z^2+6z-4`

Since the two polynomials are identical, we equate the coefficients of `z^3`, `z^2`, `z^1`and `z^0`(the constant term) on either side,

Actually,we just need to pick one equation and to solve it for `a`

Equating the constant terms,

`=>-4a=-4`

`=>a=1`

Hence the last root is `color(blue)(z=1)`