Complex Number equations from roots

AI Thread Summary
The discussion focuses on solving the polynomial equation z^4 + az^3 + bz^2 + cz + d = 0 with roots z1 = 3 + j and z2 = -5 + 5j. Participants emphasize the importance of recognizing that complex roots must occur in conjugate pairs for real-coefficient polynomials, leading to the identification of additional roots. The conversation includes hints on how to approach the problem by multiplying factors corresponding to the roots and their conjugates. One participant expresses confusion about assigning values to the coefficients a, b, c, and d after calculating a product of roots. The dialogue concludes with a reminder to consider the variable nature of z when forming the polynomial equation.
DmytriE
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Homework Statement


Determine the only real values a, b, c, and d such that the equation:
z4+az3+bz2+cz+d = 0​
has both z1 and z2 as roots.

z1 = 3 + j
z2 = -5 + 5j

Homework Equations


z = x + yj.
z = |z|ej\theta


The Attempt at a Solution


I am not sure where to begin. I can convert between Cartesian coordinates and Euler's formula but don't know where to go from there.

Any help would be greatly appreciated!
 
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Hi DmytriE! :smile:

Hint: if 3 + j is a root (of a real-coefficient equation), can you spot another root? :wink:
 
tiny-tim said:
can you spot another root? :wink:

Indeed! Now, I do. Thanks for the hint!
 
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I have calculated the different combination of roots but how do I know which variable (a,b,c,d) the values go to? When I multiplied z1, z1*, z2, and z2* I got a value of 500. Would that go into d?

It seems logical when I compare it to a second degree polynomial...
 
DmytriE said:
When I multiplied z1, z1*, z2, and z2* I got a value of 500. Would that go into d?

Yes.

But why not do it all in one go? …

Find (z - z1)(z - z1*), then (z - z2)(z - z2*), then multiply them. :wink:
 
tiny-tim said:
Yes.

But why not do it all in one go? …

Find (z - z1)(z - z1*), then (z - z2)(z - z2*), then multiply them. :wink:

Great! Thank you again. I keep forgetting that z is a variable so I was strictly looking for the number rather than the equation with the coefficients.
 
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