Complex City 4

Put $x+2=t$

The given equation becomes $(t-1)^4+(t+1)^4=4$

$t^4+6t^2+1=2$

$t^4+6t^2-1=0$

$t^2=\frac{-6 \pm \sqrt{36+4}}{2}$

Non real complex solutions of the equation are given by $t^2=\frac{-6 - \sqrt{40}}{2}$ (i.e. when $t^2$ is negative)

$(x+2)^2= -\frac{-6-\sqrt{40}}{2}$

Sum of roots = $\large -\frac{\text{ coefficient of}x}{\text{ coefficient of} x^2}=\boxed{-4}$

NOTE: This is for the question creater. As the real numbers are also complex numbers you should mention "non real complex" in your question.

Alright , here is my solution , we can see that the equation on the LHS is symmetric about the point $x = -2$ , And is strictly increasing from $-\infty \text{ to } -3$ , then constant from $- 3 \text{ to } -1$ and then again strictly increasing from $-1 \text{ to } \infty$ . Hence the graph of $(x + 1)^4 + (x + 3)^4$ cuts the line $y = 4$ only twice , once before $-3 \text{ and once after } -1$ as it is strictly increasing . Therefore , the two solutions are of the form $-2 + t$ and $-2 - t \text{ ; since symmetry about the point } x = -2$ and so , the sum of these two solutions = $-2 -t -2 +t = -4$

This solution is a little bit longer but i use more complex geometry and i think it's nice.

Let $x=a+i b$

Now because ( $x+1)^4+(x+3)^4=4$ .

It must be that $(x+1)^4$ and $(x+3)^4$ are real.

So their argument are equal to $k\pi$ .

$4 arctan(\frac{b}{a+1})=k\pi$

$4 arctan(\frac{b}{a+3})=c\pi$

$\frac{b}{a+3}=tan(\frac{c\pi}{4})$

$\frac{b}{a+1}=tan(\frac{k\pi}{4})$

And you will get the equation that says

$a=\frac{3 tan(\frac{c\pi}{4})-tan(\frac{k\pi}{4})}{tan(\frac{k\pi}{4})-tan(\frac{c\pi}{4})}$

The only two cases are for $(tan(\frac{c\pi}{4}),tan(\frac{k\pi}{4})) ={(1,-1),(-1,1)}$ , others are undefined or real.

So the answer is $a_1=a_2=-2$