# Complex Number equations from roots

DmytriE

## 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!

Homework Helper
Hi DmytriE!

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

DmytriE
can you spot another root?

Indeed! Now, I do. Thanks for the hint!

Last edited:
DmytriE
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...

Homework Helper
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.

DmytriE
Yes.

But why not do it all in one go? …

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

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.