Determine sum of all complex solutions of $x^3+3=0, \ x^4+4=0, \ x^5+5=0$

Question

Determine sum of all complex solutions of $x^3+3=0, \ x^4+4=0, \ x^5+5=0$

This problem is confusing me a lot. Any idea where my logic is wrong?

$$x^3+3 =0\iff x^3 = -3 \Rightarrow x_1=(-3)^\frac{1}{3} \text{ for } x_1\in\mathbb{R} \\ x_1 \text{ is the only real solution, and the remaining two are complex.} \\ \text{From Viete's relations: } x_1+x_2+x_3=0 \Rightarrow x_2+x_3=-x_1=3^\frac{1}{3} \\ x^4 = -4 \text{ has no real solutions, so the sum of its complex solutions is } 0 \text{ (Again, Viete)} \\ \text{For } x^5 =-5, \text{ the sum of its complex roots, applying the same logic from the first equation, is } 5^\frac{1}{5} \\ \text{Thus, the sum of all complex roots is } 3^\frac{1}{3} + 5^\frac{1}{5}.$$ However, my workbook says the sum is $0$, but gives no explanation. Any clues?

Answer 1

What you understood is that the question asks sum of all complex numbers where the imaginary part is non zero. But if it is not explicitly written, then we assume that sum of all complex solutions mean sum of all solutions as real solutions also belong to complex numbers (imaginary part being $0$)

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For your problem, sum of roots of first equation is $0$ by vietes formula. Same goes for second and third. Hence when we will add we simply do $0+0+0$ which equals to $0$.

I assumed that you're calculating the roots of individual equations and then adding all, not that you're simultaneously solving them.

Answer 2

Another way to see that the sum of the roots are $0$:

Consider $$(x^3+3)(x^4+4)(x^5+5)=0$$ Expanding gives a polynomial with no $x$-term, hence by Vietè's formulas the sum of the roots are $0$.