[FIX] Im z & Re z, given one root of... find the other two roots of...?

Question

Given thats z = 1 + i is one root of
P(z) = z^3 - 4z^2 + 6z - 4 = 0

find the other two roots.

plz help, this is what i have done so far

if z = 1 + i and is a root of P(z), then the complex conjugate is (1 - i) and therefore this is also a root.

Now this becomes [z-(1 + i)] and the other [z-(1 - i)]

therefore [z-(1 + i)] [z-(1 - i)] (z - a)

(z - 1 - i) (z - 1 + i) (z - a)

[(z - 1)^2 - (i)^2] (z - a)

[(z - 1)^2 -(-1)] (z - a)

[(z - 1)^2 + 1] (z - a)

(z^2 - 2z + 1) (z - a)

Now how do i find the value of (a) from here, and the other root, plz explain step by step!

==============================...

Question below do answer plz, Answer at the back of the specialist book says (1 - i) and 2

Does this mean the first root was given and the second root was the conjugate and the third root, a = 2 which just means z = 2 ??

Answer 1

First, you made one tiny mistake in the last step. It should be:

(z^2 - 2z + 2) (z - a)

Then, the only way I know that you could find a using the fact that you know that (z^2 - 2z + 2) is a factor is to use long division of (z^2 - 2z + 2) into (z^3 - 4z^2 + 6z - 4); the result SHOULD be z - a (or z - 2, since you know a = 2).

The alternate way, of course, is to try the positive and negative factors of 4 using synthetic division, which is the way you'd attack a problem like this most of the time.

And, yes, if you're looking for solutions over the real numbers, z = 2 is the only solution. If you are looking for solutions over the complex numbers, the solutions are (1 + i), (1 - i), and 2.

Answer 2

Thanks for deleting the question while i was in the middle of answering.

start with

P(z)=0=z^3 - 4z^2 + 6z - 4=(z-a)(z-b)(z-c)
P(z)=0=z^3 - 4z^2 + 6z - 4=(z^2- (a+b)*z + a*b)*(z-c)
P(z)=0=z^3 - 4z^2 + 6z - 4=(z^3- (a+b)*z^2 + a*b*z)+(-c*z^2+ (a+b)*c*z - a*b*c)
z^3 - 4z^2 + 6z - 4=(z^3- (a+b+c)*z^2 + (a*b+a*c+b*c)*z- a*b*c)

then a, b, c are the roots, right?

so

- (a+b+c) = -4
a*b + a*c + b*c = 6
- a*b*c=-4

you have the answer for one of the roots, call it a.
a=3 + 5i

a+b+c = 4
a*b + a*c + b*c = 6
a*b*c = 4
a=3 + 5i

3 + 5i+b+c = 4
(3 + 5i)*b + (3 + 5i)*c + b*c = 6
(3 + 5i)*b*c = 4

b = 4 - (3 + 5i) - c
(3 + 5i)*b + (3 + 5i)*c + b*c = 6
(3 + 5i)*b*c = 4

b = 1 - 5i - c
(3 + 5i)*b + (3 + 5i)*c + b*c = 6
(3 + 5i)*b*c = 4

(3 + 5i)*(1 - 5i - c) + (3 + 5i)*c + (1 - 5i - c)*c = 6
(3 + 5i)*(1 - 5i - c)*c = 4

you know what to do from here.

Answer 3

The cost of this selection is 8 instances sqrt(a million+3) = sixteen. So the fee of z is two. The process the unique complicated selection is 240 tiers, measured from the helpful actual axis. So the guidelines of the roots are 240/4, six hundred/4, 960/4, and 1320/4, or 60 tiers, a hundred and fifty tiers, 240 tiers, and 330 tiers. In Cartesian form, they seem to be a million + i sqrt(3), - sqrt(3) + i, -a million - i sqrt(3), and sqrt (3) - i. examine by using drawing something, i did not.

Answer 4

You do it correct! let us just polish it.
Assume z1= 1+j is a root of quadratic equation zz -bz +c = (z-z1)(z-z2) = zz –(z1+z2)z +z1z2 = 0; then a real b= z1+z2 means that imaginary part of z1 must be compensated by imaginary part of z2, i.e. Im[z1] = -Im[z2]; thus z2 =x2 –j;
Now c = (1+j)(x2-j) = x2+1 +j(x2 –1) must be also a real value, hence x2=1
Thus z2=1-j is complex conjugate to z1=1+j;
Thus zz –2z +2 =0 is equation with roots z1 and z2;
Now divide zzz –4zz +6z –4 by zz –2z +2 and you receive z-2; thus z3=2;
If you don’t know how to divide, then assume (z-a)(zz –2z +2) = zzz –4zz +6z –4; multiply zzz -(a+2)zz +(2a+2)z –2a = zzz –4zz +6z –4; thence
-a-2 = -4, hence a=2;
2a+2=6, hence a=2;
-2a=-4, hence a=2;
thus your equation has 2 complex conjugate roots z1=1+j and z2=1-j, and a real root z3=2;