Finding roots of an equation dealing complex numbers For the equation
$z^6
 + 6z + 20 = 0,$ z is a complex number
Putting z=x+iy is tiresome . Moreover my question demands to find the roots lying in each quadrant . Is there an easy way out?
Edit.

The actual statement is
The number of roots in 1st,2nd,3rd ,4th quadrant are

 A: Set $z=a+ib\;$ and expand
$$(a+ib)^6 + 6(a+ib) + 20 = 0 \tag{1}$$
Equating real and imaginary parts to $0$ we obtain the system
$$\begin{aligned}a^6-15a^4b^2+15a^2b^4-b^6+6a+20&=0 \quad\quad\quad\quad (*)\\
6a^5b-20a^3b^3+6ab^5+6b&=0 \quad\quad\quad\quad (**)\end{aligned}$$
Recall that the system has six solutions $(a,b)\in \mathbb{R}\times\mathbb{R}.$ Clearly $ab\neq0.$
Take the equation $(**)$ and divide it by $2b.$ We obtain $$3a^5-10a^3b^2+3ab^4+3=0 \tag{2}$$ Let us consider $(2)$ as equation in variable $a.$
Descartes rule of signs says that $(2)$ has $2$ positive and $3$ negative roots $a$ (more precisely, two $a$'s have positive real part and three have negative real part, but we know that all $a$'s are real).
Recall that all solutions of $(1)$ are in pairs  $(a_1,\pm b_1), (a_2,\pm b_2),(a_3, \pm b_3).$
Only one of $a$'s is missing in $(**)$, thus we can conclude that $(**)$ has one positive double root, one negative double root and one simple root $a.$

Therefore, one solution of the original equation $(1)$ lies in the first quadrant, two in the second, two in the third and one in the fourth.

To verify our result, let us apply Descartes rule to $(*):$

*

*if $\;20-b^6<0$ then $(*)$ has $3$ positive and $3$ negative roots $a_i,$ which is impossible

*if $\;20-b^6=0$ then $(*)$ has $2$ positive and $3$ negative roots $a_i$

*if $\;20-b^6>0$ then $(*)$ has $2$ positive and $4$ negative roots $a_i$
Our conclusion is justified. As a by-result, necessarilly $\;20-b^6\geq 0.$
