Complex Number equations from roots

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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[itex]\theta[/itex]


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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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.