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Complex polynomial help

  • #1
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Homework Statement


[/B]
Suppose q(z) = z^3 − z^2 + rz + s, is a complex polynomial with 1 + i and i as zeros. Find r and s and the third complex zero.

The Attempt at a Solution


[/B]
(z-(1+i)(z-i) = Z^2-z-1-2iz+i

(Z^2-z-1-2iz+i)(z+d) = Z^3+z^2(d-1-zi)-z(d+1+2di-i)-d(1-i)

Z^2 term

Z^2(d-1-zi)=-z^2
d-1-2i=-1
d=2i

z term

-z(d+1+2di-i)=rz
-d-1-2di+i=r
2i-1-4i^2+i=r
-i+3=r

constant term

-d(1-i)=s
-2i+2i^2=s
-2i-2=s

This is what I have done but I am when I expand the complex zeros I do not get anything close to q(z)
 

Answers and Replies

  • #2
Delta2
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For start I believe you do something wrong with the factor ##(z+d)## in line 2 of your post, shouldn't that be ##(z-d)##?
 
  • #3
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For start I believe you do something wrong with the factor ##(z+d)## in line 2 of your post, shouldn't that be ##(z-d)##?
Why would it matter if it is positive or negative?
 
  • #4
Delta2
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It matters on the sign d will have. I believe the 3rd root of your polynomial is not 2i but -2i.

if you write a polynomial as ##p(z)=(z+d)q(z)## then one of the roots of ##p(z)## is not d but -d as you can easily verify.
 
  • #5
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It matters on the sign d will have. I believe the 3rd root of your polynomial is not 2i but -2i.

if you write a polynomial as ##p(z)=(z+d)q(z)## then one of the roots of ##p(z)## is not d but -d as you can easily verify.
Thanks that works :)
 
  • #6
epenguin
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Homework Statement


[/B]
Suppose q(z) = z^3 − z^2 + rz + s, is a complex polynomial with 1 + i and i as zeros. Find r and s and the third complex zero.
I think you can achieve some shortcuts here. Equation is of exactly the same form as one for a problem you have just solved (or claimed to :oldwink: ). Help with finding Zeros of a polynomial with 1+i as a zero

In the previous equation you were given two roots and found the third, or somebody did.

Here you are given two roots which are the same as two of the roots of the previous problem multiplied by -i if I am not mistaken.

If all roots of a polynomial are multiplied by the same factor, what happens to the coefficients?

Edit: however that does not seem to work in the way I guessed. :redface:

The sum of roots must be real, so the third root must contain -2i.

The problem can be done in the same way as before.

There ought to be some greater analogy with the previous problem, which at the moment I cannot see.

 
Last edited:

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