Question 951807

The fundamental theorem of algebra implies that a polynomial equation of degree
{{{n}}} has precisely {{{n}}} solutions in the complex number system. 
These solutions can be {{{real}}} or {{{complex}}} and may be {{{repeated}}}.

recall the quadratic formula and its discriminant:
 
using the discriminant {{{b^2-4ac}}}, we are able to determine solutions which are real or complex and may be repeated 

if the discriminant is {{{less}}} than {{{zero}}}, equation has {{{two}}} {{{complex}}} solutions

{{{b^2-4ac<0}}}

like you have here {{{x^2=-1}}}=>{{{x^2+1=0}}}=>{{{a=1}}}, {{{b=0}}}, {{{c=1}}}
using {{{b^2-4ac<0}}}=>{{{0^2-4*1*1<0}}}=>{{{-4<0}}}; so, discriminant is negative and we have two complex solutions

these are {{{x^2=-1}}}=>{{{x=sqrt(-1)}}}=>{{{x=i}}}or {{{x=-i}}}


now, find all {{{n}}} complex solutions of the equation of the form {{{x^n=k}}}:

{{{x^n-k=0 }}}=> {{{a=1}}}, {{{b=0}}}, {{{c=-k}}}

{{{0^2-4*1(-k)<0}}}

{{{4k<0}}}

{{{k<0/4}}}

{{{k<0 }}} 

for any {{{k<0}}}  there will be two {{{complex}}} solutions of the equation {{{x^n=k}}}

and {{{k}}} can be any number from:

({{{-infinity<k<0 }}})