Question

Solve for all complex numbers z such that `z^4 + 4z^2 + 6 = z`?

Answer 1

The trick to solve this is to substitute `z^2=t`

Explanation:

What you want to do is turn `z^2` into `t`
The equation would then look like this:
`t^2+4t+6`

Then use the formula we typically use to solve a squared polynomial, solve the equation for t.

In this case, the number is non-existent or imaginary, I don't know
if you covered imaginary numbers in class or not (if you didn't, the solution is DNE or Does Not Exist). Assuming you did `t=-2-sqrt(2i)` and `t=-2+sqrt(2i)`

Now replace `t` with `z^2` you get that `z=+-sqrt(-2-sqrt(2i))` and `z=+-sqrt(2+sqrt(2i))`

Solving that, you should have 4 different imaginary answers for z.
Hope this was helpful, good luck.

Answer 2

Factor into a pair of quadratics...

Explanation:

Given:

`z^4+4z^2+6=z`

Subtract `z` from both sides to get the quartic into standard form:

`z^4+4z^2-z+6 = 0`

Note that since there is no `z^3` term, this quartic has a factorisation of the form:

`z^4+4z^2-z+6 = (z^2-az+b)(z^2+az+c)`

`color(white)(z^4+4z^2-z+6) = z^4+(b+c-a^2)z^2+a(b-c)z+bc`

Equating coefficients, we find:

`{ (b+c = a^2+4), (b-c=-1/a), (bc=6) :}`

Hence:

`a^4+8a^2+16 = (a^2+4)^2`

`color(white)(a^4+8a^2+16) = (b+c)^2`

`color(white)(a^4+8a^2+16) = (b-c)^2+4bc`

`color(white)(a^4+8a^2+16) = 1/a^2+24`

Subtracting `1/a^2+24` from both ends and multiplying by `a^2` we find:

`0 = (a^2)^3+8(a^2)^2-8(a^2)-1`

`color(white)(0) = (a^2-1)((a^2)^2+9(a^2)+1)`

Note that the second factor is always non-zero for real values of `a`, but the first factor gives us solutions `a=+-1`

Let:

`a=1`

Then:

`b = 1/2(a^2+4-1/a) = 1/2(1+4-1) = 2`

`c = 1/2(a^2+4+1/a) = 1/2(1+4+1) = 3`

So:

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

I will leave it as an exercise to the reader to solve the remaining quadratic equations:

`z^2-z+2 = 0`

`z^2+z+3 = 0`