Question

Find the values and roots of the equation `z^4-2z^3+7z^2-4z+10=0`?

Find the values of `a in RR` for which `ai` is a solution of
`z^4-2z^3+7z^2-4z+10=0`
Also find all the roots of this equation.

Answer 1

The roots are: `+-sqrt(2)i` and `1+-2i`

Explanation:

Given:

`z^4-2z^3+7z^2-4z+10 = 0`

If `z = ai` for some real number `a` then, the terms of even degree are real and those of odd degree imaginary...

`0 = (ai)^4-2(ai)^3+7(ai)^2-4(ai)+10`

`color(white)(0) = (a^4-7a^2+10)+2a(a^2-2)i`

`color(white)(0) = (a^2-5)(a^2-2)+2a(a^2-2)i`

`color(white)(0) = ((a^2-5)+2ai)(a^2-2)`

Hence `a=+-sqrt(2)`

So two of the roots of the original quartic are `+-sqrt(2)i`, with associated factors:

`(z-sqrt(2)i)(z+sqrt(2)i) = z^2+2`

We find:

`z^4-2x^3+7z^2-4z+10 = (z^2+2)(z^2-2z+5)`

`color(white)(z^4-2x^3+7z^2-4z+10) = (z^2+2)(z^2-2z+1+4)`

`color(white)(z^4-2x^3+7z^2-4z+10) = (z^2+2)((z-1)^2-(2i)^2)`

`color(white)(z^4-2x^3+7z^2-4z+10) = (z^2+2)(z-1-2i)(z-1+2i)`

So the other two roots are:

`z = 1+-2i`

Answer 2

`a=\pmsqrt2`
`z=\pmsqrt(2)i, 1\pmsqrt(2)i`

Explanation:

We can use the fact that the problem asks us to find values of `a` so that `ai` is a solution. Thus, if `ai` is a solution, so is the conjugate, or `-ai`.

Let us denote the other two solutions as `c\pmdi`. If we write out our polynomial, say `P(z)`, by way of the roots, we have that:

`P(z)=(z-ai)(z+ai)(z-(c-di))(z-(c+di))=0`.

We can expand this, group like terms, and find that:
`P(z)=z^4-(2c)z^3+(c^2+d^2+a^2)z^2-(2a^2c)z+(a^2c^2+a^2d^2)=0`.

Set like-coefficients equal (so set `-2z^3=-2c` etc), we find that:

`a=\pmsqrt2`
`z=\pmsqrt(2)i, 1\pmsqrt(2)i`

By graphing, we see that this solution set works.