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

2 Answers
Jan 18, 2018

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.

Jan 18, 2018

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