if x=-i is a zero of the fuction F(x)=x^3+ix^2+ix-1? WHy?how you resolve it?

Answer 1

YES, x = - i IS a zero of F(x)=x^3+ix^2+ix-1. Why? Because:

F(-i) = (-i)^3 + i (-i)^2 + 1(-i) - 1 = - (-1) i + i (+) (- 1) + i (-i) - 1

= i - i + 1 - 1 = 0.

It's not clear what you meant by "how you resolve it?" If however, you meant "How do you continue to FACTOR it", then here is what you should get:

F(x)=x^3+ix^2+ix-1 = (x + i) (x^2 + i).

Again, if you wanted F(x) = 0 to be SOLVED, the roots are:

x = - i and x = +/- sqrt (- i).

I'll leave you to express sqrt (- i) as a complex number. (It would be useful to envision the Argand diagram.)

Live long and prosper.

Answer 2

substitute -i for all of the x's in the equation.....you get:
-i = ( -i^3) +( i(-i)^2) +( i (-i) -1)
which is -i= -i^3 + (i . -i . -i) + i . -i -1

- (i^3) + i^3 + -i^2 -1
so, the first two cancel each other...and you are left with -i = (-i^2) - 1, change all signs....

so, -i^2 is the same as i^2....so, add -i to each side.....
-(-i^2) = -1 + i
so -i^2= -1 +i....