equation with imaginary number?

Question

If z = -1 + i is a root of the equation z³ + az + b = 0 where a and b are real numbers, find the values of a and b.
Show that z = -1 - i is also a root of the equation.

Answer 1

Let p(z) = z^3 + az + b.
Then, since z = -1 + i is a root, p(-1 + i) = 0.

p(-1 + i) = (-1 + i)^3 + a(-1 + i) + b.

Factoring out (-1 + i) from the first two terms,

p(-1 + i) = (-1 + i) [ (-1 + i)^2 + a ] + b
p(-1 + i) = (-1 + i) [1 - 2i + i^2 + a] + b

Keep in mind i^2 = -1, so

p(-1 + i) = (-1 + i) [1 - 2i - 1 + a] + b
p(-1 + i) = (-1 + i) [-2i + a] + b
p(-1 + i) = 2i - a - 2i^2 + ai + b
p(-1 + i) = 2i - a - 2(-1) + ai + b
p(-1 + i) = 2i - a + 2 + ai + b

Now we equate this to 0.

0 = 2i - a + 2 + ai + b. Grouping together the terms with i in them,
0 = -a + 2 + b + 2i + ai
0 = -a + 2 + b + (2 + a)i

0 + 0i = (-a + 2 + b) + (2 + a)i

Notice that 0 is really equal to 0 + 0i. That means, we can equate the left hand side and right hand side component-wise; that is, the real part of the left hand side should match the real part of the right hand side, and the coefficient of the complex number should match on both sides as well. As a result:

-a + 2 + b = 0 (or b = a - 2)
2 + a = 0

Therefore, a = -2, and b = (-2 - 2) = -4

Therefore, the values of a = -2 and b = -4. As a result, we now have the equation

z^3 - 2z - 4 = 0

2) To show that z = -1 - i is also a root of the equation, let's just tr and solve this normally.

Let p(z) = z^3 - 2z - 4
What we want to do is test factors of -4 until we get a result of 0.
p(1), p(-1), p(2), p(-2), p(4), p(-4). Our goal is to get a result of 0, so that when we get a working value r, we'll know (z - r) will be a factor.

p(1) = 1 - 2 - 4 = [non-zero]. Reject this possibility.
p(-1) = (-1)^3 - 2(-1) - 4 = -1 + 2 - 4 = [non-zero]. Reject.
p(2) = 2^3 - 2(2) - 4 = 8 - 4 - 4 = 0.

Since 2 is a root, (z - 2) is a factor.

You now perform long division with (z^3 - 2z - 4) and (z - 2).
I won't show you the details of the long division (since it's difficult to show on here), but you should get a remainder of 0 with an answer of z^2 + 2z + 2.

This means that if

z^3 - 2z - 4 = 0, then

(z - 2) (z^2 + 2z + 2) = 0

Now, we equate each factor to 0.
z - 2 = 0 (implying z = 2), and
z^2 + 2z + 2 = 0

We *could* use the quadratic equation, but instead I'm going to complete the square.

z^2 + 2z + 2 = 0
z^2 + 2z + 1 + 1 = 0
(z + 1)^2 = -1
z + 1 = +/- i
z = -1 +/- i

So our roots are: z = {2, -1 + i, -1 - i}, clearly showing how -1 - i is also a factor.

Answer 2

since it is a cubic expression it mus have 3 roots, if one of those is z=-1+i, the second root MUST be z= -1-i
a= -2
b= -4
and the third root is z = 2
use asintethic division and put at the top line
1 0 a b
divided between the root the residue must be 0 so the last sum you get that is (a-2)+(a+2)i +b=0
do the same with the second root

i have the procedure in paper if u want it

Answer 3

The first part is more algebra than I want to do right now. But it is clear that if "z" is a root of this polynomial then the complex conjugate of "z" is also a root, since conjugation is commutative with addition and multiplication operations. Also conjugation doesn't change the real numbers "a" and "b". We are constructing a polynomial that has -1+i as a root so it's conjugate -1-i must be a root.

Answer 4

For a quadratic equation with roots a and b, the equation is (x-a)(x-b)=0 or x^2-(a+b)x +ab=0 you need to write as factors and then enhance or use the above formula and write the equation as: x^2 -(a million+2i+a million-2i)x +(a million-2i)(a million+2i)=0 x^2 -2x+(a million-4i^2)=0 x^2 - 2x +5=0

Answer 5

assume z1=-1+j be solution of equation with real coefficients z^2+2pz+q=0 (eq1), then z1=-p+sqrt(p^2-q) and z2=-p-sqrt(p^2-q); thus p=1 and sqrt(1-q)=j or 1-q=-1 and q=2; so (eq1) looks z^2+2z+2=0, where z1=-1+j, z2=-1-j;
a cubic equation has always 1 real root r;
now z^3+az+b = (z-r)(z^2+2z+2) = z^3 +(2-r)z^2 +(2-2r)z -2r,
hence 2-r=0 or r=2, 2-2r=a, -2r=b; thus a=-2 and b=-4;
our equation looks z^3-2z-4=0, its roots being z1=-1+j, z2=-1-j, z3=2;