Find complex roots of quartic function $(3z + 1)(4z + 1)(6z + 1)(12z + 1) = 2$

Question

I found a math problem involving complex number

Find all complex number z such that $$(3z + 1)(4z + 1)(6z + 1)(12z + 1) = 2$$

The complex number form is z = a + bi

If I multiply all the factor to make it to the standard form of a polynomial, It would be too long to solve. Not even substituting z = a + bi to the polynomial yet.

I don't think that's a good way to solve

I want to know if there's a simpler way to solve this problem, can anyone shows me a hint?

Answer 1

Let $z=\frac{x}{12}.$

Thus, $$(3z + 1)(4z + 1)(6z + 1)(12z + 1)-2=$$ $$=\left(\frac{x}{4}+1\right)\left(\frac{x}{3}+1\right)\left(\frac{x}{2}+1\right)(x+1)-2=$$ $$=\frac{1}{24}((x+4)(x+3)(x+2)(x+1)-48)=$$ $$=\frac{1}{24}((x^2+5x+4)(x^2+5x+6)-48)=$$ $$=\frac{1}{24}(x^2+5x+12)(x^2+5x-2)=$$ $$=(12z^2+5z+1)(72z^2+30z-1).$$ Can you end it now?

Answer 2

Hint: \begin{eqnarray*} (3z+1)(4z+1)(6z+1)(12z+1)-2&=& 864z^4+720z^3+210z^2+25z-1 \\ &=& (72z^2+30z-1)(12z^2+5z+1). \end{eqnarray*}

Answer 3

Write the equation as

$$(12z + 4)(12z + 3)(12z + 2)(12z + 1) = 48$$ Recognize the arithmetic progression of 1, 2, 3, 4 and the symmetry around $\frac52$, which prompts the variable change

$$ 12z=t-\frac52\tag1$$

to reduce the equation to a quadratic one in $t^2$,

$$(t^2-\frac94)(t^2-\frac14)=48$$

Solve to get $t^2=\pm\frac{33}4$ and, via (1), the complex roots

$$z=-\frac1{24}(5\pm\sqrt{33}),\> -\frac1{24}(5\pm i\sqrt{33}) $$