• Support PF! Buy your school textbooks, materials and every day products Here!

Complex Number equations from roots

  • Thread starter DmytriE
  • Start date
  • #1
78
0

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!
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hi DmytriE! :smile:

Hint: if 3 + j is a root (of a real-coefficient equation), can you spot another root? :wink:
 
  • #3
78
0
can you spot another root? :wink:
Indeed! Now, I do. Thanks for the hint!
 
Last edited:
  • #4
78
0
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...
 
  • #5
tiny-tim
Science Advisor
Homework Helper
25,832
251
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:
 
  • #6
78
0
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.
 

Related Threads on Complex Number equations from roots

  • Last Post
Replies
1
Views
1K
Replies
9
Views
2K
  • Last Post
Replies
14
Views
2K
Replies
6
Views
1K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
2K
Replies
6
Views
7K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
11
Views
2K
Top